Introduction to Three Dimensional GeometryClass 11 Mathematics NCERT Solutions

Concept: Any point lying on the $x$-axis has its perpendicular distance from the YZ-plane as some value $x$, but its perpendicular distances from the XZ-plane and XY-plane are zero. The distance from the XZ-plane is the $y$-coordinate, and the distance from the XY-plane is the $z$-coordinate.

Solution: For any point on the $x$-axis, its coordinates are of the form $(x, 0, 0)$. Therefore, its $y$-coordinate is 0 and its $z$-coordinate is 0.

Final Answer: The $y$-coordinate is 0 and the $z$-coordinate is 0.

Concept: Any point lying in the XZ-plane has its perpendicular distance from the XZ-plane equal to zero. The perpendicular distance of a point $(x, y, z)$ from the XZ-plane is given by its $y$-coordinate.

Solution: For any point in the XZ-plane, its coordinates are of the form $(x, 0, z)$. Therefore, its $y$-coordinate must be 0.

Final Answer: Its $y$-coordinate is 0.

Concept: The octant in which a point lies is determined by the signs of its $x, y,$ and $z$ coordinates.

Solution: We determine the octant for each point by observing the signs of its coordinates:

Formula: The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by: $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

(i) Points: $(2,3,5)$ and $(4,3,1)$ Solution: $d = \sqrt{(4-2)^2 + (3-3)^2 + (1-5)^2}$ $d = \sqrt{2^2 + 0^2 + (-4)^2}$ $d = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \text{ units}$

(ii) Points: $(-3,7,2)$ and $(2,4,-1)$ Solution: $d = \sqrt{(2 - (-3))^2 + (4-7)^2 + (-1-2)^2}$ $d = \sqrt{(2+3)^2 + (-3)^2 + (-3)^2}$ $d = \sqrt{5^2 + 9 + 9} = \sqrt{25 + 9 + 9} = \sqrt{43} \text{ units}$

(iii) Points: $(-1,3,-4)$ and $(1,-3,4)$ Solution: $d = \sqrt{(1 - (-1))^2 + (-3-3)^2 + (4 - (-4))^2}$ $d = \sqrt{(1+1)^2 + (-6)^2 + (4+4)^2}$ $d = \sqrt{2^2 + (-6)^2 + 8^2} = \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26} \text{ units}$

(iv) Points: $(2,-1,3)$ and $(-2,1,3)$ Solution: $d = \sqrt{(-2-2)^2 + (1 - (-1))^2 + (3-3)^2}$ $d = \sqrt{(-4)^2 + (1+1)^2 + 0^2}$ $d = \sqrt{16 + 2^2 + 0} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \text{ units}$

To Show: The points P(-2,3,5), Q(1,2,3), and R(7,0,-1) are collinear.

Concept: Three points are collinear if they lie on the same straight line. This can be verified by showing that the sum of the distances between two pairs of points is equal to the distance between the third pair. That is, $PQ + QR = PR$.

Formula: Distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Proof: First, we calculate the distances between each pair of points.

Distance PQ: $PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2}$ $PQ = \sqrt{(3)^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

Distance QR: $QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2}$ $QR = \sqrt{(6)^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$

Distance PR: $PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2}$ $PR = \sqrt{(9)^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = \sqrt{9 \times 14} = 3\sqrt{14}$

Now, we check if the sum of the two smaller distances equals the largest distance: $PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14}$ We see that $PQ + QR = PR$.

Since the sum of the lengths of two segments is equal to the length of the third segment, the points P, Q, and R are collinear.

Given: Vertices A(0, 7, -10), B(1, 6, -6), and C(4, 9, -6).

To Verify: Triangle ABC is an isosceles triangle.

Solution: We calculate the lengths of the three sides of the triangle. $AB = \sqrt{(1-0)^2 + (6-7)^2 + (-6 - (-10))^2} = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$ $BC = \sqrt{(4-1)^2 + (9-6)^2 + (-6 - (-6))^2} = \sqrt{3^2 + 3^2 + 0^2} = \sqrt{9+9+0} = \sqrt{18} = 3\sqrt{2}$ $CA = \sqrt{(0-4)^2 + (7-9)^2 + (-10 - (-6))^2} = \sqrt{(-4)^2 + (-2)^2 + (-4)^2} = \sqrt{16+4+16} = \sqrt{36} = 6$ Since $AB = BC = 3\sqrt{2}$, two sides of the triangle are equal.

Conclusion: The triangle is an isosceles triangle. Hence verified.

Given: Vertices A(0, 7, 10), B(-1, 6, 6), and C(-4, 9, 6).

To Verify: Triangle ABC is a right-angled triangle.

Solution: We calculate the squares of the lengths of the three sides. $AB^2 = (-1-0)^2 + (6-7)^2 + (6-10)^2 = (-1)^2 + (-1)^2 + (-4)^2 = 1+1+16 = 18$ $BC^2 = (-4 - (-1))^2 + (9-6)^2 + (6-6)^2 = (-3)^2 + 3^2 + 0^2 = 9+9+0 = 18$ $CA^2 = (0 - (-4))^2 + (7-9)^2 + (10-6)^2 = 4^2 + (-2)^2 + 4^2 = 16+4+16 = 36$ Now we check if the Pythagorean theorem holds: $AB^2 + BC^2 = CA^2$. $18 + 18 = 36$ Since $AB^2 + BC^2 = CA^2$, the triangle satisfies the Pythagorean theorem.

Conclusion: The triangle is a right-angled triangle, with the right angle at vertex B. Hence verified.

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

To Verify: The quadrilateral ABCD is a parallelogram.

Concept: A quadrilateral is a parallelogram if and only if its diagonals bisect each other. This means the midpoint of diagonal AC must be the same as the midpoint of diagonal BD.

Solution: Midpoint of diagonal AC: $M_{AC} = \left( \frac{-1+4}{2}, \frac{2+(-7)}{2}, \frac{1+8}{2} \right) = \left( \frac{3}{2}, \frac{-5}{2}, \frac{9}{2} \right)$ Midpoint of diagonal BD: $M_{BD} = \left( \frac{1+2}{2}, \frac{-2+(-3)}{2}, \frac{5+4}{2} \right) = \left( \frac{3}{2}, \frac{-5}{2}, \frac{9}{2} \right)$ Since the midpoint of AC is the same as the midpoint of BD, the diagonals bisect each other.

Conclusion: The given vertices form a parallelogram. Hence verified.

Given: Let the points be A(1, 2, 3) and B(3, 2, -1). Let P(x, y, z) be any point on the required set.

Condition: The point P is equidistant from A and B, which means PA = PB. This implies $PA^2 = PB^2$.

Formula: The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.

Solution: Using the distance formula for $PA^2$ and $PB^2$: $PA^2 = (x-1)^2 + (y-2)^2 + (z-3)^2$ $PB^2 = (x-3)^2 + (y-2)^2 + (z-(-1))^2 = (x-3)^2 + (y-2)^2 + (z+1)^2$

Setting $PA^2 = PB^2$: $(x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z+1)^2$

The term $(y-2)^2$ is on both sides, so it can be cancelled: $(x-1)^2 + (z-3)^2 = (x-3)^2 + (z+1)^2$

Expanding the squared terms: $(x^2 - 2x + 1) + (z^2 - 6z + 9) = (x^2 - 6x + 9) + (z^2 + 2z + 1)$

The terms $x^2$ and $z^2$ cancel from both sides: $-2x + 1 - 6z + 9 = -6x + 9 + 2z + 1$ $-2x - 6z + 10 = -6x + 2z + 10$

The constant term 10 cancels from both sides: $-2x - 6z = -6x + 2z$

Now, we group the terms with variables on one side: $-2x + 6x - 6z - 2z = 0$ $4x - 8z = 0$

Dividing the entire equation by 4: $x - 2z = 0$

This is the equation of a plane that is the perpendicular bisector of the line segment AB.

Final Answer: The equation of the set of points is $x - 2z = 0$.

Given: Points A(4, 0, 0) and B(-4, 0, 0). Let P(x, y, z) be any point in the set.

Condition: The sum of the distances from P to A and B is 10. That is, PA + PB = 10.

Solution: Using the distance formula: $PA = \sqrt{(x-4)^2 + (y-0)^2 + (z-0)^2} = \sqrt{(x-4)^2 + y^2 + z^2}$ $PB = \sqrt{(x-(-4))^2 + (y-0)^2 + (z-0)^2} = \sqrt{(x+4)^2 + y^2 + z^2}$

According to the condition: $\sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x+4)^2 + y^2 + z^2} = 10$

Isolate one of the square root terms: $\sqrt{(x-4)^2 + y^2 + z^2} = 10 - \sqrt{(x+4)^2 + y^2 + z^2}$

Square both sides: $(x-4)^2 + y^2 + z^2 = \left(10 - \sqrt{(x+4)^2 + y^2 + z^2}\right)^2$ $(x^2 - 8x + 16) + y^2 + z^2 = 100 - 20\sqrt{(x+4)^2 + y^2 + z^2} + ((x+4)^2 + y^2 + z^2)$ $x^2 - 8x + 16 + y^2 + z^2 = 100 - 20\sqrt{(x+4)^2 + y^2 + z^2} + (x^2 + 8x + 16 + y^2 + z^2)$

Cancel common terms ($x^2, y^2, z^2, 16$) from both sides: $-8x = 100 + 8x - 20\sqrt{(x+4)^2 + y^2 + z^2}$

Isolate the remaining square root term: $20\sqrt{(x+4)^2 + y^2 + z^2} = 100 + 16x$

Divide the entire equation by 4 to simplify: $5\sqrt{(x+4)^2 + y^2 + z^2} = 25 + 4x$

Square both sides again to eliminate the square root: $25 \left[ (x+4)^2 + y^2 + z^2 \right] = (25 + 4x)^2$ $25 \left[ x^2 + 8x + 16 + y^2 + z^2 \right] = 625 + 200x + 16x^2$

Expand the left side: $25x^2 + 200x + 400 + 25y^2 + 25z^2 = 625 + 200x + 16x^2$

Cancel the $200x$ term from both sides and rearrange: $25x^2 - 16x^2 + 25y^2 + 25z^2 = 625 - 400$ $9x^2 + 25y^2 + 25z^2 = 225$

This is the required equation. It can also be written in standard form by dividing by 225: $\frac{9x^2}{225} + \frac{25y^2}{225} + \frac{25z^2}{225} = 1$ $\frac{x^2}{25} + \frac{y^2}{9} + \frac{z^2}{9} = 1$

Final Answer: The equation of the set of points P is $9x^2 + 25y^2 + 25z^2 = 225$.

Given: Vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4), and C(-1, 1, 2).

To Find: The coordinates of the fourth vertex D(x, y, z).

Concept: In a parallelogram, the diagonals bisect each other. This means the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

Formula: The midpoint of a line segment with endpoints $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by: $M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right)$

Solution: Let the coordinates of the fourth vertex be D(x, y, z).

Midpoint of diagonal AC: $M_{AC} = \left( \frac{3+(-1)}{2}, \frac{-1+1}{2}, \frac{2+2}{2} \right) = \left( \frac{2}{2}, \frac{0}{2}, \frac{4}{2} \right) = (1, 0, 2)$

Midpoint of diagonal BD: $M_{BD} = \left( \frac{1+x}{2}, \frac{2+y}{2}, \frac{-4+z}{2} \right)$

Since $M_{AC} = M_{BD}$, we can equate their corresponding coordinates:

For the x-coordinate: $\frac{1+x}{2} = 1 \implies 1+x = 2 \implies x = 1$

For the y-coordinate: $\frac{2+y}{2} = 0 \implies 2+y = 0 \implies y = -2$

For the z-coordinate: $\frac{-4+z}{2} = 2 \implies -4+z = 4 \implies z = 8$

Thus, the coordinates of the fourth vertex D are (1, -2, 8).

Final Answer: The coordinates of the fourth vertex D are (1, -2, 8).

Given: The vertices of a triangle are A(0, 0, 6), B(0, 4, 0), and C(6, 0, 0).

To Find: The lengths of the three medians of the triangle.

Concept: A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

Solution: First, we find the midpoints of the sides BC, AC, and AB. Let D, E, and F be the midpoints of BC, AC, and AB, respectively.

Midpoint D of BC: $D = \left( \frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2} \right) = (3, 2, 0)$

Midpoint E of AC: $E = \left( \frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2} \right) = (3, 0, 3)$

Midpoint F of AB: $F = \left( \frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2} \right) = (0, 2, 3)$

Now, we find the lengths of the medians AD, BE, and CF using the distance formula.

Length of median AD (from vertex A to midpoint D): $AD = \sqrt{(3-0)^2 + (2-0)^2 + (0-6)^2}$ $AD = \sqrt{3^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

Length of median BE (from vertex B to midpoint E): $BE = \sqrt{(3-0)^2 + (0-4)^2 + (3-0)^2}$ $BE = \sqrt{3^2 + (-4)^2 + 3^2} = \sqrt{9 + 16 + 9} = \sqrt{34}$

Length of median CF (from vertex C to midpoint F): $CF = \sqrt{(0-6)^2 + (2-0)^2 + (3-0)^2}$ $CF = \sqrt{(-6)^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$

Final Answer: The lengths of the medians are 7, $\sqrt{34}$, and 7 units.

Given: The vertices of triangle PQR are P(2a, 2, 6), Q(-4, 3b, -10), and R(8, 14, 2c). The centroid of the triangle is the origin, G(0, 0, 0).

To Find: The values of a, b, and c.

Formula: The coordinates of the centroid G of a triangle with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ are given by: $G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$

Solution: Using the formula for the centroid and equating it to the coordinates of the origin (0, 0, 0):

For the x-coordinate: $\frac{2a + (-4) + 8}{3} = 0$ $\frac{2a + 4}{3} = 0$ $2a + 4 = 0$ $2a = -4$ $a = -2$

For the y-coordinate: $\frac{2 + 3b + 14}{3} = 0$ $\frac{3b + 16}{3} = 0$ $3b + 16 = 0$ $3b = -16$ $b = -\frac{16}{3}$

For the z-coordinate: $\frac{6 + (-10) + 2c}{3} = 0$ $\frac{2c - 4}{3} = 0$ $2c - 4 = 0$ $2c = 4$ $c = 2$

Final Answer: The values are $a = -2$, $b = -\frac{16}{3}$, and $c = 2$.

Given: Points A(3, 4, 5) and B(-1, 3, -7). Let P(x, y, z) be any point in the set.

Condition: $PA^2 + PB^2 = k^2$, where $k$ is a constant.

Formula: The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.

Solution: First, we express $PA^2$ and $PB^2$ in terms of the coordinates of P(x, y, z).

$PA^2 = (x-3)^2 + (y-4)^2 + (z-5)^2$ $PB^2 = (x-(-1))^2 + (y-3)^2 + (z-(-7))^2 = (x+1)^2 + (y-3)^2 + (z+7)^2$

Now, we substitute these expressions into the given condition $PA^2 + PB^2 = k^2$: $(x-3)^2 + (y-4)^2 + (z-5)^2 + (x+1)^2 + (y-3)^2 + (z+7)^2 = k^2$

Expand each squared term: $(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 - 10z + 25) + (x^2 + 2x + 1) + (y^2 - 6y + 9) + (z^2 + 14z + 49) = k^2$

Group like terms: $(x^2 + x^2) + (y^2 + y^2) + (z^2 + z^2) + (-6x + 2x) + (-8y - 6y) + (-10z + 14z) + (9 + 16 + 25 + 1 + 9 + 49) = k^2$ $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = k^2$

To get the final equation, we can move the constant term to the right side: $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = k^2 - 109$

This is the equation of a sphere.

Final Answer: The equation of the set of points P is $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = k^2 - 109$.