Three Dimensional GeometryClass 12 Mathematics NCERT Solutions

Given: The angles made by the line with the x, y, and z-axes are $\alpha = 90^{\circ}$, $\beta = 135^{\circ}$, and $\gamma = 45^{\circ}$ respectively.

To Find: The direction cosines of the line.

Formula: The direction cosines ($l, m, n$) of a line making angles $\alpha, \beta, \gamma$ with the coordinate axes are given by: $l = \cos \alpha$ $m = \cos \beta$ $n = \cos \gamma$

Solution: We have: $l = \cos 90^{\circ} = 0$ $m = \cos 135^{\circ} = \cos (180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} = -\frac{1}{\sqrt{2}}$ $n = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$

Final Answer: The direction cosines of the line are $0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

Given: A line makes equal angles with the coordinate axes. Let this angle be $\alpha$. So, $\alpha = \beta = \gamma$.

To Find: The direction cosines of the line.

Formula: The direction cosines are $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$. Also, we know that $l^2 + m^2 + n^2 = 1$.

Solution: Since the angles are equal, the direction cosines are also equal. Let $l = m = n = \cos \alpha$.

Using the relation $l^2 + m^2 + n^2 = 1$: $(\cos \alpha)^2 + (\cos \alpha)^2 + (\cos \alpha)^2 = 1$ $3 \cos^2 \alpha = 1$ $\cos^2 \alpha = \frac{1}{3}$ $\cos \alpha = \pm \frac{1}{\sqrt{3}}$

So, the direction cosines are $l = m = n = \pm \frac{1}{\sqrt{3}}$.

Final Answer: The direction cosines of the line are $\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}$.

Given: The direction ratios of a line are $a = -18, b = 12, c = -4$.

To Find: The direction cosines of the line.

Formula: If $a, b, c$ are the direction ratios of a line, then its direction cosines ($l, m, n$) are given by: $l = \frac{a}{\sqrt{a^2+b^2+c^2}}$ $m = \frac{b}{\sqrt{a^2+b^2+c^2}}$ $n = \frac{c}{\sqrt{a^2+b^2+c^2}}$

Solution: First, we calculate the magnitude: $\sqrt{a^2+b^2+c^2} = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$ $= \sqrt{324 + 144 + 16}$ $= \sqrt{484} = 22$

Now, we find the direction cosines: $l = \frac{-18}{22} = -\frac{9}{11}$ $m = \frac{12}{22} = \frac{6}{11}$ $n = \frac{-4}{22} = -\frac{2}{11}$

Final Answer: The direction cosines of the line are $-\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}$.

Given: Three points: A(2, 3, 4), B(-1, -2, 1), and C(5, 8, 7).

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

Method: Three points are collinear if the direction ratios of the line segment joining any two pairs of points are proportional.

Solution: Step 1: Find the direction ratios of the line segment AB. Let the points be $\mathrm{A}(x_1, y_1, z_1) = (2, 3, 4)$ and $\mathrm{B}(x_2, y_2, z_2) = (-1, -2, 1)$. The direction ratios of AB are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. $a_1 = -1 - 2 = -3$ $b_1 = -2 - 3 = -5$ $c_1 = 1 - 4 = -3$ So, the direction ratios of AB are $(-3, -5, -3)$.

Step 2: Find the direction ratios of the line segment BC. Let the points be $\mathrm{B}(x_1, y_1, z_1) = (-1, -2, 1)$ and $\mathrm{C}(x_2, y_2, z_2) = (5, 8, 7)$. The direction ratios of BC are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. $a_2 = 5 - (-1) = 6$ $b_2 = 8 - (-2) = 10$ $c_2 = 7 - 1 = 6$ So, the direction ratios of BC are $(6, 10, 6)$.

Step 3: Check for proportionality. We compare the direction ratios of AB and BC: $\frac{a_2}{a_1} = \frac{6}{-3} = -2$ $\frac{b_2}{b_1} = \frac{10}{-5} = -2$ $\frac{c_2}{c_1} = \frac{6}{-3} = -2$

Since $\frac{a_2}{a_1} = \frac{b_2}{b_1} = \frac{c_2}{c_1}$, the direction ratios are proportional. This means that the line segment AB is parallel to the line segment BC.

Since B is a common point to both line segments AB and BC, the points A, B, and C must lie on the same line.

Hence Proved: The points (2, 3, 4), (-1, -2, 1), and (5, 8, 7) are collinear.

Given: The vertices of a triangle are A(3, 5, -4), B(-1, 1, 2), and C(-5, -5, -2).

To Find: The direction cosines of the sides AB, BC, and CA.

1. For side AB: The direction ratios of AB are: $a = -1 - 3 = -4$ $b = 1 - 5 = -4$ $c = 2 - (-4) = 6$

Length of AB = $\sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}$.

The direction cosines of AB are: $l = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$ $m = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$ $n = \frac{6}{2\sqrt{17}} = \frac{3}{\sqrt{17}}$

2. For side BC: The direction ratios of BC are: $a = -5 - (-1) = -4$ $b = -5 - 1 = -6$ $c = -2 - 2 = -4$

Length of BC = $\sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}$.

The direction cosines of BC are: $l = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$ $m = \frac{-6}{2\sqrt{17}} = -\frac{3}{\sqrt{17}}$ $n = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$

3. For side CA: The direction ratios of CA are: $a = 3 - (-5) = 8$ $b = 5 - (-5) = 10$ $c = -4 - (-2) = -2$

Length of CA = $\sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168} = 2\sqrt{42}$.

The direction cosines of CA are: $l = \frac{8}{2\sqrt{42}} = \frac{4}{\sqrt{42}}$ $m = \frac{10}{2\sqrt{42}} = \frac{5}{\sqrt{42}}$ $n = \frac{-2}{2\sqrt{42}} = -\frac{1}{\sqrt{42}}$

Final Answer: Direction cosines of side AB: $(-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}})$ Direction cosines of side BC: $(-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}})$ Direction cosines of side CA: $(\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}})$

Given: Direction cosines of three lines: Line 1 ($L_1$): $(l_1, m_1, n_1) = (\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13})$ Line 2 ($L_2$): $(l_2, m_2, n_2) = (\frac{4}{13}, \frac{12}{13}, \frac{3}{13})$ Line 3 ($L_3$): $(l_3, m_3, n_3) = (\frac{3}{13}, \frac{-4}{13}, \frac{12}{13})$

To Show: The three lines are mutually perpendicular.

Condition for Perpendicularity: Two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are perpendicular if $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

Proof: 1. Check if $L_1 \perp L_2$: $l_1 l_2 + m_1 m_2 + n_1 n_2 = (\frac{12}{13})(\frac{4}{13}) + (\frac{-3}{13})(\frac{12}{13}) + (\frac{-4}{13})(\frac{3}{13})$ $= \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = \frac{48 - 36 - 12}{169} = \frac{0}{169} = 0$ Therefore, $L_1$ is perpendicular to $L_2$.

2. Check if $L_2 \perp L_3$: $l_2 l_3 + m_2 m_3 + n_2 n_3 = (\frac{4}{13})(\frac{3}{13}) + (\frac{12}{13})(\frac{-4}{13}) + (\frac{3}{13})(\frac{12}{13})$ $= \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = \frac{12 - 48 + 36}{169} = \frac{0}{169} = 0$ Therefore, $L_2$ is perpendicular to $L_3$.

3. Check if $L_3 \perp L_1$: $l_3 l_1 + m_3 m_1 + n_3 n_1 = (\frac{3}{13})(\frac{12}{13}) + (\frac{-4}{13})(\frac{-3}{13}) + (\frac{12}{13})(\frac{-4}{13})$ $= \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = \frac{36 + 12 - 48}{169} = \frac{0}{169} = 0$ Therefore, $L_3$ is perpendicular to $L_1$.

Since each pair of lines is perpendicular, the three lines are mutually perpendicular.

Given: Line 1 ($L_1$) passes through points A(1, -1, 2) and B(3, 4, -2). Line 2 ($L_2$) passes through points C(0, 3, 2) and D(3, 5, 6).

To Show: Line $L_1$ is perpendicular to Line $L_2$.

Condition for Perpendicularity: Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Solution: Step 1: Find the direction ratios of Line $L_1$. Direction ratios of $L_1$ are given by the differences in coordinates of points A and B. $a_1 = 3 - 1 = 2$ $b_1 = 4 - (-1) = 5$ $c_1 = -2 - 2 = -4$ So, the direction ratios of $L_1$ are $(2, 5, -4)$.

Step 2: Find the direction ratios of Line $L_2$. Direction ratios of $L_2$ are given by the differences in coordinates of points C and D. $a_2 = 3 - 0 = 3$ $b_2 = 5 - 3 = 2$ $c_2 = 6 - 2 = 4$ So, the direction ratios of $L_2$ are $(3, 2, 4)$.

Step 3: Check the condition for perpendicularity. $a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(3) + (5)(2) + (-4)(4)$ $= 6 + 10 - 16$ $= 16 - 16 = 0$

Since the sum is 0, the lines are perpendicular.

Given: Line 1 ($L_1$) passes through points A(4, 7, 8) and B(2, 3, 4). Line 2 ($L_2$) passes through points C(-1, -2, 1) and D(1, 2, 5).

To Show: Line $L_1$ is parallel to Line $L_2$.

Condition for Parallelism: Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are parallel if their direction ratios are proportional, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Solution: Step 1: Find the direction ratios of Line $L_1$. Direction ratios of $L_1$ are given by the differences in coordinates of points A and B. $a_1 = 2 - 4 = -2$ $b_1 = 3 - 7 = -4$ $c_1 = 4 - 8 = -4$ So, the direction ratios of $L_1$ are $(-2, -4, -4)$.

Step 2: Find the direction ratios of Line $L_2$. Direction ratios of $L_2$ are given by the differences in coordinates of points C and D. $a_2 = 1 - (-1) = 2$ $b_2 = 2 - (-2) = 4$ $c_2 = 5 - 1 = 4$ So, the direction ratios of $L_2$ are $(2, 4, 4)$.

Step 3: Check the condition for parallelism. We check the ratio of the corresponding direction ratios: $\frac{a_1}{a_2} = \frac{-2}{2} = -1$ $\frac{b_1}{b_2} = \frac{-4}{4} = -1$ $\frac{c_1}{c_2} = \frac{-4}{4} = -1$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -1$, the direction ratios are proportional. Therefore, the lines are parallel.

Given: A line passes through the point A(1, 2, 3). The position vector of point A is $\vec{a} = 1 \hat{i} + 2 \hat{j} + 3 \hat{k}$. The line is parallel to the vector $\vec{b} = 3 \hat{i} + 2 \hat{j} - 2 \hat{k}$.

To Find: The vector and Cartesian equations of the line.

Formula: Vector equation of a line: $\vec{r} = \vec{a} + \lambda \vec{b}$ Cartesian equation of a line: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Solution: 1. Vector Equation: Substituting the given vectors $\vec{a}$ and $\vec{b}$ into the formula: $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ This is the required vector equation of the line, where $\lambda$ is a real number.

2. Cartesian Equation: The line passes through the point $(x_1, y_1, z_1) = (1, 2, 3)$. The direction ratios of the line are the components of the parallel vector $\vec{b}$, so $(a, b, c) = (3, 2, -2)$.

Substituting these values into the Cartesian formula: $\frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z - 3}{-2}$ This is the required Cartesian equation of the line.

Final Answer: Vector equation: $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ Cartesian equation: $\frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z - 3}{-2}$

Given: The position vector of the point through which the line passes is $\vec{a} = 2 \hat{i} - \hat{j} + 4 \hat{k}$. This corresponds to the point $(x_1, y_1, z_1) = (2, -1, 4)$. The direction vector of the line is $\vec{b} = \hat{i} + 2 \hat{j} - \hat{k}$. This corresponds to direction ratios $(a, b, c) = (1, 2, -1)$.

To Find: The vector and Cartesian equations of the line.

Formula: Vector equation: $\vec{r} = \vec{a} + \lambda \vec{b}$ Cartesian equation: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Solution: 1. Vector Equation: Substituting the given vectors $\vec{a}$ and $\vec{b}$ into the formula: $\vec{r} = (2 \hat{i} - \hat{j} + 4 \hat{k}) + \lambda(\hat{i} + 2 \hat{j} - \hat{k})$ This is the required vector equation, where $\lambda$ is a real number.

2. Cartesian Equation: Using the point $(x_1, y_1, z_1) = (2, -1, 4)$ and direction ratios $(a, b, c) = (1, 2, -1)$: $\frac{x - 2}{1} = \frac{y - (-1)}{2} = \frac{z - 4}{-1}$ $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$ This is the required Cartesian equation.

Final Answer: Vector equation: $\vec{r} = (2 \hat{i} - \hat{j} + 4 \hat{k}) + \lambda(\hat{i} + 2 \hat{j} - \hat{k})$ Cartesian equation: $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$

Given: The required line passes through the point $(x_1, y_1, z_1) = (-2, 4, -5)$. The required line is parallel to the given line: $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

To Find: The Cartesian equation of the required line.

Concept: Parallel lines have the same or proportional direction ratios.

Solution: The direction ratios of the given line are $(a, b, c) = (3, 5, 6)$. Since the required line is parallel to the given line, its direction ratios will also be $(3, 5, 6)$.

The required line passes through the point $(-2, 4, -5)$.

Using the Cartesian equation formula $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$: $\frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6}$ $\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}$

Final Answer: The Cartesian equation of the line is $\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}$.

Given: The Cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$.

To Find: The vector form of the equation of the line.

Formula: The vector form of a line is $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is a vector parallel to the line.

Solution: Comparing the given Cartesian equation with the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, we can identify the point and direction ratios.

The equation can be written as $\frac{x-5}{3} = \frac{y-(-4)}{7} = \frac{z-6}{2}$.

The line passes through the point $(x_1, y_1, z_1) = (5, -4, 6)$. The position vector of this point is $\vec{a} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k}$.

The direction ratios of the line are $(a, b, c) = (3, 7, 2)$. The vector parallel to the line is $\vec{b} = 3 \hat{i} + 7 \hat{j} + 2 \hat{k}$.

Substituting $\vec{a}$ and $\vec{b}$ into the vector equation formula: $\vec{r} = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + \lambda(3 \hat{i} + 7 \hat{j} + 2 \hat{k})$

Final Answer: The vector form of the line is $\vec{r} = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + \lambda(3 \hat{i} + 7 \hat{j} + 2 \hat{k})$.

Formula: The angle $\theta$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by: $\cos \theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} \right|$

(i) Solution: Given lines: $L_1: \vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + \lambda(3 \hat{i} + 2 \hat{j} + 6 \hat{k})$ $L_2: \vec{r} = 7 \hat{i} - 6 \hat{k} + \mu(\hat{i} + 2 \hat{j} + 2 \hat{k})$

Here, $\vec{b}_1 = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and $\vec{b}_2 = \hat{i} + 2 \hat{j} + 2 \hat{k}$.

Calculations: $\vec{b}_1 \cdot \vec{b}_2 = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19$ $|\vec{b}_1| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$ $|\vec{b}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Angle: $\cos \theta = \left| \frac{19}{7 \times 3} \right| = \frac{19}{21}$ $\theta = \cos^{-1}(\frac{19}{21})$

Final Answer (i): The angle between the lines is $\cos^{-1}(\frac{19}{21})$.

(ii) Solution: Given lines: $L_1: \vec{r} = 3 \hat{i} + \hat{j} - 2 \hat{k} + \lambda(\hat{i} - \hat{j} - 2 \hat{k})$ $L_2: \vec{r} = 2 \hat{i} - \hat{j} - 56 \hat{k} + \mu(3 \hat{i} - 5 \hat{j} - 4 \hat{k})$

Here, $\vec{b}_1 = \hat{i} - \hat{j} - 2 \hat{k}$ and $\vec{b}_2 = 3 \hat{i} - 5 \hat{j} - 4 \hat{k}$.

Calculations: $\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16$ $|\vec{b}_1| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$ $|\vec{b}_2| = \sqrt{3^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$

Angle: $\cos \theta = \left| \frac{16}{\sqrt{6} \times 5\sqrt{2}} \right| = \frac{16}{5\sqrt{12}} = \frac{16}{5 \times 2\sqrt{3}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$ $\theta = \cos^{-1}(\frac{8}{5\sqrt{3}})$

Final Answer (ii): The angle between the lines is $\cos^{-1}(\frac{8}{5\sqrt{3}})$.

Formula: The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by: $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$

(i) Solution: Given lines: $L_1: \frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$ $L_2: \frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$

Direction ratios are: For $L_1$: $(a_1, b_1, c_1) = (2, 5, -3)$ For $L_2$: $(a_2, b_2, c_2) = (-1, 8, 4)$

Calculations: $a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(-1) + (5)(8) + (-3)(4) = -2 + 40 - 12 = 26$ $\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$ $\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$

Angle: $\cos \theta = \left| \frac{26}{9\sqrt{38}} \right| = \frac{26}{9\sqrt{38}}$ $\theta = \cos^{-1}(\frac{26}{9\sqrt{38}})$

Final Answer (i): The angle between the lines is $\cos^{-1}(\frac{26}{9\sqrt{38}})$.

(ii) Solution: Given lines: $L_1: \frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ $L_2: \frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$

Direction ratios are: For $L_1$: $(a_1, b_1, c_1) = (2, 2, 1)$ For $L_2$: $(a_2, b_2, c_2) = (4, 1, 8)$

Calculations: $a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18$ $\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$ $\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$

Angle: $\cos \theta = \left| \frac{18}{3 \times 9} \right| = \frac{18}{27} = \frac{2}{3}$ $\theta = \cos^{-1}(\frac{2}{3})$

Final Answer (ii): The angle between the lines is $\cos^{-1}(\frac{2}{3})$.

Given: Two lines are at right angles (perpendicular). $L_1: \frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ $L_2: \frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$

To Find: The value of $p$.

Condition for Perpendicularity: Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Solution: Step 1: Rewrite the equations in standard form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$.

For line $L_1$: $\frac{1-x}{3} \implies \frac{-(x-1)}{3} \implies \frac{x-1}{-3}$ $\frac{7y-14}{2p} \implies \frac{7(y-2)}{2p} \implies \frac{y-2}{2p/7}$ $\frac{z-3}{2}$ So, the standard form for $L_1$ is $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$. Direction ratios of $L_1$ are $(a_1, b_1, c_1) = (-3, \frac{2p}{7}, 2)$.

For line $L_2$: $\frac{7-7x}{3p} \implies \frac{-7(x-1)}{3p} \implies \frac{x-1}{-3p/7}$ $\frac{y-5}{1}$ $\frac{6-z}{5} \implies \frac{-(z-6)}{5} \implies \frac{z-6}{-5}$ So, the standard form for $L_2$ is $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$. Direction ratios of $L_2$ are $(a_2, b_2, c_2) = (-\frac{3p}{7}, 1, -5)$.

Step 2: Apply the condition for perpendicularity. $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ $(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0$ $\frac{9p}{7} + \frac{2p}{7} - 10 = 0$ $\frac{11p}{7} = 10$ $11p = 70$ $p = \frac{70}{11}$

Final Answer: The value of $p$ is $\frac{70}{11}$.

Given: Two lines: $L_1: \frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ $L_2: \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$

To Show: The lines are perpendicular.

Condition for Perpendicularity: Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Solution: Step 1: Identify the direction ratios of both lines. For line $L_1$, the direction ratios are $(a_1, b_1, c_1) = (7, -5, 1)$. For line $L_2$, the direction ratios are $(a_2, b_2, c_2) = (1, 2, 3)$.