Q1: Represent graphically a displacement of 40 km, 30^° east of north.

To represent the displacement graphically: 1. Draw the coordinate axes: Draw two perpendicular lines representing the cardinal directions. Label the vertical line as North-South (N-S) and the horizontal line as East-West (E-W), with the origin O as the starting point. 2. Determine the direction:…

Q1: Compute the magnitude of the following vectors: veca=hati+hatj+k; vecb=2 hati-7 hatj-3 hatk; vecc=frac1√(3) hati+frac1√(3) hatj-frac1√(3) hatk

Formula: The magnitude of a vector vecr = xhati + yhatj + zhatk is given by vecr = √(x^2 + y^2 + z^2). For vector veca=hati+hatj+hatk: veca = √(1^2 + 1^2 + 1^2) = √(1+1+1) = √(3) For vector vecb=2 hati-7 hatj-3 hatk: vecb = √(2^2 + (-7)^2 + (-3)^2) = √(4+49+9) = √(62) For vector vecc=frac1√(3)…

Vector Algebra Class 12 Mathematics NCERT Solutions

Q1EXERCISE 10.1

Represent graphically a displacement of $40 \text{ km}, 30^{\circ}$ east of north.

Solution

To represent the displacement graphically:

Draw the coordinate axes: Draw two perpendicular lines representing the cardinal directions. Label the vertical line as North-South (N-S) and the horizontal line as East-West (E-W), with the origin O as the starting point.
Determine the direction: The direction is $30^{\circ}$ east of north. This means we start from the North direction (positive Y-axis) and turn $30^{\circ}$ towards the East (positive X-axis).
Draw the vector: From the origin O, draw a line segment OP in this direction. The angle between the North line (ON) and OP should be $30^{\circ}$ .
Represent the magnitude: The length of the line segment OP should represent the magnitude of the displacement, which is 40 km. We can use a scale, for example, 1 cm = 10 km. So, the length of OP would be 4 cm.
Indicate the vector: Place an arrowhead at point P to show the direction of displacement. The vector $\overrightarrow{OP}$ represents the displacement of 40 km, $30^{\circ}$ east of north.

Description of the graph: The graph would show the standard N-S and E-W axes. A vector

\overrightarrow{OP}

starts at the origin O and goes into the first quadrant. The angle between the vector and the North axis is

30^{\circ}

. The length of the vector is labeled as 40 km.

Q2EXERCISE 10.1

Classify the following measures as scalars and vectors.

(i)

10 kg

(ii)

2 meters north-west

(iii)

$40^{\circ}$

(iv)

40 watt

(v)

$10^{-19}$ coulomb

(vi)

$20 \text{ m/s}^{2}$

Solution

Classification:

A scalar quantity has only magnitude. A vector quantity has both magnitude and direction.

(i)

10 kg: This is a measure of mass, which has only magnitude. Answer: Scalar

(ii)

2 meters north-west: This is a measure of displacement, which has both magnitude (2 meters) and direction (north-west). Answer: Vector

(iii)

$40^{\circ}$ : This is a measure of an angle, which has only magnitude. Answer: Scalar

(iv)

40 watt: This is a measure of power, which has only magnitude. Answer: Scalar

(v)

$10^{-19}$ coulomb: This is a measure of electric charge, which has only magnitude. Answer: Scalar

(vi)

$20 \text{ m/s}^{2}$ : This is a measure of acceleration, which has both magnitude (

20 \text{ m/s}^{2}

) and a direction (though not specified here, acceleration is inherently a directed quantity). Answer: Vector

Q3EXERCISE 10.1

Classify the following as scalar and vector quantities.

(i)

time period

(ii)

distance

(iii)

force

(iv)

velocity

(v)

work done

Solution

Classification:

(i)

time period: Time has only magnitude. Answer: Scalar

(ii)

distance: Distance is the length of a path and has only magnitude. Answer: Scalar

(iii)

force: Force has both magnitude (e.g., Newtons) and a direction in which it is applied. Answer: Vector

(iv)

velocity: Velocity is speed in a specific direction. It has both magnitude and direction. Answer: Vector

(v)

work done: Work is the dot product of force and displacement, which results in a scalar quantity. It has only magnitude. Answer: Scalar

Q4EXERCISE 10.1

In a square, identify the following vectors.

(i)

Coinitial

(ii)

Equal

(iii)

Collinear but not equal

Solution

Description of the vectors in the square: Let the vertices of the square be A (bottom-left), B (bottom-right), C (top-right), and D (top-left). The vectors are represented by the sides of the square as follows:

Vector $\vec{a}$ is from D to C (pointing right).
Vector $\vec{b}$ is from C to B (pointing down).
Vector $\vec{c}$ is from A to B (pointing right).
Vector $\vec{d}$ is from A to D (pointing up).

(Note: The solution is based on a common representation of vectors in a square for this problem, as the figure is not provided in the text.)

(i) Coinitial vectors: Coinitial vectors are vectors that have the same starting point.

Vectors $\vec{c}$ and $\vec{d}$ both start from point A. Answer: $\vec{c}$ and $\vec{d}$ are coinitial.

(ii) Equal vectors: Equal vectors have the same magnitude and the same direction.

Vectors $\vec{a}$ and $\vec{c}$ both point to the right and have the same length (side of the square). Answer: $\vec{a}$ and $\vec{c}$ are equal.

(iii) Collinear but not equal vectors: Collinear vectors are parallel to the same line, but they are not equal if they have different directions or different magnitudes.

Vectors $\vec{b}$ and $\vec{d}$ are parallel (both are vertical). However, $\vec{b}$ points down and $\vec{d}$ points up, so their directions are opposite. They have the same magnitude. Answer: $\vec{b}$ and $\vec{d}$ are collinear but not equal.

Q5EXERCISE 10.1

Answer the following as true or false.

(i)

$\vec{a}$ and $-\vec{a}$ are collinear.

(ii)

Two collinear vectors are always equal in magnitude.

(iii)

Two vectors having same magnitude are collinear.

(iv)

Two collinear vectors having the same magnitude are equal.

Solution

(i)

$\vec{a}$ and $-\vec{a}$ are collinear. Answer: True. Reason: The vector

-\vec{a}

has the same magnitude as

\vec{a}

but the opposite direction. Since they are parallel to the same line, they are collinear by definition.

(ii)

Two collinear vectors are always equal in magnitude. Answer: False. Reason: Collinear vectors are parallel. Their magnitudes can be different. For example,

\vec{a} = \hat{i}

and

\vec{b} = 2\hat{i}

are collinear, but

|\vec{a}| = 1

and

|\vec{b}| = 2

.

(iii)

Two vectors having same magnitude are collinear. Answer: False. Reason: Two vectors can have the same magnitude but point in different directions. For example,

\vec{a} = \hat{i}

and

\vec{b} = \hat{j}

both have magnitude 1, but they are perpendicular, not collinear.

(iv)

Two collinear vectors having the same magnitude are equal. Answer: False. Reason: Equal vectors must have the same magnitude and the same direction. Two collinear vectors with the same magnitude can have opposite directions. For example,

\vec{a} = \hat{i}

and

\vec{b} = -\hat{i}

are collinear and have the same magnitude (1), but they are not equal because their directions are opposite.

Q1EXERCISE 10.2

Compute the magnitude of the following vectors: $\vec{a}=\hat{i}+\hat{j}+k ; \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$

Solution

Formula: The magnitude of a vector

\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}

is given by

|\vec{r}| = \sqrt{x^2 + y^2 + z^2}

.

For vector $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ :

|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1+1+1} = \sqrt{3}

For vector $\vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k}$ :

|\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4+49+9} = \sqrt{62}

For vector $\vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$ :

|\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(-\frac{1}{\sqrt{3}}\right)^2}

= \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{\frac{3}{3}} = \sqrt{1} = 1

Final Answer:

|\vec{a}| = \sqrt{3}

|\vec{b}| = \sqrt{62}

|\vec{c}| = 1

Q2EXERCISE 10.2

Write two different vectors having same magnitude.

Solution

To Find: Two different vectors with the same magnitude.

Solution: Let us choose two vectors,

\vec{a}

and

\vec{b}

. Let

\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}

. Its magnitude is:

|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}

Now, let's create a different vector

\vec{b}

by rearranging the components or changing their signs. For example: Let

\vec{b} = 3\hat{i} + \hat{j} + 2\hat{k}

. Its magnitude is:

|\vec{b}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}

Clearly,

\vec{a} \neq \vec{b}

, but

|\vec{a}| = |\vec{b}| = \sqrt{14}

.

Final Answer: Two such vectors are

\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}

and

\vec{b} = 3\hat{i} + \hat{j} + 2\hat{k}

.

Q3EXERCISE 10.2

Write two different vectors having same direction.

Solution

To Find: Two different vectors with the same direction.

Solution: Two vectors have the same direction if one is a positive scalar multiple of the other. Let

\vec{a}

be any vector, for example,

\vec{a} = \hat{i} + \hat{j} + \hat{k}

.

To get another vector

\vec{b}

with the same direction, we can multiply

\vec{a}

by any positive scalar

\lambda \neq 1

. Let's choose

\lambda = 2

.

\vec{b} = 2\vec{a} = 2(\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} + 2\hat{j} + 2\hat{k}

Here, the vectors

\vec{a}

and

\vec{b}

are different because their magnitudes are different (

|\vec{a}| = \sqrt{3}

and

|\vec{b}| = 2\sqrt{3}

), but their direction is the same.

Final Answer: Two such vectors are

\vec{a} = \hat{i} + \hat{j} + \hat{k}

and

\vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k}

.

Q4EXERCISE 10.2

Find the values of $x$ and $y$ so that the vectors $2 \hat{i}+3 \hat{j}$ and $x \hat{i}+y \hat{j}$ are equal.

Solution

Given: Two vectors

\vec{a} = 2\hat{i} + 3\hat{j}

and

\vec{b} = x\hat{i} + y\hat{j}

are equal.

Condition for Equal Vectors: Two vectors are equal if and only if their corresponding scalar components are equal.

Solution: For

\vec{a} = \vec{b}

, we must have: The component of

\hat{i}

in

\vec{a}

equals the component of

\hat{i}

in

\vec{b}

.

2 = x

The component of

\hat{j}

in

\vec{a}

equals the component of

\hat{j}

in

\vec{b}

.

3 = y

Final Answer:

x = 2

and

y = 3

.

Q5EXERCISE 10.2

Find the scalar and vector components of the vector with initial point $(2,1)$ and terminal point ( $-5,7$ ).

Solution

Given: Initial point P = (2, 1) Terminal point Q = (-5, 7)

To Find: Scalar and vector components of the vector

\overrightarrow{PQ}

.

Solution: The vector joining P to Q is given by:

\overrightarrow{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}

Here,

(x_1, y_1) = (2, 1)

and

(x_2, y_2) = (-5, 7)

.

\overrightarrow{PQ} = (-5 - 2)\hat{i} + (7 - 1)\hat{j}

\overrightarrow{PQ} = -7\hat{i} + 6\hat{j}

Scalar Components: The scalar components are the coefficients of

\hat{i}

and

\hat{j}

. Scalar components are -7 and 6.

Vector Components: The vector components are the terms of the vector expression. Vector components are

-7\hat{i}

and

6\hat{j}

.

Final Answer: The scalar components are -7 and 6. The vector components are

-7\hat{i}

and

6\hat{j}

.

Q6EXERCISE 10.2

Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$ .

Solution

Given:

\vec{a}=\hat{i}-2 \hat{j}+\hat{k}

\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}

\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}

To Find: The sum

\vec{a} + \vec{b} + \vec{c}

.

Solution: The sum of vectors is found by adding their corresponding scalar components.

\vec{a} + \vec{b} + \vec{c} = (1-2+1)\hat{i} + (-2+4-6)\hat{j} + (1+5-7)\hat{k}

= (0)\hat{i} + (-4)\hat{j} + (-1)\hat{k}

= 0\hat{i} - 4\hat{j} - \hat{k}

Final Answer: The sum of the vectors is

-4\hat{j} - \hat{k}

.

Q7EXERCISE 10.2

Find the unit vector in the direction of the vector $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ .

Solution

Given: Vector

\vec{a} = \hat{i} + \hat{j} + 2\hat{k}

.

To Find: The unit vector in the direction of

\vec{a}

, denoted as

\hat{a}

.

Formula:

\hat{a} = \frac{\vec{a}}{|\vec{a}|}

Solution: First, find the magnitude of

\vec{a}

.

|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}

Now, find the unit vector.

\hat{a} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}}

= \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}

Final Answer: The unit vector is

\frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}

.

Q8EXERCISE 10.2

Find the unit vector in the direction of vector $\overrightarrow{\mathrm{PQ}}$ , where P and Q are the points $(1,2,3)$ and (4, 5, 6), respectively.

Solution

Given: Initial point P = (1, 2, 3) Terminal point Q = (4, 5, 6)

To Find: The unit vector in the direction of

\overrightarrow{PQ}

.

Solution: Step 1: Find the vector $\overrightarrow{PQ}$ .

\overrightarrow{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

\overrightarrow{PQ} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k}

\overrightarrow{PQ} = 3\hat{i} + 3\hat{j} + 3\hat{k}

Step 2: Find the magnitude of $\overrightarrow{PQ}$ .

|\overrightarrow{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}

Step 3: Find the unit vector. The unit vector in the direction of

\overrightarrow{PQ}

is

\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}

.

= \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}}

= \frac{3( \hat{i} + \hat{j} + \hat{k})}{3\sqrt{3}}

= \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}

Final Answer: The unit vector is

\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}

.

Q9EXERCISE 10.2

For given vectors, $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ , find the unit vector in the direction of the vector $\vec{a}+\vec{b}$ .

Solution

Given:

\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}

\vec{b} = -\hat{i} + \hat{j} - \hat{k}

To Find: The unit vector in the direction of

\vec{a} + \vec{b}

.

Solution: Step 1: Find the sum vector $\vec{c} = \vec{a} + \vec{b}$ .

\vec{c} = (2\hat{i} - \hat{j} + 2\hat{k}) + (-\hat{i} + \hat{j} - \hat{k})

\vec{c} = (2-1)\hat{i} + (-1+1)\hat{j} + (2-1)\hat{k}

\vec{c} = 1\hat{i} + 0\hat{j} + 1\hat{k} = \hat{i} + \hat{k}

Step 2: Find the magnitude of $\vec{c}$ .

|\vec{c}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1+1} = \sqrt{2}

Step 3: Find the unit vector. The unit vector in the direction of

\vec{c}

is

\hat{c} = \frac{\vec{c}}{|\vec{c}|}

.

\hat{c} = \frac{\hat{i} + \hat{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}

Final Answer: The unit vector is

\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}

.

Q10EXERCISE 10.2

Find a vector in the direction of vector $5 \hat{i}-\hat{j}+2 \hat{k}$ which has magnitude 8 units.

Solution

Given: A direction vector

\vec{a} = 5\hat{i} - \hat{j} + 2\hat{k}

. Required magnitude is 8 units.

To Find: A vector of magnitude 8 in the direction of

\vec{a}

.

Solution: Step 1: Find the unit vector in the direction of $\vec{a}$ . First, find the magnitude of

\vec{a}

.

|\vec{a}| = \sqrt{5^2 + (-1)^2 + 2^2} = \sqrt{25 + 1 + 4} = \sqrt{30}

The unit vector is

\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}}

.

Step 2: Find the required vector. The vector with magnitude 8 in the direction of

\vec{a}

is

8\hat{a}

.

8\hat{a} = 8 \left( \frac{5\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{30}} \right)

= \frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}

Final Answer: The required vector is

\frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}

.

Q11EXERCISE 10.2

Show that the vectors $2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+6 \hat{j}-8 \hat{k}$ are collinear.

Solution

Given:

\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}

\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}

To Show: The vectors

\vec{a}

and

\vec{b}

are collinear.

Condition for Collinearity: Two vectors are collinear if one is a scalar multiple of the other, i.e.,

\vec{b} = \lambda \vec{a}

for some scalar

\lambda

.

Proof: Let's try to express

\vec{b}

in terms of

\vec{a}

.

\vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}

We can factor out -2 from each component:

\vec{b} = -2(2\hat{i}) - 2(-3\hat{j}) - 2(4\hat{k})

\vec{b} = -2(2\hat{i} - 3\hat{j} + 4\hat{k})

This shows that

\vec{b} = -2\vec{a}

.

Since

\vec{b}

is a scalar multiple of

\vec{a}

(with

\lambda = -2

), the vectors are collinear.

Hence Proved.

Q12EXERCISE 10.2

Find the direction cosines of the vector $\hat{i}+2 \hat{j}+3 \hat{k}$ .

Solution

Given: Vector

\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}

.

To Find: The direction cosines of

\vec{r}

.

Formula: If

\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}

, its direction cosines

(l, m, n)

are given by:

l = \frac{x}{|\vec{r}|}, m = \frac{y}{|\vec{r}|}, n = \frac{z}{|\vec{r}|}

.

Solution: Step 1: Find the magnitude of $\vec{r}$ .

|\vec{r}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}

Step 2: Calculate the direction cosines. Here,

x=1, y=2, z=3

.

l = \frac{1}{\sqrt{14}}

m = \frac{2}{\sqrt{14}}

n = \frac{3}{\sqrt{14}}

Final Answer: The direction cosines are

\left( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right)

.

Q13EXERCISE 10.2

Find the direction cosines of the vector joining the points $\mathrm{A}(1,2,-3)$ and $\mathrm{B}(-1,-2,1)$ , directed from A to B .

Solution

Given: Initial point A = (1, 2, -3) Terminal point B = (-1, -2, 1)

To Find: The direction cosines of the vector

\overrightarrow{AB}

.

Solution: Step 1: Find the vector $\overrightarrow{AB}$ .

\overrightarrow{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

\overrightarrow{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k}

\overrightarrow{AB} = -2\hat{i} - 4\hat{j} + 4\hat{k}

Step 2: Find the magnitude of $\overrightarrow{AB}$ .

|\overrightarrow{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6

Step 3: Calculate the direction cosines. The direction ratios are

a=-2, b=-4, c=4

. The direction cosines

(l, m, n)

are:

l = \frac{a}{|\overrightarrow{AB}|} = \frac{-2}{6} = -\frac{1}{3}

m = \frac{b}{|\overrightarrow{AB}|} = \frac{-4}{6} = -\frac{2}{3}

n = \frac{c}{|\overrightarrow{AB}|} = \frac{4}{6} = \frac{2}{3}

Final Answer: The direction cosines are

\left( -\frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \right)

.

Q14EXERCISE 10.2

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $\mathrm{OX}, \mathrm{OY}$ and OZ .

Solution

Given: Vector

\vec{r} = \hat{i} + \hat{j} + \hat{k}

.

To Show: The vector is equally inclined to the axes OX, OY, and OZ.

Condition: A vector is equally inclined to the axes if its direction angles

\alpha, \beta, \gamma

are equal. This is true if its direction cosines

(l, m, n)

, where

l=\cos\alpha, m=\cos\beta, n=\cos\gamma

, are equal.

Proof: Step 1: Find the magnitude of $\vec{r}$ .

|\vec{r}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}

Step 2: Find the direction cosines. The components of the vector are

x=1, y=1, z=1

.

l = \cos\alpha = \frac{x}{|\vec{r}|} = \frac{1}{\sqrt{3}}

m = \cos\beta = \frac{y}{|\vec{r}|} = \frac{1}{\sqrt{3}}

n = \cos\gamma = \frac{z}{|\vec{r}|} = \frac{1}{\sqrt{3}}

Since

l = m = n

, we have

\cos\alpha = \cos\beta = \cos\gamma

. As the angles are typically considered in the range

[0, \pi]

, this implies

\alpha = \beta = \gamma

.

Therefore, the vector

\hat{i} + \hat{j} + \hat{k}

is equally inclined to the axes OX, OY, and OZ.

Hence Proved.

Q15EXERCISE 10.2

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio 2 : 1

(i)

internally

(ii)

externally

Solution

Given: Position vector of P,

\overrightarrow{OP} = \vec{p} = \hat{i} + 2\hat{j} - \hat{k}

. Position vector of Q,

\overrightarrow{OQ} = \vec{q} = -\hat{i} + \hat{j} + \hat{k}

. Ratio

m:n = 2:1

.

To Find: Position vector of R,

\overrightarrow{OR}

.

(i) Internally:

Formula: For internal division,

\overrightarrow{OR} = \frac{m\vec{q} + n\vec{p}}{m+n}

.

Solution:

\overrightarrow{OR} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2+1}

= \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} + \hat{i} + 2\hat{j} - \hat{k}}{3}

= \frac{(-2+1)\hat{i} + (2+2)\hat{j} + (2-1)\hat{k}}{3}

= \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}

(ii) Externally:

Formula: For external division,

\overrightarrow{OR} = \frac{m\vec{q} - n\vec{p}}{m-n}

.

Solution:

\overrightarrow{OR} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2-1}

= \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{1}

= (-2-1)\hat{i} + (2-2)\hat{j} + (2+1)\hat{k}

= -3\hat{i} + 0\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{k}

Final Answer:

(i)

Internally:

\overrightarrow{OR} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}

(ii)

Externally:

\overrightarrow{OR} = -3\hat{i} + 3\hat{k}

Q16EXERCISE 10.2

Find the position vector of the mid point of the vector joining the points $\mathrm{P}(2,3,4)$ and $\mathrm{Q}(4,1,-2)$ .

Solution

Given: Point P = (2, 3, 4) Point Q = (4, 1, -2)

To Find: The position vector of the midpoint R of the line segment PQ.

Solution: Step 1: Write the position vectors of P and Q. Position vector of P,

\overrightarrow{OP} = \vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}

. Position vector of Q,

\overrightarrow{OQ} = \vec{q} = 4\hat{i} + \hat{j} - 2\hat{k}

.

Step 2: Use the midpoint formula. The position vector of the midpoint R is given by:

\overrightarrow{OR} = \frac{\vec{p} + \vec{q}}{2}

= \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} - 2\hat{k})}{2}

= \frac{(2+4)\hat{i} + (3+1)\hat{j} + (4-2)\hat{k}}{2}

= \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2}

= 3\hat{i} + 2\hat{j} + \hat{k}

Final Answer: The position vector of the midpoint is

3\hat{i} + 2\hat{j} + \hat{k}

.

Q17EXERCISE 10.2

Show that the points $\mathrm{A}, \mathrm{B}$ and C with position vectors, $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}$ , $\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$ , respectively form the vertices of a right angled triangle.

Solution

Given: Position vector of A,

\overrightarrow{OA} = \vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}

Position vector of B,

\overrightarrow{OB} = \vec{b} = 2\hat{i} - \hat{j} + \hat{k}

Position vector of C,

\overrightarrow{OC} = \vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}

To Show: Points A, B, and C form the vertices of a right-angled triangle.

Method: We will find the vectors representing the sides of the triangle and then find their magnitudes. We will then check if the Pythagorean theorem holds.

Solution: Step 1: Find the vectors of the sides.

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k})

= (2-3)\hat{i} + (-1+4)\hat{j} + (1+4)\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}

\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k})

= (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}

\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k})

= (3-1)\hat{i} + (-4+3)\hat{j} + (-4+5)\hat{k} = 2\hat{i} - \hat{j} + \hat{k}

Step 2: Find the squares of the magnitudes of the sides.

|\overrightarrow{AB}|^2 = (-1)^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35

|\overrightarrow{BC}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41

|\overrightarrow{CA}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6

Step 3: Check the Pythagorean theorem. We observe that:

|\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2 = 35 + 6 = 41

And

|\overrightarrow{BC}|^2 = 41

. So,

|\overrightarrow{AB}|^2 + |\overrightarrow{CA}|^2 = |\overrightarrow{BC}|^2

.

Since the sum of the squares of two sides is equal to the square of the third side, the triangle is a right-angled triangle, with the right angle at vertex A.

Hence Proved.

Q18EXERCISE 10.2

In triangle ABC, which of the following is not true:

(A)

\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}

(B)

\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}

(C)

\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}

(D)

\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}

Solution

Analysis of the options based on the Triangle Law of Vector Addition:

The triangle law of vector addition states that for a triangle ABC, the sum of vectors from A to B and B to C is the vector from A to C.

\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}

Also, the vector in the opposite direction is the negative of the vector:

\overrightarrow{CA} = -\overrightarrow{AC}

Let's check each option:

(A) $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$ Using the triangle law,

\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}

. Substituting this into the equation:

\overrightarrow{AC} + \overrightarrow{CA} = \overrightarrow{AC} + (-\overrightarrow{AC}) = \overrightarrow{0}

This statement is True. This represents the law for a closed loop of vectors.

(B) $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}$ From the triangle law,

\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}

. Rearranging this gives:

\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC} = \overrightarrow{0}

This statement is True.

(C) $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}$ This is identical to option (B) and is therefore also True.

(D) $\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$ We know that

-\overrightarrow{CB} = \overrightarrow{BC}

. So the equation becomes:

\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{0}

This is the same as option (A), which is True.

Conclusion: All the given options (A), (B), (C), and (D) are correct statements derived from the triangle law of vector addition. The question asks which statement is 'not true'. This indicates an error in the question as presented in the textbook, as there is no incorrect option among the choices.

If we assume a common typo in option (C) to be

\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{AC} = \overrightarrow{0}

, then:

(\overrightarrow{AB} + \overrightarrow{BC}) + \overrightarrow{AC} = \overrightarrow{AC} + \overrightarrow{AC} = 2\overrightarrow{AC} = \overrightarrow{0}

This would only be true if A and C are the same point, which is not a triangle. In this case, this modified option would be the one that is 'not true'. However, based on the provided text, no option is incorrect.

Q19EXERCISE 10.2

If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:

(A)

\vec{b}=\lambda \vec{a}

, for some scalar

\lambda

(B)

\vec{a}= \pm \vec{b}

(C)

the respective components of

\vec{a}

and

\vec{b}

are not proportional

(D)

both the vectors

\vec{a}

and

\vec{b}

have same direction, but different magnitudes.

Solution

Analysis of the options for two collinear vectors $\vec{a}$ and $\vec{b}$ :

Two vectors are collinear if they are parallel to the same line, irrespective of their magnitudes and directions.

(A) $\vec{b}=\lambda \vec{a}$ , for some scalar $\lambda$ This is the mathematical definition of collinear vectors. This statement is correct.

(B) $\vec{a}= \pm \vec{b}$ This implies that

|\vec{a}| = |\vec{b}|

. This is only true for collinear vectors that have the same magnitude. It is not true for all collinear vectors (e.g., if

\vec{b} = 2\vec{a}

). Therefore, this statement as a general rule for all collinear vectors is incorrect.

(C) the respective components of $\vec{a}$ and $\vec{b}$ are not proportional If

\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

and

\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}

are collinear, then

\vec{b} = \lambda\vec{a}

. This means

b_1 = \lambda a_1, b_2 = \lambda a_2, b_3 = \lambda a_3

. So,

\frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} = \lambda

, which means the components are proportional. The statement that the components are not proportional is a direct contradiction of a necessary property of collinear vectors. This statement is fundamentally incorrect.

(D) both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes. This statement is incorrect for two reasons:

Collinear vectors can have opposite directions (if $\lambda < 0$ ).
Collinear vectors can have the same magnitude (if $|\lambda| = 1$ ). Therefore, this statement as a general rule is incorrect.

Conclusion: The question asks which statements are incorrect. Statements (B), (C), and (D) are all incorrect descriptions of the general properties of any two collinear vectors. However, in multiple-choice questions, we often look for the most fundamentally incorrect statement. Statement (C) contradicts the very definition of collinearity in component form. The proportionality of components is a necessary condition. Thus, the statement that they are 'not proportional' is the most definitive incorrect statement.

Final Answer: (C) the respective components of

\vec{a}

and

\vec{b}

are not proportional

Q1EXERCISE 10.3

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 2 , respectively having $\vec{a} \cdot \vec{b}=\sqrt{6}$ .

Solution

Given: Magnitude of

\vec{a}

,

|\vec{a}| = \sqrt{3}

. Magnitude of

\vec{b}

,

|\vec{b}| = 2

. Scalar product,

\vec{a} \cdot \vec{b} = \sqrt{6}

.

To Find: The angle

\theta

between vectors

\vec{a}

and

\vec{b}

.

Formula: The angle

\theta

between two vectors

\vec{a}

and

\vec{b}

is given by:

\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}

Solution: Substitute the given values into the formula:

\cos \theta = \frac{\sqrt{6}}{(\sqrt{3})(2)} = \frac{\sqrt{2} \times \sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

Now, we find the angle

\theta

for which

\cos \theta = \frac{1}{\sqrt{2}}

.

\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \text{ or } 45^{\circ}

Final Answer: The angle between the vectors is

\frac{\pi}{4}

.

Q2EXERCISE 10.3

Find the angle between the vectors $\hat{i}-2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$

Solution

Given: Let

\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}

. Let

\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}

.

To Find: The angle

\theta

between vectors

\vec{a}

and

\vec{b}

.

Formula:

\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}

Solution: Step 1: Calculate the scalar product $\vec{a} \cdot \vec{b}$ .

\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10

Step 2: Calculate the magnitudes $|\vec{a}|$ and $|\vec{b}|$ .

|\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}

|\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}

Step 3: Calculate $\cos \theta$ .

\cos \theta = \frac{10}{\sqrt{14} \times \sqrt{14}} = \frac{10}{14} = \frac{5}{7}

Step 4: Find the angle $\theta$ .

\theta = \cos^{-1}\left(\frac{5}{7}\right)

Final Answer: The angle between the vectors is

\cos^{-1}\left(\frac{5}{7}\right)

.

Q3EXERCISE 10.3

Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$ .

Solution

Given: Let

\vec{a} = \hat{i} - \hat{j}

. Let

\vec{b} = \hat{i} + \hat{j}

.

To Find: The projection of vector

\vec{a}

on vector

\vec{b}

.

Formula: The projection of

\vec{a}

on

\vec{b}

is given by:

\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Solution: Step 1: Calculate the scalar product $\vec{a} \cdot \vec{b}$ .

\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0

Step 2: Calculate the magnitude $|\vec{b}|$ .

|\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{1+1} = \sqrt{2}

Step 3: Calculate the projection.

\text{proj}_{\vec{b}} \vec{a} = \frac{0}{\sqrt{2}} = 0

Final Answer: The projection of the vector

\hat{i}-\hat{j}

on the vector

\hat{i}+\hat{j}

is 0.

Q4EXERCISE 10.3

Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $7 \hat{i}-\hat{j}+8 \hat{k}$ .

Solution

Given: Let

\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}

. Let

\vec{b} = 7\hat{i} - \hat{j} + 8\hat{k}

.

To Find: The projection of vector

\vec{a}

on vector

\vec{b}

.

Formula:

\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Solution: Step 1: Calculate the scalar product $\vec{a} \cdot \vec{b}$ .

\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60

Step 2: Calculate the magnitude $|\vec{b}|$ .

|\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{49 + 1 + 64} = \sqrt{114}

Step 3: Calculate the projection.

\text{proj}_{\vec{b}} \vec{a} = \frac{60}{\sqrt{114}}

Final Answer: The projection of the vector

\hat{i}+3 \hat{j}+7 \hat{k}

on the vector

7 \hat{i}-\hat{j}+8 \hat{k}

is

\frac{60}{\sqrt{114}}

.

Q5EXERCISE 10.3

Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \quad \frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k}), \quad \frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$ Also, show that they are mutually perpendicular to each other.

Solution

Given: Let

\vec{a} = \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) = \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}

Let

\vec{b} = \frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k}) = \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}

Let

\vec{c} = \frac{1}{7}(6\hat{i} + 2\hat{j} - 3\hat{k}) = \frac{6}{7}\hat{i} + \frac{2}{7}\hat{j} - \frac{3}{7}\hat{k}

Part 1: Show that each vector is a unit vector. A vector is a unit vector if its magnitude is 1.

|\vec{a}| = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2} = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{4+9+36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1

.

|\vec{b}| = \sqrt{\left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2} = \sqrt{\frac{9}{49} + \frac{36}{49} + \frac{4}{49}} = \sqrt{\frac{9+36+4}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1

.

|\vec{c}| = \sqrt{\left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2 + \left(-\frac{3}{7}\right)^2} = \sqrt{\frac{36}{49} + \frac{4}{49} + \frac{9}{49}} = \sqrt{\frac{36+4+9}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1

.

Thus,

\vec{a}, \vec{b},

and

\vec{c}

are all unit vectors.

Part 2: Show that they are mutually perpendicular. Two vectors are perpendicular if their scalar (dot) product is 0.

\vec{a} \cdot \vec{b} = \left(\frac{2}{7}\right)\left(\frac{3}{7}\right) + \left(\frac{3}{7}\right)\left(-\frac{6}{7}\right) + \left(\frac{6}{7}\right)\left(\frac{2}{7}\right) = \frac{6}{49} - \frac{18}{49} + \frac{12}{49} = \frac{6-18+12}{49} = \frac{0}{49} = 0

. So,

\vec{a} \perp \vec{b}

.

\vec{b} \cdot \vec{c} = \left(\frac{3}{7}\right)\left(\frac{6}{7}\right) + \left(-\frac{6}{7}\right)\left(\frac{2}{7}\right) + \left(\frac{2}{7}\right)\left(-\frac{3}{7}\right) = \frac{18}{49} - \frac{12}{49} - \frac{6}{49} = \frac{18-12-6}{49} = \frac{0}{49} = 0

. So,

\vec{b} \perp \vec{c}

.

\vec{c} \cdot \vec{a} = \left(\frac{6}{7}\right)\left(\frac{2}{7}\right) + \left(\frac{2}{7}\right)\left(\frac{3}{7}\right) + \left(-\frac{3}{7}\right)\left(\frac{6}{7}\right) = \frac{12}{49} + \frac{6}{49} - \frac{18}{49} = \frac{12+6-18}{49} = \frac{0}{49} = 0

. So,

\vec{c} \perp \vec{a}

.

Since the vectors are perpendicular to each other in pairs, they are mutually perpendicular.

Hence Proved.

Q6EXERCISE 10.3

Find $|\vec{a}|$ and $|\vec{b}|$ , if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|$ .

Solution

Given:

$(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$
$|\vec{a}|=8|\vec{b}|$

To Find:

|\vec{a}|

and

|\vec{b}|

.

Solution: Step 1: Simplify the first equation. Using the distributive property of the dot product:

(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b}) = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b}

Since

\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

and

\vec{v} \cdot \vec{v} = |\vec{v}|^2

:

= |\vec{a}|^2 - |\vec{b}|^2

So, we have:

|\vec{a}|^2 - |\vec{b}|^2 = 8 \quad (*)

Step 2: Substitute the second equation into the simplified first equation. Substitute

|\vec{a}|=8|\vec{b}|

into equation (*):

(8|\vec{b}|)^2 - |\vec{b}|^2 = 8

64|\vec{b}|^2 - |\vec{b}|^2 = 8

63|\vec{b}|^2 = 8

|\vec{b}|^2 = \frac{8}{63}

|\vec{b}| = \sqrt{\frac{8}{63}} = \sqrt{\frac{4 \times 2}{9 \times 7}} = \frac{2\sqrt{2}}{3\sqrt{7}}

We can rationalize the denominator:

|\vec{b}| = \frac{2\sqrt{2}}{3\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{2\sqrt{14}}{21}

Step 3: Find $|\vec{a}|$ . Using the relation

|\vec{a}|=8|\vec{b}|_

:

|\vec{a}| = 8 \times \frac{2\sqrt{14}}{21} = \frac{16\sqrt{14}}{21}

Final Answer:

|\vec{b}| = \frac{2\sqrt{14}}{21}

|\vec{a}| = \frac{16\sqrt{14}}{21}

Q7EXERCISE 10.3

Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$ .

Solution

To Evaluate:

(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})

Solution: We use the distributive property of the scalar product:

(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b}) = 3\vec{a} \cdot (2\vec{a} + 7\vec{b}) - 5\vec{b} \cdot (2\vec{a} + 7\vec{b})

= (3\vec{a} \cdot 2\vec{a}) + (3\vec{a} \cdot 7\vec{b}) - (5\vec{b} \cdot 2\vec{a}) - (5\vec{b} \cdot 7\vec{b})

= 6(\vec{a} \cdot \vec{a}) + 21(\vec{a} \cdot \vec{b}) - 10(\vec{b} \cdot \vec{a}) - 35(\vec{b} \cdot \vec{b})

Using the properties

\vec{v} \cdot \vec{v} = |\vec{v}|^2

and

\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

(commutative property):

= 6|\vec{a}|^2 + 21(\vec{a} \cdot \vec{b}) - 10(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2

= 6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2

Final Answer: The product evaluates to

6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) - 35|\vec{b}|^2

.

Q8EXERCISE 10.3

Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$ , having the same magnitude and such that the angle between them is $60^{\circ}$ and their scalar product is $\frac{1}{2}$ .

Solution

Given:

$|\vec{a}| = |\vec{b}|$
Angle between them, $\theta = 60^{\circ}$
Scalar product, $\vec{a} \cdot \vec{b} = \frac{1}{2}$

To Find: The magnitude of vectors

\vec{a}

and

\vec{b}

.

Formula:

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

Solution: Substitute the given values into the formula:

\frac{1}{2} = |\vec{a}| |\vec{b}| \cos 60^{\circ}

Since

|\vec{a}| = |\vec{b}|

, let's denote this common magnitude by

m

. So,

|\vec{a}| = |\vec{b}| = m

. We also know that

\cos 60^{\circ} = \frac{1}{2}

.

\frac{1}{2} = (m)(m) \left(\frac{1}{2}\right)

\frac{1}{2} = \frac{1}{2} m^2

m^2 = 1

Since magnitude must be non-negative, we take the positive square root:

m = 1

Therefore,

|\vec{a}| = 1

and

|\vec{b}| = 1

.

Final Answer: The magnitude of each vector is 1.

Q9EXERCISE 10.3

Find $|\vec{x}|$ , if for a unit vector $\vec{a}$ , $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$ .

Solution

Given:

$\vec{a}$ is a unit vector, which means $|\vec{a}| = 1$ .
$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$ .

To Find:

|\vec{x}|

.

Solution: Start with the given equation:

(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a}) = 12

Expand the dot product using the distributive property:

\vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} = 12

Since

\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}

and

\vec{v} \cdot \vec{v} = |\vec{v}|^2

:

|\vec{x}|^2 - |\vec{a}|^2 = 12

Now, substitute the given value

|\vec{a}| = 1

:

|\vec{x}|^2 - 1^2 = 12

|\vec{x}|^2 - 1 = 12

|\vec{x}|^2 = 13

Since magnitude must be non-negative:

|\vec{x}| = \sqrt{13}

Final Answer:

|\vec{x}| = \sqrt{13}

.

Q10EXERCISE 10.3

If $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{c}$ , then find the value of $\lambda$ .

Solution

Given:

\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}

\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}

\vec{c} = 3\hat{i} + \hat{j}

Condition:

(\vec{a} + \lambda\vec{b}) \perp \vec{c}

.

To Find: The value of

\lambda

.

Solution: Step 1: Find the vector $\vec{d} = \vec{a} + \lambda\vec{b}$ .

\vec{d} = (2\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k})

= 2\hat{i} + 2\hat{j} + 3\hat{k} - \lambda\hat{i} + 2\lambda\hat{j} + \lambda\hat{k}

Group the components:

\vec{d} = (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}

Step 2: Use the condition for perpendicularity. Two vectors are perpendicular if their dot product is zero. So,

\vec{d} \cdot \vec{c} = 0

.

\vec{c} = 3\hat{i} + 1\hat{j} + 0\hat{k}

((2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}) \cdot (3\hat{i} + 1\hat{j} + 0\hat{k}) = 0

(2 - \lambda)(3) + (2 + 2\lambda)(1) + (3 + \lambda)(0) = 0

6 - 3\lambda + 2 + 2\lambda + 0 = 0

8 - \lambda = 0

\lambda = 8

Final Answer: The value of

\lambda

is 8.

Q11EXERCISE 10.3

Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$ , for any two nonzero vectors $\vec{a}$ and $\vec{b}$ .

Solution

To Show: The vectors

|\vec{a}| \vec{b}+|\vec{b}| \vec{a}

and

|\vec{a}| \vec{b}-|\vec{b}| \vec{a}

are perpendicular.

Condition for Perpendicularity: Two vectors are perpendicular if their scalar (dot) product is zero.

Proof: Let

\vec{p} = |\vec{a}| \vec{b}+|\vec{b}| \vec{a}

and

\vec{q} = |\vec{a}| \vec{b}-|\vec{b}| \vec{a}

. We need to show that

\vec{p} \cdot \vec{q} = 0

.

\vec{p} \cdot \vec{q} = (|\vec{a}| \vec{b}+|\vec{b}| \vec{a}) \cdot (|\vec{a}| \vec{b}-|\vec{b}| \vec{a})

This is in the form

(X+Y) \cdot (X-Y) = X \cdot X - Y \cdot Y = |X|^2 - |Y|^2

. Let

X = |\vec{a}|\vec{b}

and

Y = |\vec{b}|\vec{a}

.

\vec{p} \cdot \vec{q} = (|\vec{a}| \vec{b}) \cdot (|\vec{a}| \vec{b}) - (|\vec{b}| \vec{a}) \cdot (|\vec{b}| \vec{a})

Using the property

(k\vec{u}) \cdot (m\vec{v}) = km(\vec{u} \cdot \vec{v})

:

= |\vec{a}|^2 (\vec{b} \cdot \vec{b}) - |\vec{b}|^2 (\vec{a} \cdot \vec{a})

Using the property

\vec{v} \cdot \vec{v} = |\vec{v}|^2

:

= |\vec{a}|^2 |\vec{b}|^2 - |\vec{b}|^2 |\vec{a}|^2

= 0

Since the dot product of the two vectors is zero, they are perpendicular to each other.

Hence Proved.

Q12EXERCISE 10.3

If $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0$ , then what can be concluded about the vector $\vec{b}$ ?

Solution

Given:

$\vec{a} \cdot \vec{a} = 0$
$\vec{a} \cdot \vec{b} = 0$

Analysis:

From the first condition,

\vec{a} \cdot \vec{a} = 0

. We know that for any vector

\vec{v}

,

\vec{v} \cdot \vec{v} = |\vec{v}|^2

. So,

|\vec{a}|^2 = 0

. This implies that the magnitude of

\vec{a}

is 0, which means

\vec{a}

is the zero vector.

\vec{a} = \overrightarrow{0}

Now consider the second condition,

\vec{a} \cdot \vec{b} = 0

. Since we have concluded that

\vec{a} = \overrightarrow{0}

, this equation becomes:

\overrightarrow{0} \cdot \vec{b} = 0

The dot product of the zero vector with any vector is always zero. This equation holds true for any vector

\vec{b}

.

Conclusion: There are no restrictions on the vector

\vec{b}

. It can be any vector in the space.

Final Answer: If

\vec{a} \cdot \vec{a}=0

and

\vec{a} \cdot \vec{b}=0

, then

\vec{a}

must be the zero vector, and

\vec{b}

can be any vector.

Q13EXERCISE 10.3

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ , find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ .

Solution

Given:

$\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1$ .
$\vec{a} + \vec{b} + \vec{c} = \overrightarrow{0}$ .

To Find: The value of

\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}

.

Solution: Start with the given equation:

\vec{a} + \vec{b} + \vec{c} = \overrightarrow{0}

Take the dot product of this equation with itself:

(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \overrightarrow{0} \cdot \overrightarrow{0}

Expand the left side:

\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} + \vec{c}\cdot\vec{b} + \vec{c}\cdot\vec{c} = 0

Using the properties

\vec{v} \cdot \vec{v} = |\vec{v}|^2

and

\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}

:

|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) + 2(\vec{b} \cdot \vec{c}) + 2(\vec{c} \cdot \vec{a}) = 0

|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0

Substitute the given magnitudes

|\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1

:

1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0

3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0

2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3

\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}

Final Answer: The value of

\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}

is

-\frac{3}{2}

.

Q14EXERCISE 10.3

If either vector $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$ , then $\vec{a} \cdot \vec{b}=0$ . But the converse need not be true. Justify your answer with an example.

Solution

Statement: If either vector

\vec{a}=\overrightarrow{0}

or

\vec{b}=\overrightarrow{0}

, then

\vec{a} \cdot \vec{b}=0

. This statement is true by the definition of the scalar product.

Converse Statement: If

\vec{a} \cdot \vec{b}=0

, then either

\vec{a}=\overrightarrow{0}

or

\vec{b}=\overrightarrow{0}

.

To Show: The converse statement is not necessarily true.

Justification: The definition of the scalar product is

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

, where

\theta

is the angle between the vectors.

For

\vec{a} \cdot \vec{b} = 0

, we have three possibilities:

$|\vec{a}| = 0$ , which means $\vec{a} = \overrightarrow{0}$ .
$|\vec{b}| = 0$ , which means $\vec{b} = \overrightarrow{0}$ .
$\cos \theta = 0$ , which means $\theta = 90^{\circ}$ (or $\frac{\pi}{2}$ radians). This implies that the vectors $\vec{a}$ and $\vec{b}$ are perpendicular (orthogonal) to each other.

The converse statement only considers the first two possibilities and omits the third.

Example: Let's take two non-zero vectors that are perpendicular to each other. Let

\vec{a} = \hat{i}

and

\vec{b} = \hat{j}

.

Here,

\vec{a} \neq \overrightarrow{0}

and

\vec{b} \neq \overrightarrow{0}

. Now, let's calculate their dot product:

\vec{a} \cdot \vec{b} = \hat{i} \cdot \hat{j} = (1)(0) + (0)(1) + (0)(0) = 0

In this case,

\vec{a} \cdot \vec{b} = 0

, but neither

\vec{a}

nor

\vec{b}

is the zero vector.

This example demonstrates that the converse is not true. Two non-zero vectors can have a dot product of zero if they are perpendicular.

Q15EXERCISE 10.3

If the vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ of a triangle ABC are $(1,2,3),(-1,0,0),(0,1,2)$ , respectively, then find $\angle \mathrm{ABC}$ . [ $\angle \mathrm{ABC}$ is the angle between the vectors $\overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{BC}}$ ].

Solution

Given: Vertices of triangle ABC: A = (1, 2, 3) B = (-1, 0, 0) C = (0, 1, 2)

To Find:

\angle ABC

, which is the angle between vectors

\overrightarrow{BA}

and

\overrightarrow{BC}

.

Solution: Step 1: Find the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$ . Position vectors are:

\overrightarrow{OA} = \hat{i} + 2\hat{j} + 3\hat{k}

\overrightarrow{OB} = -\hat{i} + 0\hat{j} + 0\hat{k}

\overrightarrow{OC} = 0\hat{i} + \hat{j} + 2\hat{k}

Vector

\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB}

:

\overrightarrow{BA} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (-\hat{i}) = (1 - (-1))\hat{i} + (2-0)\hat{j} + (3-0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k}

Vector

\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}

:

\overrightarrow{BC} = (\hat{j} + 2\hat{k}) - (-\hat{i}) = (0 - (-1))\hat{i} + (1-0)\hat{j} + (2-0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}

Step 2: Use the dot product formula for the angle. Let

\theta = \angle ABC

. Then:

\cos \theta = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|}

Step 3: Calculate the required components. Dot product

\overrightarrow{BA} \cdot \overrightarrow{BC}

:

\overrightarrow{BA} \cdot \overrightarrow{BC} = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10

Magnitude

|\overrightarrow{BA}|

:

|\overrightarrow{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4+4+9} = \sqrt{17}

Magnitude

|\overrightarrow{BC}|

:

|\overrightarrow{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1+1+4} = \sqrt{6}

Step 4: Calculate $\cos \theta$ .

\cos \theta = \frac{10}{\sqrt{17} \sqrt{6}} = \frac{10}{\sqrt{102}}

Step 5: Find the angle.

\theta = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right)

Final Answer:

\angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right)

.

Q16EXERCISE 10.3

Show that the points $\mathrm{A}(1,2,7), \mathrm{B}(2,6,3)$ and $\mathrm{C}(3,10,-1)$ are collinear.

Solution

Given: Points A(1, 2, 7), B(2, 6, 3), and C(3, 10, -1).

To Show: The points A, B, and C are collinear.

Method: Three points are collinear if the vector between the first and second points is a scalar multiple of the vector between the second and third points. That is, if

\overrightarrow{BC} = \lambda \overrightarrow{AB}

for some scalar

\lambda

.

Solution: Step 1: Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ .

\overrightarrow{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} + (z_B - z_A)\hat{k}

= (2-1)\hat{i} + (6-2)\hat{j} + (3-7)\hat{k}

= \hat{i} + 4\hat{j} - 4\hat{k}

\overrightarrow{BC} = (x_C - x_B)\hat{i} + (y_C - y_B)\hat{j} + (z_C - z_B)\hat{k}

= (3-2)\hat{i} + (10-6)\hat{j} + (-1-3)\hat{k}

= \hat{i} + 4\hat{j} - 4\hat{k}

Step 2: Compare the vectors. We can see that

\overrightarrow{AB} = \overrightarrow{BC}

. This can be written as

\overrightarrow{BC} = 1 \cdot \overrightarrow{AB}

.

Since

\overrightarrow{BC}

is a scalar multiple of

\overrightarrow{AB}

(with

\lambda=1

), the vectors are parallel and share a common point B. Therefore, the points A, B, and C lie on the same straight line.

Alternative Method (using magnitudes):

|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1+16+16} = \sqrt{33}

|\overrightarrow{BC}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1+16+16} = \sqrt{33}

Find vector

\overrightarrow{AC} = (3-1)\hat{i} + (10-2)\hat{j} + (-1-7)\hat{k} = 2\hat{i} + 8\hat{j} - 8\hat{k}

|\overrightarrow{AC}| = \sqrt{2^2 + 8^2 + (-8)^2} = \sqrt{4+64+64} = \sqrt{132} = \sqrt{4 \times 33} = 2\sqrt{33}

Check if

|\overrightarrow{AB}| + |\overrightarrow{BC}| = |\overrightarrow{AC}|

.

\sqrt{33} + \sqrt{33} = 2\sqrt{33}

. This is true. Since the sum of the lengths of two segments equals the length of the third, the points are collinear.

Hence Proved.

Q17EXERCISE 10.3

Show that the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$ form the vertices of a right angled triangle.

Solution

Given: Let the position vectors of the vertices A, B, and C be:

\overrightarrow{OA} = 2\hat{i} - \hat{j} + \hat{k}

\overrightarrow{OB} = \hat{i} - 3\hat{j} - 5\hat{k}

\overrightarrow{OC} = 3\hat{i} - 4\hat{j} - 4\hat{k}

To Show: The points A, B, and C form the vertices of a right-angled triangle.

Method: We will find the vectors representing the sides of the triangle. A triangle is right-angled if the dot product of any two of its side vectors is zero (meaning those two sides are perpendicular).

Solution: Step 1: Find the vectors of the sides.

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k})

= (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}

\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k})

= (3-1)\hat{i} + (-4+3)\hat{j} + (-4+5)\hat{k} = 2\hat{i} - \hat{j} + \hat{k}

\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k})

= (2-3)\hat{i} + (-1+4)\hat{j} + (1+4)\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}

Step 2: Check for perpendicularity using the dot product. Let's check the dot product of pairs of side vectors.

\overrightarrow{AB} \cdot \overrightarrow{BC} = (-\hat{i} - 2\hat{j} - 6\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k}) = (-1)(2) + (-2)(-1) + (-6)(1) = -2 + 2 - 6 = -6 \neq 0

.

\overrightarrow{BC} \cdot \overrightarrow{CA} = (2\hat{i} - \hat{j} + \hat{k}) \cdot (-\hat{i} + 3\hat{j} + 5\hat{k}) = (2)(-1) + (-1)(3) + (1)(5) = -2 - 3 + 5 = 0

.

Since

\overrightarrow{BC} \cdot \overrightarrow{CA} = 0

, the vectors

\overrightarrow{BC}

and

\overrightarrow{CA}

are perpendicular. This means the angle at vertex C is

90^{\circ}

.

Therefore, the given vectors form the vertices of a right-angled triangle.

Hence Proved.

Q18EXERCISE 10.3

If $\vec{a}$ is a nonzero vector of magnitude ' $a$ ' and $\lambda$ a nonzero scalar, then $\lambda \vec{a}$ is unit vector if

(A)

\lambda=1

(B)

\lambda=-1

(C)

a=|\lambda|

(D)

a=1 /|\lambda|

Solution

Given:

$\vec{a}$ is a nonzero vector with magnitude $|\vec{a}| = a$ .
$\lambda$ is a nonzero scalar.
$\lambda \vec{a}$ is a unit vector.

To Find: The condition that must be satisfied.

Solution: By definition, a unit vector has a magnitude of 1. So, we are given that

|\lambda \vec{a}| = 1

.

Using the property of scalar multiplication and magnitudes,

|k \vec{v}| = |k| |\vec{v}|

, we have:

|\lambda \vec{a}| = |\lambda| |\vec{a}| = 1

We are given that the magnitude of

\vec{a}

is

a

. So,

|\vec{a}| = a

. Substituting this into the equation:

|\lambda| a = 1

Since we need to find the condition on

a

, we can rearrange the equation:

a = \frac{1}{|\lambda|}

This matches option (D).

Final Answer: (D)

a=1 /|\lambda|

Q1EXERCISE 10.4

Find $|\vec{a} \times \vec{b}|$ , if $\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$ .

Solution

Given:

\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}

\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}

To Find: The magnitude of the cross product,

|\vec{a} \times \vec{b}|

.

Solution: Step 1: Calculate the cross product $\vec{a} \times \vec{b}$ . The cross product is calculated using the determinant:

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -7 & 7 \ 3 & -2 & 2 \end{vmatrix}

Expanding the determinant:

= \hat{i}((-7)(2) - (7)(-2)) - \hat{j}((1)(2) - (7)(3)) + \hat{k}((1)(-2) - (-7)(3))

= \hat{i}(-14 - (-14)) - \hat{j}(2 - 21) + \hat{k}(-2 - (-21))

= \hat{i}(-14 + 14) - \hat{j}(-19) + \hat{k}(-2 + 21)

= 0\hat{i} + 19\hat{j} + 19\hat{k}

Step 2: Calculate the magnitude of the resulting vector.

|\vec{a} \times \vec{b}| = |19\hat{j} + 19\hat{k}| = \sqrt{0^2 + 19^2 + 19^2}

= \sqrt{2 \times 19^2} = 19\sqrt{2}

Final Answer:

|\vec{a} \times \vec{b}| = 19\sqrt{2}

.

Q2EXERCISE 10.4

Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$ .

Solution

Given:

\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}

\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}

To Find: A unit vector perpendicular to both

(\vec{a} + \vec{b})

and

(\vec{a} - \vec{b})

.

Solution: Step 1: Calculate the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ . Let

\vec{p} = \vec{a} + \vec{b}

:

\vec{p} = (3\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} - 2\hat{k}) = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j}

Let

\vec{q} = \vec{a} - \vec{b}

:

\vec{q} = (3\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} - 2\hat{k}) = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 4\hat{k}

Step 2: Find a vector perpendicular to both $\vec{p}$ and $\vec{q}$ . A vector perpendicular to both

\vec{p}

and

\vec{q}

is given by their cross product,

\vec{n} = \vec{p} \times \vec{q}

.

\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 4 & 0 \ 2 & 0 & 4 \end{vmatrix}

= \hat{i}((4)(4) - (0)(0)) - \hat{j}((4)(4) - (0)(2)) + \hat{k}((4)(0) - (4)(2))

= \hat{i}(16) - \hat{j}(16) + \hat{k}(-8) = 16\hat{i} - 16\hat{j} - 8\hat{k}

Step 3: Find the unit vector in the direction of $\vec{n}$ . First, find the magnitude of

\vec{n}

.

|\vec{n}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24

The required unit vector is

\hat{n} = \frac{\vec{n}}{|\vec{n}|}

.

\hat{n} = \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} = \frac{16}{24}\hat{i} - \frac{16}{24}\hat{j} - \frac{8}{24}\hat{k}

= \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}

Final Answer: The unit vector is

\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}

.

Q3EXERCISE 10.4

If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}, \frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$ , then find $\theta$ and hence, the components of $\vec{a}$ .

Solution

Given:

$\vec{a}$ is a unit vector, so $|\vec{a}| = 1$ .
Angle with $\hat{i}$ is $\alpha = \frac{\pi}{3}$ .
Angle with $\hat{j}$ is $\beta = \frac{\pi}{4}$ .
Angle with $\hat{k}$ is $\gamma = \theta$ , and $\theta$ is acute ( $0 < \theta < \frac{\pi}{2}$ ).

To Find: The angle

\theta

and the components of

\vec{a}

.

Formula: The direction cosines

(l, m, n)

of a vector are related by the identity

l^2 + m^2 + n^2 = 1

, where

l=\cos\alpha, m=\cos\beta, n=\cos\gamma

.

Solution: Step 1: Find the direction cosines $l$ and $m$ .

l = \cos\alpha = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}

m = \cos\beta = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

n = \cos\gamma = \cos\theta

Step 2: Use the identity to find $n = \cos\theta$ .

l^2 + m^2 + n^2 = 1

\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + (\cos\theta)^2 = 1

\frac{1}{4} + \frac{1}{2} + \cos^2\theta = 1

\frac{1+2}{4} + \cos^2\theta = 1

\frac{3}{4} + \cos^2\theta = 1

\cos^2\theta = 1 - \frac{3}{4} = \frac{1}{4}

\cos\theta = \pm\sqrt{\frac{1}{4}} = \pm\frac{1}{2}

Since it is given that

\theta

is an acute angle,

\cos\theta

must be positive. So,

\cos\theta = \frac{1}{2}

.

Step 3: Find the angle $\theta$ .

\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}

Step 4: Find the components of $\vec{a}$ . For a unit vector, its components are its direction cosines.

\vec{a} = l\hat{i} + m\hat{j} + n\hat{k}

\vec{a} = \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k}

The components are the scalar coefficients:

\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}

.

Final Answer:

\theta = \frac{\pi}{3}

The components of

\vec{a}

are

\left( \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \right)

.

Q4EXERCISE 10.4

Show that $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$

Solution

To Prove:

(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})

Proof: Starting with the Left Hand Side (LHS):

LHS = (\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})

Using the distributive property of the vector product:

LHS = \vec{a} \times (\vec{a} + \vec{b}) - \vec{b} \times (\vec{a} + \vec{b})

LHS = (\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{b})

Now, we use two properties of the cross product:

The cross product of any vector with itself is the zero vector: $\vec{v} \times \vec{v} = \overrightarrow{0}$ .
The cross product is anti-commutative: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$ .

Applying these properties:

\vec{a} \times \vec{a} = \overrightarrow{0}

\vec{b} \times \vec{b} = \overrightarrow{0}

-(\vec{b} \times \vec{a}) = +(\vec{a} \times \vec{b})

Substitute these back into the expression:

LHS = \overrightarrow{0} + (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b}) - \overrightarrow{0}

LHS = 2(\vec{a} \times \vec{b})

Since LHS = RHS,

(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})

Hence Proved.

Q5EXERCISE 10.4

Find $\lambda$ and $\mu$ if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$ .

Solution

Given:

(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}

To Find: The values of

\lambda

and

\mu

.

Condition: The cross product of two non-zero vectors is the zero vector if and only if the vectors are collinear (parallel). Let

\vec{a} = 2\hat{i} + 6\hat{j} + 27\hat{k}

and

\vec{b} = \hat{i} + \lambda\hat{j} + \mu\hat{k}

. For

\vec{a}

and

\vec{b}

to be collinear, their corresponding components must be proportional.

Solution: If

\vec{a}

and

\vec{b}

are collinear, then there exists a scalar

k

such that

\vec{a} = k\vec{b}

.

2\hat{i} + 6\hat{j} + 27\hat{k} = k(\hat{i} + \lambda\hat{j} + \mu\hat{k})

2\hat{i} + 6\hat{j} + 27\hat{k} = k\hat{i} + k\lambda\hat{j} + k\mu\hat{k}

Comparing the coefficients of

\hat{i}, \hat{j},

and

\hat{k}

:

$2 = k$
$6 = k\lambda$
$27 = k\mu$

From (1), we have

k=2

. Substitute

k=2

into (2):

6 = 2\lambda \implies \lambda = \frac{6}{2} = 3

Substitute

k=2

into (3):

27 = 2\mu \implies \mu = \frac{27}{2}

Alternative Method (Proportionality): For collinear vectors with components

(a_1, a_2, a_3)

and

(b_1, b_2, b_3)

, we have:

\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}

\frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu}

From

\frac{2}{1} = \frac{6}{\lambda}

:

2\lambda = 6 \implies \lambda = 3

From

\frac{2}{1} = \frac{27}{\mu}

:

2\mu = 27 \implies \mu = \frac{27}{2}

Final Answer:

\lambda = 3

and

\mu = \frac{27}{2}

.

Q6EXERCISE 10.4

Given that $\vec{a} \cdot \vec{b}=0$ and $\vec{a} \times \vec{b}=\overrightarrow{0}$ . What can you conclude about the vectors $\vec{a}$ and $\vec{b}$ ?

Solution

Given:

$\vec{a} \cdot \vec{b} = 0$
$\vec{a} \times \vec{b} = \overrightarrow{0}$

Analysis of the conditions:

From condition 1: $\vec{a} \cdot \vec{b} = 0$ The dot product is defined as

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta

. For this to be zero, at least one of the following must be true:

$|\vec{a}| = 0$ , which means $\vec{a} = \overrightarrow{0}$ .
$|\vec{b}| = 0$ , which means $\vec{b} = \overrightarrow{0}$ .
$\cos\theta = 0$ , which means $\vec{a}$ and $\vec{b}$ are perpendicular (assuming they are non-zero).

From condition 2: $\vec{a} \times \vec{b} = \overrightarrow{0}$ The magnitude of the cross product is

|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta

. For the cross product to be the zero vector, its magnitude must be zero. This means at least one of the following must be true:

$|\vec{a}| = 0$ , which means $\vec{a} = \overrightarrow{0}$ .
$|\vec{b}| = 0$ , which means $\vec{b} = \overrightarrow{0}$ .
$\sin\theta = 0$ , which means $\vec{a}$ and $\vec{b}$ are parallel (collinear) (assuming they are non-zero).

Conclusion: We have two conditions that must be simultaneously true. If we assume that both

\vec{a}

and

\vec{b}

are non-zero vectors, then:

Condition 1 implies they must be perpendicular.
Condition 2 implies they must be parallel.

Two non-zero vectors cannot be both parallel and perpendicular at the same time. This is a contradiction.

Therefore, the assumption that both vectors are non-zero must be false. This means at least one of the vectors must be the zero vector.

Final Answer: If

\vec{a} \cdot \vec{b}=0

and

\vec{a} \times \vec{b}=\overrightarrow{0}

, then either

\vec{a} = \overrightarrow{0}

or

\vec{b} = \overrightarrow{0}

(or both).

Q7EXERCISE 10.4

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$ , $c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$ . Then show that $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$ .

Solution

To Prove:

\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}

(Distributive property of vector product over addition).

Given:

\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}

\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}

Proof:

Step 1: Calculate the Left Hand Side (LHS). First, find

\vec{b} + \vec{c}

:

\vec{b} + \vec{c} = (b_1 + c_1)\hat{i} + (b_2 + c_2)\hat{j} + (b_3 + c_3)\hat{k}

Now, calculate

\vec{a} \times (\vec{b} + \vec{c})

using the determinant form:

LHS = \vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1+c_1 & b_2+c_2 & b_3+c_3 \end{vmatrix}

Using a property of determinants which states that if an element of a row is a sum of two terms, the determinant can be split into the sum of two determinants:

LHS = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} + \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ c_1 & c_2 & c_3 \end{vmatrix}

Step 2: Relate this to the Right Hand Side (RHS). The first determinant is the expression for

\vec{a} \times \vec{b}

. The second determinant is the expression for

\vec{a} \times \vec{c}

.

So, we have:

LHS = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})

This is exactly the RHS.

Therefore,

\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}

.

Hence Proved.

Q8EXERCISE 10.4

If either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$ , then $\vec{a} \times \vec{b}=\overrightarrow{0}$ . Is the converse true? Justify your answer with an example.

Solution

Statement: If either

\vec{a}=\overrightarrow{0}

or

\vec{b}=\overrightarrow{0}

, then

\vec{a} \times \vec{b}=\overrightarrow{0}

. This statement is true by the definition of the vector product.

Converse Statement: If

\vec{a} \times \vec{b}=\overrightarrow{0}

, then either

\vec{a}=\overrightarrow{0}

or

\vec{b}=\overrightarrow{0}

.

To Show: The converse statement is not necessarily true.

Justification: The magnitude of the vector product is given by

|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta

, where

\theta

is the angle between the vectors.

For

\vec{a} \times \vec{b} = \overrightarrow{0}

, its magnitude must be zero, so

|\vec{a}| |\vec{b}| \sin \theta = 0

. This implies one of three possibilities:

$|\vec{a}| = 0$ , which means $\vec{a} = \overrightarrow{0}$ .
$|\vec{b}| = 0$ , which means $\vec{b} = \overrightarrow{0}$ .
$\sin \theta = 0$ , which means $\theta = 0$ or $\theta = \pi$ . This implies that the vectors $\vec{a}$ and $\vec{b}$ are parallel (collinear) to each other.

The converse statement only considers the first two possibilities and omits the third.

Example: Let's take two non-zero vectors that are parallel. Let

\vec{a} = \hat{i} + \hat{j} + \hat{k}

and

\vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k}

.

Here,

\vec{a} \neq \overrightarrow{0}

and

\vec{b} \neq \overrightarrow{0}

. Note that

\vec{b} = 2\vec{a}

, so the vectors are parallel and the angle between them is

\theta = 0

.

Now, let's calculate their cross product:

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 2 & 2 \end{vmatrix}

Since the second and third rows are proportional, the determinant is zero.

= \hat{i}(1 \cdot 2 - 1 \cdot 2) - \hat{j}(1 \cdot 2 - 1 \cdot 2) + \hat{k}(1 \cdot 2 - 1 \cdot 2) = 0\hat{i} - 0\hat{j} + 0\hat{k} = \overrightarrow{0}

In this case,

\vec{a} \times \vec{b} = \overrightarrow{0}

, but neither

\vec{a}

nor

\vec{b}

is the zero vector.

This example demonstrates that the converse is not true. Two non-zero vectors can have a cross product of zero if they are collinear.

Q9EXERCISE 10.4

Find the area of the triangle with vertices $\mathrm{A}(1,1,2), \mathrm{B}(2,3,5)$ and $\mathrm{C}(1,5,5)$ .

Solution

Given: Vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).

To Find: The area of triangle ABC.

Formula: The area of a triangle with vertices A, B, and C is given by:

\text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|

Solution: Step 1: Find the vectors of two adjacent sides, $\overrightarrow{AB}$ and $\overrightarrow{AC}$ .

\overrightarrow{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} + (z_B - z_A)\hat{k}

= (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}

\overrightarrow{AC} = (x_C - x_A)\hat{i} + (y_C - y_A)\hat{j} + (z_C - z_A)\hat{k}

= (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}

Step 2: Calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ .

\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 0 & 4 & 3 \end{vmatrix}

= \hat{i}((2)(3) - (3)(4)) - \hat{j}((1)(3) - (3)(0)) + \hat{k}((1)(4) - (2)(0))

= \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0)

= -6\hat{i} - 3\hat{j} + 4\hat{k}

Step 3: Find the magnitude of the cross product.

|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}

Step 4: Calculate the area of the triangle.

\text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}\sqrt{61}

Final Answer: The area of the triangle is

\frac{\sqrt{61}}{2}

square units.

Q10EXERCISE 10.4

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$ .

Solution

Given: Adjacent sides of the parallelogram are the vectors:

\vec{a} = \hat{i} - \hat{j} + 3\hat{k}

\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}

To Find: The area of the parallelogram.

Formula: The area of a parallelogram with adjacent sides

\vec{a}

and

\vec{b}

is given by the magnitude of their cross product:

\text{Area} = |\vec{a} \times \vec{b}|

Solution: Step 1: Calculate the cross product $\vec{a} \times \vec{b}$ .

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 3 \ 2 & -7 & 1 \end{vmatrix}

= \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2))

= \hat{i}(-1 - (-21)) - \hat{j}(1 - 6) + \hat{k}(-7 - (-2))

= \hat{i}(-1 + 21) - \hat{j}(-5) + \hat{k}(-7 + 2)

= 20\hat{i} + 5\hat{j} - 5\hat{k}

Step 2: Find the magnitude of the cross product.

|\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450}

= \sqrt{225 \times 2} = 15\sqrt{2}

Final Answer: The area of the parallelogram is

15\sqrt{2}

square units.

Q11EXERCISE 10.4

Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$ , then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is

(A)

\pi / 6

(B)

\pi / 4

(C)

\pi / 3

(D)

\pi / 2

Solution

Given:

$|\vec{a}| = 3$
$|\vec{b}| = \frac{\sqrt{2}}{3}$
$\vec{a} \times \vec{b}$ is a unit vector, which means $|\vec{a} \times \vec{b}| = 1$ .

To Find: The angle

\theta

between

\vec{a}

and

\vec{b}

.

Formula: The magnitude of the cross product is given by:

|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta

Solution: Substitute the given values into the formula:

1 = (3) \left(\frac{\sqrt{2}}{3}\right) \sin\theta

1 = \sqrt{2} \sin\theta

\sin\theta = \frac{1}{\sqrt{2}}

For the angle

\theta

in the range

0 \le \theta \le \pi

, the value of

\theta

for which

\sin\theta = \frac{1}{\sqrt{2}}

is

\theta = \frac{\pi}{4}

.

This corresponds to option (B).

Final Answer: (B)

\pi / 4

Q12EXERCISE 10.4

Area of a rectangle having vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ , respectively is

(A)

\frac{1}{2}

(B)

1

(C)

2

(D)

4

Solution

Given: Position vectors of the vertices of a rectangle:

\overrightarrow{OA} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}

\overrightarrow{OB} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}

\overrightarrow{OC} = \hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}

\overrightarrow{OD} = -\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}

To Find: The area of the rectangle ABCD.

Formula: The area of a rectangle is the product of the lengths of its adjacent sides. We can find the vectors for two adjacent sides, say

\overrightarrow{AB}

and

\overrightarrow{AD}

, and then the area is

|\overrightarrow{AB}| \times |\overrightarrow{AD}|

. Alternatively, the area of the parallelogram formed by adjacent sides

\overrightarrow{AB}

and

\overrightarrow{AD}

is

|\overrightarrow{AB} \times \overrightarrow{AD}|

.

Solution: Step 1: Find the vectors for two adjacent sides. Let's find

\overrightarrow{AB}

and

\overrightarrow{AD}

.

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \left(\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\right) - \left(-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\right)

= (1 - (-1))\hat{i} + \left(\frac{1}{2} - \frac{1}{2}\right)\hat{j} + (4-4)\hat{k} = 2\hat{i}

\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = \left(-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}\right) - \left(-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\right)

= (-1 - (-1))\hat{i} + \left(-\frac{1}{2} - \frac{1}{2}\right)\hat{j} + (4-4)\hat{k} = -\hat{j}

Step 2: Calculate the area. Method 1: Using magnitudes of sides. Length of side AB =

|\overrightarrow{AB}| = |2\hat{i}| = \sqrt{2^2} = 2

. Length of side AD =

|\overrightarrow{AD}| = |-\hat{j}| = \sqrt{(-1)^2} = 1

. Area = Length

\times

Width =

2 \times 1 = 2

.

Method 2: Using cross product. Area =

|\overrightarrow{AB} \times \overrightarrow{AD}|

\overrightarrow{AB} \times \overrightarrow{AD} = (2\hat{i}) \times (-\hat{j}) = -2(\hat{i} \times \hat{j}) = -2\hat{k}

|\overrightarrow{AB} \times \overrightarrow{AD}| = |-2\hat{k}| = \sqrt{(-2)^2} = 2

Both methods give the area as 2 square units. This corresponds to option (C).

Final Answer: (C) 2

Q1Miscellaneous Exercise on Chapter 10

Write down a unit vector in XY-plane, making an angle of $30^{\circ}$ with the positive direction of $x$ -axis.

Solution

To Find: A unit vector in the XY-plane at an angle of

30^{\circ}

with the positive x-axis.

Formula: A unit vector

\hat{r}

in the XY-plane that makes an angle

\theta

with the positive direction of the x-axis is given by:

\hat{r} = (\cos\theta)\hat{i} + (\sin\theta)\hat{j}

Solution: Given angle

\theta = 30^{\circ}

. We need to find the values of

\cos 30^{\circ}

and

\sin 30^{\circ}

.

\cos 30^{\circ} = \frac{\sqrt{3}}{2}

\sin 30^{\circ} = \frac{1}{2}

Substitute these values into the formula:

\hat{r} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}

Verification: The magnitude of this vector should be 1.

|\hat{r}| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{\frac{4}{4}} = \sqrt{1} = 1

So, it is a unit vector.

Final Answer: The required unit vector is

\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}

.

Q2Miscellaneous Exercise on Chapter 10

Find the scalar components and magnitude of the vector joining the points $\mathrm{P}\left(x_{1}, y_{1}, z_{1}\right)$ and $\mathrm{Q}\left(x_{2}, y_{2}, z_{2}\right)$ .

Solution

Given: Initial point P =

(x_1, y_1, z_1)

Terminal point Q =

(x_2, y_2, z_2)

To Find:

Scalar components of the vector $\overrightarrow{PQ}$ .
Magnitude of the vector $\overrightarrow{PQ}$ .

Solution: Step 1: Find the vector $\overrightarrow{PQ}$ . The vector joining point P to point Q is given by:

\overrightarrow{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

Step 2: Identify the scalar components. The scalar components of a vector are the coefficients of

\hat{i}, \hat{j},

and

\hat{k}

. So, the scalar components are

(x_2 - x_1)

,

(y_2 - y_1)

, and

(z_2 - z_1)

.

Step 3: Calculate the magnitude. The magnitude of the vector

\overrightarrow{PQ}

is given by the distance formula:

|\overrightarrow{PQ}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Final Answer:

The scalar components are: $(x_2 - x_1), (y_2 - y_1), (z_2 - z_1)$ .
The magnitude is: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ .

Q3Miscellaneous Exercise on Chapter 10

A girl walks 4 km towards west, then she walks 3 km in a direction $30^{\circ}$ east of north and stops. Determine the girl's displacement from her initial point of departure.

Solution

Given: Let the initial point of departure be the origin O(0, 0). Let the positive x-axis be East and the positive y-axis be North.

First displacement,

\vec{d_1}

: 4 km towards West. This corresponds to a vector of length 4 along the negative x-axis.

\vec{d_1} = -4\hat{i}

Second displacement,

\vec{d_2}

: 3 km in a direction

30^{\circ}

east of north. This direction means starting from North (positive y-axis) and turning

30^{\circ}

towards East (positive x-axis). The angle with the positive y-axis is

30^{\circ}

. The angle with the positive x-axis is

90^{\circ} - 30^{\circ} = 60^{\circ}

. The vector

\vec{d_2}

has magnitude 3 and makes an angle of

60^{\circ}

with the positive x-axis.

\vec{d_2} = |\vec{d_2}|(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j})

= 3\left(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right) = \frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}

To Find: The girl's total displacement,

\vec{D}

.

Solution: The total displacement is the vector sum of the individual displacements.

\vec{D} = \vec{d_1} + \vec{d_2}

\vec{D} = (-4\hat{i}) + \left(\frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}\right)

= \left(-4 + \frac{3}{2}\right)\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}

= \left(-\frac{8}{2} + \frac{3}{2}\right)\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}

= -\frac{5}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}

Final Answer: The girl's displacement from her initial point of departure is the vector

-\frac{5}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}

km.

Q4Miscellaneous Exercise on Chapter 10

If $\vec{a}=\vec{b}+\vec{c}$ , then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.

Solution

Question: If

\vec{a}=\vec{b}+\vec{c}

, is it true that

|\vec{a}|=|\vec{b}|+|\vec{c}|

?

Answer: No, it is not always true.

Justification: The relationship between the magnitudes is governed by the triangle inequality of vector addition, which states:

|\vec{b} + \vec{c}| \le |\vec{b}| + |\vec{c}|

Since

\vec{a} = \vec{b} + \vec{c}

, this means

|\vec{a}| \le |\vec{b}| + |\vec{c}|

.

The equality

|\vec{a}| = |\vec{b}| + |\vec{c}|

holds only in the specific case where the vectors

\vec{b}

and

\vec{c}

are collinear and point in the same direction (i.e., the angle between them is

0^{\circ}

). In all other cases, the inequality is strict (

|\vec{a}| < |\vec{b}| + |\vec{c}|

).

Counterexample: Let

\vec{b} = \hat{i}

and

\vec{c} = \hat{j}

. Then

\vec{a} = \vec{b} + \vec{c} = \hat{i} + \hat{j}

.

Now, let's find the magnitudes:

|\vec{a}| = |\hat{i} + \hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2}

.

|\vec{b}| = |\hat{i}| = 1

.

|\vec{c}| = |\hat{j}| = 1

.

Let's check if the equality holds: Is

|\vec{a}| = |\vec{b}| + |\vec{c}|

? Is

\sqrt{2} = 1 + 1

?

\sqrt{2} = 2

, which is false. (Since

\sqrt{2} \approx 1.414

)

This counterexample shows that the statement is not generally true.

Q5Miscellaneous Exercise on Chapter 10

Find the value of $x$ for which $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector.

Solution

Given: The vector

\vec{v} = x(\hat{i} + \hat{j} + \hat{k})

is a unit vector.

To Find: The value of

x

.

Condition: A vector is a unit vector if its magnitude is 1. So,

|\vec{v}| = 1

.

Solution:

|x(\hat{i} + \hat{j} + \hat{k})| = 1

Using the property

|k\vec{u}| = |k||\vec{u}|

, where

k

is a scalar:

|x| |\hat{i} + \hat{j} + \hat{k}| = 1

First, calculate the magnitude of

(\hat{i} + \hat{j} + \hat{k})

:

|\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}

Now substitute this back into the equation:

|x| \sqrt{3} = 1

|x| = \frac{1}{\sqrt{3}}

This implies that

x

can be either positive or negative:

x = \pm \frac{1}{\sqrt{3}}

Final Answer: The value of

x

is

\pm \frac{1}{\sqrt{3}}

.

Q6Miscellaneous Exercise on Chapter 10

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ .

Solution

Given:

\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}

\vec{b} = \hat{i} - 2\hat{j} + \hat{k}

Required magnitude = 5 units.

To Find: A vector of magnitude 5 parallel to the resultant of

\vec{a}

and

\vec{b}

.

Solution: Step 1: Find the resultant vector $\vec{R}$ . The resultant is the sum of the vectors.

\vec{R} = \vec{a} + \vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) + (\hat{i} - 2\hat{j} + \hat{k})

= (2+1)\hat{i} + (3-2)\hat{j} + (-1+1)\hat{k}

= 3\hat{i} + \hat{j} + 0\hat{k} = 3\hat{i} + \hat{j}

Step 2: Find the unit vector in the direction of $\vec{R}$ . First, find the magnitude of

\vec{R}

.

|\vec{R}| = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}

The unit vector is

\hat{R} = \frac{\vec{R}}{|\vec{R}|} = \frac{3\hat{i} + \hat{j}}{\sqrt{10}}

.

Step 3: Find the required vector. The vector with magnitude 5 parallel to

\vec{R}

is

5\hat{R}

.

5\hat{R} = 5 \left( \frac{3\hat{i} + \hat{j}}{\sqrt{10}} \right) = \frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j}

Final Answer: The required vector is

\frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j}

.

Q7Miscellaneous Exercise on Chapter 10

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-2 \hat{j}+\hat{k}$ , find a unit vector parallel to the vector $2 \vec{a}-\vec{b}+3 \vec{c}$ .

Solution

Given:

\vec{a} = \hat{i} + \hat{j} + \hat{k}

\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}

\vec{c} = \hat{i} - 2\hat{j} + \hat{k}

To Find: A unit vector parallel to

2\vec{a} - \vec{b} + 3\vec{c}

.

Solution: Step 1: Calculate the vector $\vec{d} = 2\vec{a} - \vec{b} + 3\vec{c}$ .

2\vec{a} = 2(\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} + 2\hat{j} + 2\hat{k}

-\vec{b} = -(2\hat{i} - \hat{j} + 3\hat{k}) = -2\hat{i} + \hat{j} - 3\hat{k}

3\vec{c} = 3(\hat{i} - 2\hat{j} + \hat{k}) = 3\hat{i} - 6\hat{j} + 3\hat{k}

Now, add these vectors:

\vec{d} = (2\hat{i} + 2\hat{j} + 2\hat{k}) + (-2\hat{i} + \hat{j} - 3\hat{k}) + (3\hat{i} - 6\hat{j} + 3\hat{k})

= (2-2+3)\hat{i} + (2+1-6)\hat{j} + (2-3+3)\hat{k}

= 3\hat{i} - 3\hat{j} + 2\hat{k}

Step 2: Find the unit vector in the direction of $\vec{d}$ . First, find the magnitude of

\vec{d}

.

|\vec{d}| = \sqrt{3^2 + (-3)^2 + 2^2} = \sqrt{9 + 9 + 4} = \sqrt{22}

The unit vector is

\hat{d} = \frac{\vec{d}}{|\vec{d}|}

.

\hat{d} = \frac{3\hat{i} - 3\hat{j} + 2\hat{k}}{\sqrt{22}} = \frac{3}{\sqrt{22}}\hat{i} - \frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}

Final Answer: The required unit vector is

\frac{1}{\sqrt{22}}(3\hat{i} - 3\hat{j} + 2\hat{k})

.

Q8Miscellaneous Exercise on Chapter 10

Show that the points $\mathrm{A}(1,-2,-8), \mathrm{B}(5,0,-2)$ and $\mathrm{C}(11,3,7)$ are collinear, and find the ratio in which B divides AC.

Solution

Given: Points A(1, -2, -8), B(5, 0, -2), and C(11, 3, 7).

Part 1: Show that the points are collinear.

Method: We will find the vectors

\overrightarrow{AB}

and

\overrightarrow{BC}

and show that one is a scalar multiple of the other.

Solution:

\overrightarrow{AB} = (5-1)\hat{i} + (0-(-2))\hat{j} + (-2-(-8))\hat{k} = 4\hat{i} + 2\hat{j} + 6\hat{k}

\overrightarrow{BC} = (11-5)\hat{i} + (3-0)\hat{j} + (7-(-2))\hat{k} = 6\hat{i} + 3\hat{j} + 9\hat{k}

Now, let's check for proportionality between the components: Ratio of x-components:

\frac{6}{4} = \frac{3}{2}

Ratio of y-components:

\frac{3}{2}

Ratio of z-components:

\frac{9}{6} = \frac{3}{2}

Since the ratios are equal, we can write:

\overrightarrow{BC} = \frac{3}{2} \overrightarrow{AB}

As

\overrightarrow{BC}

is a scalar multiple of

\overrightarrow{AB}

, the vectors are parallel. Since they share the common point B, the points A, B, and C are collinear.

Part 2: Find the ratio in which B divides AC.

Since B lies on the line segment AC, it divides AC internally. Let the ratio be

k:1

. By the section formula, the coordinates of B are given by:

B = \left( \frac{k \cdot x_C + 1 \cdot x_A}{k+1}, \frac{k \cdot y_C + 1 \cdot y_A}{k+1}, \frac{k \cdot z_C + 1 \cdot z_A}{k+1} \right)

We are given B = (5, 0, -2). Let's use the x-coordinate to find

k

:

5 = \frac{k(11) + 1(1)}{k+1}

5(k+1) = 11k + 1

5k + 5 = 11k + 1

4 = 6k

k = \frac{4}{6} = \frac{2}{3}

The ratio is

k:1

, which is

\frac{2}{3}:1

, or

2:3

.

Verification using y-coordinate:

0 = \frac{k(3) + 1(-2)}{k+1} = \frac{3k-2}{k+1}

$$3k-2=0 \implies k = 2/3$. The ratio is consistent.

Final Answer: The points are collinear. B divides AC internally in the ratio 2:3.

Q9Miscellaneous Exercise on Chapter 10

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ( $2 \vec{a}+\vec{b}$ ) and ( $\vec{a}-3 \vec{b}$ ) externally in the ratio $1: 2$ . Also, show that P is the mid point of the line segment RQ .

Solution

Given: Position vector of P,

\overrightarrow{OP} = 2\vec{a} + \vec{b}

Position vector of Q,

\overrightarrow{OQ} = \vec{a} - 3\vec{b}

Point R divides PQ externally in the ratio

m:n = 1:2

.

Part 1: Find the position vector of R, $\overrightarrow{OR}$ .

Formula: For external division,

\overrightarrow{OR} = \frac{m\overrightarrow{OQ} - n\overrightarrow{OP}}{m-n}

.

Solution:

\overrightarrow{OR} = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1-2}

= \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1}

= \frac{-3\vec{a} - 5\vec{b}}{-1}

= 3\vec{a} + 5\vec{b}

Part 2: Show that P is the midpoint of the line segment RQ.

To Show: The position vector of the midpoint of RQ is equal to

\overrightarrow{OP}

.

Formula: The position vector of the midpoint of a line segment is the average of the position vectors of its endpoints. Position vector of midpoint of RQ =

\frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2}

.

Proof:

\frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2}

= \frac{(3+1)\vec{a} + (5-3)\vec{b}}{2}

= \frac{4\vec{a} + 2\vec{b}}{2}

= 2\vec{a} + \vec{b}

This is equal to the given position vector of P,

\overrightarrow{OP}

. Since the position vector of the midpoint of RQ is the same as the position vector of P, P is the midpoint of RQ.

Final Answer:

The position vector of R is $\overrightarrow{OR} = 3\vec{a} + 5\vec{b}$ .
It is shown that P is the midpoint of the line segment RQ.

Q10Miscellaneous Exercise on Chapter 10

The two adjacent sides of a parallelogram are $2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\hat{i}-2 \hat{j}-3 \hat{k}$ . Find the unit vector parallel to its diagonal. Also, find its area.

Solution

Given: The adjacent sides of a parallelogram are:

\vec{a} = 2\hat{i} - 4\hat{j} + 5\hat{k}

\vec{b} = \hat{i} - 2\hat{j} - 3\hat{k}

Part 1: Find the unit vector parallel to its diagonal.

Solution: The diagonals of a parallelogram with adjacent sides

\vec{a}

and

\vec{b}

are given by

\vec{d_1} = \vec{a} + \vec{b}

and

\vec{d_2} = \vec{a} - \vec{b}

. The question asks for "its diagonal" (singular), which usually refers to the main diagonal represented by the sum.

Let's find the diagonal

\vec{d_1} = \vec{a} + \vec{b}

.

\vec{d_1} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + (\hat{i} - 2\hat{j} - 3\hat{k})

= (2+1)\hat{i} + (-4-2)\hat{j} + (5-3)\hat{k}

= 3\hat{i} - 6\hat{j} + 2\hat{k}

Now, find the unit vector parallel to

\vec{d_1}

. First, find the magnitude

|\vec{d_1}|

.

|\vec{d_1}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7

The unit vector is

\hat{d_1} = \frac{\vec{d_1}}{|\vec{d_1}|} = \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{7} = \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}

.

Part 2: Find its area.

Formula: The area of a parallelogram is given by the magnitude of the cross product of its adjacent sides, Area =

|\vec{a} \times \vec{b}|

.

Solution:

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -4 & 5 \ 1 & -2 & -3 \end{vmatrix}

= \hat{i}((-4)(-3) - (5)(-2)) - \hat{j}((2)(-3) - (5)(1)) + \hat{k}((2)(-2) - (-4)(1))

= \hat{i}(12 - (-10)) - \hat{j}(-6 - 5) + \hat{k}(-4 - (-4))

= \hat{i}(22) - \hat{j}(-11) + \hat{k}(0) = 22\hat{i} + 11\hat{j}

Now, find the magnitude of this vector:

\text{Area} = |22\hat{i} + 11\hat{j}| = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605}

= \sqrt{121 \times 5} = 11\sqrt{5}

Final Answer:

The unit vector parallel to the diagonal is $\frac{1}{7}(3\hat{i} - 6\hat{j} + 2\hat{k})$ .
The area of the parallelogram is $11\sqrt{5}$ square units.

Q11Miscellaneous Exercise on Chapter 10

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ .

Solution

To Show: The direction cosines of a vector equally inclined to the coordinate axes are

\pm\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)

.

Proof: Let a vector be equally inclined to the axes OX, OY, and OZ. This means the angles it makes with the positive directions of the axes are equal. Let these angles be

\alpha, \beta,

and

\gamma

. So,

\alpha = \beta = \gamma

.

The direction cosines of the vector are

l, m, n

, where:

l = \cos\alpha

m = \cos\beta

n = \cos\gamma

Since

\alpha = \beta = \gamma

, it follows that

\cos\alpha = \cos\beta = \cos\gamma

, which means

l = m = n

.

We know the fundamental identity for direction cosines:

l^2 + m^2 + n^2 = 1

Substitute

l=m=n

into this identity:

l^2 + l^2 + l^2 = 1

3l^2 = 1

l^2 = \frac{1}{3}

l = \pm\sqrt{\frac{1}{3}} = \pm\frac{1}{\sqrt{3}}

Since

l=m=n

, we have:

l = m = n = \pm\frac{1}{\sqrt{3}}

Therefore, the direction cosines are

\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)

or

\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)

. This can be written compactly as

\pm\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)

.

Hence Proved.

Q12Miscellaneous Exercise on Chapter 10

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$ . Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ , and $\vec{c} \cdot \vec{d}=15$ .

Solution

Given:

\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}

\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}

\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}

Conditions for $\vec{d}$ :

$\vec{d}$ is perpendicular to $\vec{a}$ ( $\vec{d} \cdot \vec{a} = 0$ ).
$\vec{d}$ is perpendicular to $\vec{b}$ ( $\vec{d} \cdot \vec{b} = 0$ ).
$\vec{c} \cdot \vec{d} = 15$ .

Solution: Step 1: Find a vector that is perpendicular to both $\vec{a}$ and $\vec{b}$ . A vector perpendicular to both

\vec{a}

and

\vec{b}

is parallel to their cross product,

\vec{a} \times \vec{b}

. Let

\vec{d} = \lambda (\vec{a} \times \vec{b})

for some scalar

\lambda

.

First, calculate

\vec{a} \times \vec{b}

:

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 4 & 2 \ 3 & -2 & 7 \end{vmatrix}

= \hat{i}((4)(7) - (2)(-2)) - \hat{j}((1)(7) - (2)(3)) + \hat{k}((1)(-2) - (4)(3))

= \hat{i}(28 - (-4)) - \hat{j}(7 - 6) + \hat{k}(-2 - 12)

= \hat{i}(32) - \hat{j}(1) + \hat{k}(-14) = 32\hat{i} - \hat{j} - 14\hat{k}

So,

\vec{d} = \lambda (32\hat{i} - \hat{j} - 14\hat{k})

.

Step 2: Use the third condition, $\vec{c} \cdot \vec{d} = 15$ , to find $\lambda$ .

\vec{c} \cdot [\lambda (32\hat{i} - \hat{j} - 14\hat{k})] = 15

\lambda [\vec{c} \cdot (32\hat{i} - \hat{j} - 14\hat{k})] = 15

\lambda [(2\hat{i} - \hat{j} + 4\hat{k}) \cdot (32\hat{i} - \hat{j} - 14\hat{k})] = 15

\lambda [(2)(32) + (-1)(-1) + (4)(-14)] = 15

\lambda [64 + 1 - 56] = 15

\lambda [9] = 15

\lambda = \frac{15}{9} = \frac{5}{3}

Step 3: Find the vector $\vec{d}$ . Substitute the value of

\lambda

back into the expression for

\vec{d}

.

\vec{d} = \frac{5}{3} (32\hat{i} - \hat{j} - 14\hat{k})

= \frac{160}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{70}{3}\hat{k}

Final Answer: The vector

\vec{d}

is

\frac{1}{3}(160\hat{i} - 5\hat{j} - 70\hat{k})

.

Q13Miscellaneous Exercise on Chapter 10

The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to one. Find the value of $\lambda$ .

Solution

Given: Let

\vec{p} = \hat{i} + \hat{j} + \hat{k}

. Let

\vec{a} = 2\hat{i} + 4\hat{j} - 5\hat{k}

. Let

\vec{b} = \lambda\hat{i} + 2\hat{j} + 3\hat{k}

.

Let

\vec{s}

be the sum of

\vec{a}

and

\vec{b}

. Let

\hat{s}

be the unit vector along

\vec{s}

. The scalar product

\vec{p} \cdot \hat{s} = 1

.

To Find: The value of

\lambda

.

Solution: Step 1: Find the sum vector $\vec{s}$ .

\vec{s} = \vec{a} + \vec{b} = (2\hat{i} + 4\hat{j} - 5\hat{k}) + (\lambda\hat{i} + 2\hat{j} + 3\hat{k})

= (2+\lambda)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k}

= (2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k}

Step 2: Find the unit vector $\hat{s}$ . First, find the magnitude of

\vec{s}

.

|\vec{s}| = \sqrt{(2+\lambda)^2 + 6^2 + (-2)^2} = \sqrt{(2+\lambda)^2 + 36 + 4} = \sqrt{(2+\lambda)^2 + 40}

The unit vector is

\hat{s} = \frac{\vec{s}}{|\vec{s}|} = \frac{(2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2+\lambda)^2 + 40}}

.

Step 3: Use the condition $\vec{p} \cdot \hat{s} = 1$ .

\vec{p} \cdot \frac{\vec{s}}{|\vec{s}|} = 1 \implies \vec{p} \cdot \vec{s} = |\vec{s}|

Calculate

\vec{p} \cdot \vec{s}

:

\vec{p} \cdot \vec{s} = (\hat{i} + \hat{j} + \hat{k}) \cdot ((2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k})

= (1)(2+\lambda) + (1)(6) + (1)(-2) = 2 + \lambda + 6 - 2 = \lambda + 6

Now, set

\vec{p} \cdot \vec{s} = |\vec{s}|

:

\lambda + 6 = \sqrt{(2+\lambda)^2 + 40}

Step 4: Solve for $\lambda$ . Square both sides of the equation:

(\lambda + 6)^2 = (2+\lambda)^2 + 40

\lambda^2 + 12\lambda + 36 = (4 + 4\lambda + \lambda^2) + 40

\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44

Subtract

\lambda^2

from both sides:

12\lambda + 36 = 4\lambda + 44

12\lambda - 4\lambda = 44 - 36

8\lambda = 8

\lambda = 1

Final Answer: The value of

\lambda

is 1.

Q14Miscellaneous Exercise on Chapter 10

If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a}+\vec{b}+\vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$ .

Solution

Given:

$\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular. This means $\vec{a} \cdot \vec{b} = 0$ , $\vec{b} \cdot \vec{c} = 0$ , and $\vec{c} \cdot \vec{a} = 0$ .
They have equal magnitudes. Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$ for some constant $k$ .

To Show: The vector

\vec{v} = \vec{a} + \vec{b} + \vec{c}

is equally inclined to

\vec{a}, \vec{b},

and

\vec{c}

.

Method: We need to show that the angles between

\vec{v}

and

\vec{a}

,

\vec{v}

and

\vec{b}

, and

\vec{v}

and

\vec{c}

are equal. We can do this by showing that the cosines of these angles are equal.

Proof: Let

\theta_1

be the angle between

\vec{v}

and

\vec{a}

. Let

\theta_2

be the angle between

\vec{v}

and

\vec{b}

. Let

\theta_3

be the angle between

\vec{v}

and

\vec{c}

.

Step 1: Calculate the dot products.

\vec{v} \cdot \vec{a} = (\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a} = \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{a} + \vec{c}\cdot\vec{a} = |\vec{a}|^2 + 0 + 0 = k^2

\vec{v} \cdot \vec{b} = (\vec{a} + \vec{b} + \vec{c}) \cdot \vec{b} = \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{b} = 0 + |\vec{b}|^2 + 0 = k^2

\vec{v} \cdot \vec{c} = (\vec{a} + \vec{b} + \vec{c}) \cdot \vec{c} = \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{c} = 0 + 0 + |\vec{c}|^2 = k^2

Step 2: Calculate the magnitude of $\vec{v}$ .

|\vec{v}|^2 = \vec{v} \cdot \vec{v} = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})

= |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})

= k^2 + k^2 + k^2 + 2(0 + 0 + 0) = 3k^2

So,

|\vec{v}| = \sqrt{3k^2} = k\sqrt{3}

.

Step 3: Calculate the cosines of the angles.

\cos\theta_1 = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}| |\vec{a}|} = \frac{k^2}{(k\sqrt{3})(k)} = \frac{k^2}{k^2\sqrt{3}} = \frac{1}{\sqrt{3}}

\cos\theta_2 = \frac{\vec{v} \cdot \vec{b}}{|\vec{v}| |\vec{b}|} = \frac{k^2}{(k\sqrt{3})(k)} = \frac{k^2}{k^2\sqrt{3}} = \frac{1}{\sqrt{3}}

\cos\theta_3 = \frac{\vec{v} \cdot \vec{c}}{|\vec{v}| |\vec{c}|} = \frac{k^2}{(k\sqrt{3})(k)} = \frac{k^2}{k^2\sqrt{3}} = \frac{1}{\sqrt{3}}

Since

\cos\theta_1 = \cos\theta_2 = \cos\theta_3

, the angles

\theta_1, \theta_2, \theta_3

are equal. Therefore, the vector

\vec{a} + \vec{b} + \vec{c}

is equally inclined to

\vec{a}, \vec{b},

and

\vec{c}

.

Hence Proved.

Q15Miscellaneous Exercise on Chapter 10

Prove that $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$ , if and only if $\vec{a}, \vec{b}$ are perpendicular, given $\vec{a} \neq \overrightarrow{0}, \vec{b} \neq \overrightarrow{0}$ .

Solution

To Prove:

(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2} \iff \vec{a} \perp \vec{b}

. This is an "if and only if" statement, so we must prove both directions.

Part 1: Prove that if $\vec{a} \perp \vec{b}$ , then $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$ .

Assumption:

\vec{a}

and

\vec{b}

are perpendicular. This means

\vec{a} \cdot \vec{b} = 0

.

Proof: Consider the Left Hand Side (LHS):

LHS = (\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})

Expand using the distributive property:

= \vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b}

Using

\vec{v}\cdot\vec{v} = |\vec{v}|^2

and

\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}

:

= |\vec{a}|^2 + 2(\vec{a}\cdot\vec{b}) + |\vec{b}|^2

Now, use the assumption that

\vec{a} \cdot \vec{b} = 0

:

= |\vec{a}|^2 + 2(0) + |\vec{b}|^2

= |\vec{a}|^2 + |\vec{b}|^2 = RHS

Thus, the first direction is proved.

Part 2: Prove that if $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$ , then $\vec{a} \perp \vec{b}$ .

Assumption:

(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}

.

Proof: Start with the LHS of the assumption and expand it:

(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + 2(\vec{a}\cdot\vec{b}) + |\vec{b}|^2

Now, set this equal to the RHS of the assumption:

|\vec{a}|^2 + 2(\vec{a}\cdot\vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2

Subtract

|\vec{a}|^2

and

|\vec{b}|^2

from both sides:

2(\vec{a}\cdot\vec{b}) = 0

\vec{a}\cdot\vec{b} = 0

Since it is given that

\vec{a} \neq \overrightarrow{0}

and

\vec{b} \neq \overrightarrow{0}

, the condition

\vec{a} \cdot \vec{b} = 0

implies that the vectors

\vec{a}

and

\vec{b}

are perpendicular. Thus, the second direction is proved.

Since both directions are proved, the statement is true.

Hence Proved.

Q16Miscellaneous Exercise on Chapter 10

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$ , then $\vec{a} \cdot \vec{b} \geq 0$ only when

(A)

0<\theta<\frac{\pi}{2}

(B)

0 \leq \theta \leq \frac{\pi}{2}

(C)

0<\theta<\pi

(D)

0 \leq \theta \leq \pi

Solution

Given:

$\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$ , where $0 \le \theta \le \pi$ .
The condition is $\vec{a} \cdot \vec{b} \ge 0$ .

To Find: The range of

\theta

that satisfies the condition.

Formula: The scalar product is defined as

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta

.

Solution: We are given

\vec{a} \cdot \vec{b} \ge 0

. Substituting the formula:

|\vec{a}| |\vec{b}| \cos\theta \ge 0

Since magnitudes

|\vec{a}|

and

|\vec{b}|

are always non-negative, for this inequality to hold, we must have:

\cos\theta \ge 0

The cosine function is non-negative (positive or zero) in the first quadrant. The standard range for the angle between two vectors is

0 \le \theta \le \pi

. Within this range,

\cos\theta \ge 0

when the angle

\theta

is between 0 and

\frac{\pi}{2}

, inclusive. So,

0 \le \theta \le \frac{\pi}{2}

.

This corresponds to option (B).

Final Answer: (B)

0 \leq \theta \leq \frac{\pi}{2}

Q17Miscellaneous Exercise on Chapter 10

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if

(A)

\theta=\frac{\pi}{4}

(B)

\theta=\frac{\pi}{3}

(C)

\theta=\frac{\pi}{2}

(D)

\theta=\frac{2 \pi}{3}

Solution

Given:

$\vec{a}$ and $\vec{b}$ are unit vectors, so $|\vec{a}| = 1$ and $|\vec{b}| = 1$ .
$\theta$ is the angle between $\vec{a}$ and $\vec{b}$ .
$\vec{a} + \vec{b}$ is also a unit vector, so $|\vec{a} + \vec{b}| = 1$ .

To Find: The value of

\theta

.

Solution: Start with the condition

|\vec{a} + \vec{b}| = 1

. Square both sides:

|\vec{a} + \vec{b}|^2 = 1^2 = 1

We know that

|\vec{v}|^2 = \vec{v} \cdot \vec{v}

. So:

(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 1

Expand the dot product:

\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} = 1

|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = 1

Substitute the definition of the dot product,

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta

:

|\vec{a}|^2 + 2|\vec{a}| |\vec{b}| \cos\theta + |\vec{b}|^2 = 1

Now substitute the given magnitudes

|\vec{a}| = 1

and

|\vec{b}| = 1

:

1^2 + 2(1)(1)\cos\theta + 1^2 = 1

1 + 2\cos\theta + 1 = 1

2 + 2\cos\theta = 1

2\cos\theta = 1 - 2 = -1

\cos\theta = -\frac{1}{2}

For the angle

\theta

in the range

0 \le \theta \le \pi

, the value of

\theta

for which

\cos\theta = -\frac{1}{2}

is

\theta = \frac{2\pi}{3}

.

This corresponds to option (D).

Final Answer: (D)

\theta=\frac{2 \pi}{3}

Q18Miscellaneous Exercise on Chapter 10

The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$ is

(A)

0

(B)

-1

(C)

1

(D)

3

Solution

To Evaluate:

\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})

Solution: We evaluate each term separately using the properties of cross products of the standard unit vectors

\hat{i}, \hat{j}, \hat{k}

.

Term 1: $\hat{i} \cdot(\hat{j} \times \hat{k})$ We know that

\hat{j} \times \hat{k} = \hat{i}

. So, the term becomes

\hat{i} \cdot \hat{i}

. Since

\hat{i}

is a unit vector,

\hat{i} \cdot \hat{i} = |\hat{i}|^2 = 1^2 = 1

.

Term 2: $\hat{j} \cdot(\hat{i} \times \hat{k})$ We know that

\hat{k} \times \hat{i} = \hat{j}

, so

\hat{i} \times \hat{k} = -(\hat{k} \times \hat{i}) = -\hat{j}

. So, the term becomes

\hat{j} \cdot (-\hat{j})

. This is equal to

-(\hat{j} \cdot \hat{j}) = -|\hat{j}|^2 = -1^2 = -1

.

Term 3: $\hat{k} \cdot(\hat{i} \times \hat{j})$ We know that

\hat{i} \times \hat{j} = \hat{k}

. So, the term becomes

\hat{k} \cdot \hat{k}

. Since

\hat{k}

is a unit vector,

\hat{k} \cdot \hat{k} = |\hat{k}|^2 = 1^2 = 1

.

Sum of the terms: Total value = (Term 1) + (Term 2) + (Term 3)

= 1 + (-1) + 1 = 1

This corresponds to option (C).

Final Answer: (C) 1

Q19Miscellaneous Exercise on Chapter 10

If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$ , then $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$ when $\theta$ is equal to

(A)

0

(B)

\frac{\pi}{4}

(C)

\frac{\pi}{2}

(D)

\pi

Solution

Given:

$\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$ .
The condition is $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ .

To Find: The value of

\theta

.

Formulas:

Scalar product: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$ .
Vector product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$ .

Solution: Substitute the formulas into the given condition:

||\vec{a}| |\vec{b}| \cos\theta| = |\vec{a}| |\vec{b}| \sin\theta

Since

|\vec{a}|

and

|\vec{b}|

are non-negative magnitudes, we can simplify (assuming the vectors are non-zero):

|\vec{a}| |\vec{b}| |\cos\theta| = |\vec{a}| |\vec{b}| \sin\theta

|\cos\theta| = \sin\theta

The angle

\theta

between two vectors is in the range

0 \le \theta \le \pi

. In this range,

\sin\theta \ge 0

. So, the equation is well-defined.

We need to find

\theta

in

[0, \pi]

for which

|\cos\theta| = \sin\theta

.

Case 1: $0 \le \theta \le \frac{\pi}{2}$ In this interval,

\cos\theta \ge 0

, so

|\cos\theta| = \cos\theta

. The equation becomes

\cos\theta = \sin\theta

. Dividing by

\cos\theta

(assuming

\cos\theta \neq 0

):

\tan\theta = 1

This gives

\theta = \frac{\pi}{4}

. This value is in our assumed interval.

Case 2: $\frac{\pi}{2} < \theta \le \pi$ In this interval,

\cos\theta < 0

, so

|\cos\theta| = -\cos\theta

. The equation becomes

-\cos\theta = \sin\theta

. Dividing by

\cos\theta

:

\tan\theta = -1

This gives

\theta = \frac{3\pi}{4}

. This value is in our assumed interval.

Looking at the options provided: (A) 0 (B)

\frac{\pi}{4}

(C)

\frac{\pi}{2}

(D)

\pi

Only

\frac{\pi}{4}

is listed as an option.

Let's check the options directly:

If $\theta = 0$ , $|\vec{a}\cdot\vec{b}| = |\vec{a}||\vec{b}|$ , but $|\vec{a}\times\vec{b}|=0$ . Not equal.
If $\theta = \frac{\pi}{4}$ , $|\vec{a}\cdot\vec{b}| = |\vec{a}||\vec{b}|\cos(\frac{\pi}{4}) = \frac{|\vec{a}||\vec{b}|}{\sqrt{2}}$ . Also, $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin(\frac{\pi}{4}) = \frac{|\vec{a}||\vec{b}|}{\sqrt{2}}$ . They are equal.
If $\theta = \frac{\pi}{2}$ , $|\vec{a}\cdot\vec{b}| = 0$ , but $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|$ . Not equal.
If $\theta = \pi$ , $|\vec{a}\cdot\vec{b}| = |\vec{a}||\vec{b}|$ , but $|\vec{a}\times\vec{b}|=0$ . Not equal.

Thus, the correct option is (B).

Final Answer: (B)

\frac{\pi}{4}

Vector AlgebraClass 12 Mathematics NCERT Solutions

Represent graphically a displacement of 40 km,30∘40 \text{ km}, 30^{\circ}40 km,30∘ east of north.

Solution

Classify the following measures as scalars and vectors.

10 kg

2 meters north-west

40∘40^{\circ}40∘

40 watt

10−1910^{-19}10−19 coulomb

20 m/s220 \text{ m/s}^{2}20 m/s2

Solution

Classify the following as scalar and vector quantities.

time period

distance

force

velocity

work done

Solution

In a square, identify the following vectors.

Coinitial

Equal

Collinear but not equal

Solution

Answer the following as true or false.

a⃗\vec{a}a and −a⃗-\vec{a}−a are collinear.

Two collinear vectors are always equal in magnitude.

Two vectors having same magnitude are collinear.

Two collinear vectors having the same magnitude are equal.

Solution

Solution

Write two different vectors having same magnitude.

Solution

Write two different vectors having same direction.

Solution

Find the values of xxx and yyy so that the vectors 2i^+3j^2 \hat{i}+3 \hat{j}2i^+3j^​ and xi^+yj^x \hat{i}+y \hat{j}xi^+yj^​ are equal.

Solution

Find the scalar and vector components of the vector with initial point (2,1)(2,1)(2,1) and terminal point ( −5,7-5,7−5,7 ).

Solution

Find the sum of the vectors a⃗=i^−2j^+k^,b⃗=−2i^+4j^+5k^\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}a=i^−2j^​+k^,b=−2i^+4j^​+5k^ and c⃗=i^−6j^−7k^\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}c=i^−6j^​−7k^.

Solution

Find the unit vector in the direction of the vector a⃗=i^+j^+2k^\vec{a}=\hat{i}+\hat{j}+2 \hat{k}a=i^+j^​+2k^.

Solution

Find the unit vector in the direction of vector PQ→\overrightarrow{\mathrm{PQ}}PQ​, where P and Q are the points (1,2,3)(1,2,3)(1,2,3) and (4, 5, 6), respectively.

Solution

For given vectors, a⃗=2i^−j^+2k^\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}a=2i^−j^​+2k^ and b⃗=−i^+j^−k^\vec{b}=-\hat{i}+\hat{j}-\hat{k}b=−i^+j^​−k^, find the unit vector in the direction of the vector a⃗+b⃗\vec{a}+\vec{b}a+b.

Solution

Find a vector in the direction of vector 5i^−j^+2k^5 \hat{i}-\hat{j}+2 \hat{k}5i^−j^​+2k^ which has magnitude 8 units.

Solution

Show that the vectors 2i^−3j^+4k^2 \hat{i}-3 \hat{j}+4 \hat{k}2i^−3j^​+4k^ and −4i^+6j^−8k^-4 \hat{i}+6 \hat{j}-8 \hat{k}−4i^+6j^​−8k^ are collinear.

Solution

Find the direction cosines of the vector i^+2j^+3k^\hat{i}+2 \hat{j}+3 \hat{k}i^+2j^​+3k^.

Solution

Find the direction cosines of the vector joining the points A(1,2,−3)\mathrm{A}(1,2,-3)A(1,2,−3) and B(−1,−2,1)\mathrm{B}(-1,-2,1)B(−1,−2,1), directed from A to B .

Solution

Show that the vector i^+j^+k^\hat{i}+\hat{j}+\hat{k}i^+j^​+k^ is equally inclined to the axes OX,OY\mathrm{OX}, \mathrm{OY}OX,OY and OZ .

Solution

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are i^+2j^−k^\hat{i}+2 \hat{j}-\hat{k}i^+2j^​−k^ and −i^+j^+k^-\hat{i}+\hat{j}+\hat{k}−i^+j^​+k^ respectively, in the ratio 2 : 1

internally

externally

Solution

Find the position vector of the mid point of the vector joining the points P(2,3,4)\mathrm{P}(2,3,4)P(2,3,4) and Q(4,1,−2)\mathrm{Q}(4,1,-2)Q(4,1,−2).

Solution

Solution

In triangle ABC, which of the following is not true:

Solution

If a⃗\vec{a}a and b⃗\vec{b}b are two collinear vectors, then which of the following are incorrect:

Solution

Find the angle between two vectors a⃗\vec{a}a and b⃗\vec{b}b with magnitudes 3\sqrt{3}3​ and 2 , respectively having a⃗⋅b⃗=6\vec{a} \cdot \vec{b}=\sqrt{6}a⋅b=6​.

Solution

Find the angle between the vectors i^−2j^+3k^\hat{i}-2 \hat{j}+3 \hat{k}i^−2j^​+3k^ and 3i^−2j^+k^3 \hat{i}-2 \hat{j}+\hat{k}3i^−2j^​+k^

Solution

Find the projection of the vector i^−j^\hat{i}-\hat{j}i^−j^​ on the vector i^+j^\hat{i}+\hat{j}i^+j^​.

Solution

Find the projection of the vector i^+3j^+7k^\hat{i}+3 \hat{j}+7 \hat{k}i^+3j^​+7k^ on the vector 7i^−j^+8k^7 \hat{i}-\hat{j}+8 \hat{k}7i^−j^​+8k^.

Solution

Solution

Find ∣a⃗∣|\vec{a}|∣a∣ and ∣b⃗∣|\vec{b}|∣b∣, if (a⃗+b⃗)⋅(a⃗−b⃗)=8(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8(a+b)⋅(a−b)=8 and ∣a⃗∣=8∣b⃗∣|\vec{a}|=8|\vec{b}|∣a∣=8∣b∣.

Solution

Evaluate the product (3a⃗−5b⃗)⋅(2a⃗+7b⃗)(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})(3a−5b)⋅(2a+7b).

Solution

Represent graphically a displacement of $40 \text{ km}, 30^{\circ}$ east of north.

$40^{\circ}$

$10^{-19}$ coulomb

$20 \text{ m/s}^{2}$

$\vec{a}$ and $-\vec{a}$ are collinear.

Find the values of $x$ and $y$ so that the vectors $2 \hat{i}+3 \hat{j}$ and $x \hat{i}+y \hat{j}$ are equal.

Find the scalar and vector components of the vector with initial point $(2,1)$ and terminal point ( $-5,7$ ).

Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$ .

Find the unit vector in the direction of the vector $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ .

Find the unit vector in the direction of vector $\overrightarrow{\mathrm{PQ}}$ , where P and Q are the points $(1,2,3)$ and (4, 5, 6), respectively.

For given vectors, $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ , find the unit vector in the direction of the vector $\vec{a}+\vec{b}$ .

Find a vector in the direction of vector $5 \hat{i}-\hat{j}+2 \hat{k}$ which has magnitude 8 units.

Show that the vectors $2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+6 \hat{j}-8 \hat{k}$ are collinear.

Find the direction cosines of the vector $\hat{i}+2 \hat{j}+3 \hat{k}$ .

Find the direction cosines of the vector joining the points $\mathrm{A}(1,2,-3)$ and $\mathrm{B}(-1,-2,1)$ , directed from A to B .

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $\mathrm{OX}, \mathrm{OY}$ and OZ .

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio 2 : 1

Find the position vector of the mid point of the vector joining the points $\mathrm{P}(2,3,4)$ and $\mathrm{Q}(4,1,-2)$ .

If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 2 , respectively having $\vec{a} \cdot \vec{b}=\sqrt{6}$ .

Find the angle between the vectors $\hat{i}-2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$

Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$ .

Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $7 \hat{i}-\hat{j}+8 \hat{k}$ .

Find $|\vec{a}|$ and $|\vec{b}|$ , if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|$ .

Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})$ .

Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$ , having the same magnitude and such that the angle between them is $60^{\circ}$ and their scalar product is $\frac{1}{2}$ .

Find $|\vec{x}|$ , if for a unit vector $\vec{a}$ , $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$ .

Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$ , for any two nonzero vectors $\vec{a}$ and $\vec{b}$ .

If $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0$ , then what can be concluded about the vector $\vec{b}$ ?

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ , find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ .

If either vector $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$ , then $\vec{a} \cdot \vec{b}=0$ . But the converse need not be true. Justify your answer with an example.

Show that the points $\mathrm{A}(1,2,7), \mathrm{B}(2,6,3)$ and $\mathrm{C}(3,10,-1)$ are collinear.

Show that the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$ form the vertices of a right angled triangle.

If $\vec{a}$ is a nonzero vector of magnitude ' $a$ ' and $\lambda$ a nonzero scalar, then $\lambda \vec{a}$ is unit vector if

Find $|\vec{a} \times \vec{b}|$ , if $\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$ .

Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$ .

If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}, \frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$ , then find $\theta$ and hence, the components of $\vec{a}$ .

Show that $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$

Find $\lambda$ and $\mu$ if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$ .

Given that $\vec{a} \cdot \vec{b}=0$ and $\vec{a} \times \vec{b}=\overrightarrow{0}$ . What can you conclude about the vectors $\vec{a}$ and $\vec{b}$ ?

If either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$ , then $\vec{a} \times \vec{b}=\overrightarrow{0}$ . Is the converse true? Justify your answer with an example.

Find the area of the triangle with vertices $\mathrm{A}(1,1,2), \mathrm{B}(2,3,5)$ and $\mathrm{C}(1,5,5)$ .

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$ .

Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$ , then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is

Write down a unit vector in XY-plane, making an angle of $30^{\circ}$ with the positive direction of $x$ -axis.

Find the scalar components and magnitude of the vector joining the points $\mathrm{P}\left(x_{1}, y_{1}, z_{1}\right)$ and $\mathrm{Q}\left(x_{2}, y_{2}, z_{2}\right)$ .

A girl walks 4 km towards west, then she walks 3 km in a direction $30^{\circ}$ east of north and stops. Determine the girl's displacement from her initial point of departure.

If $\vec{a}=\vec{b}+\vec{c}$ , then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.

Find the value of $x$ for which $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector.

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ .

Show that the points $\mathrm{A}(1,-2,-8), \mathrm{B}(5,0,-2)$ and $\mathrm{C}(11,3,7)$ are collinear, and find the ratio in which B divides AC.

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ( $2 \vec{a}+\vec{b}$ ) and ( $\vec{a}-3 \vec{b}$ ) externally in the ratio $1: 2$ . Also, show that P is the mid point of the line segment RQ .

The two adjacent sides of a parallelogram are $2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\hat{i}-2 \hat{j}-3 \hat{k}$ . Find the unit vector parallel to its diagonal. Also, find its area.

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ .