Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.2 | Set 2

Question 13. Prove that the tetrahedron with vertices at points O(0,0,0), A(0,1,1), B(1,0,1) and C(1,1,0) is a regular one.

Solution:

Given: The points O(0,0,0), A(0,1,1), B(1,0,1) and C(1,1,0).

A regular tetrahedron has all equal sides and diagonals.

We know that the distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(0 − 1)^2 + (1 − 0)^2 + (1 −1)^2}

= \sqrt{2}

BC = \sqrt{(1 − 1)^2 + (0 −1)^2 + (1 −0)^2}

= \sqrt{2}

CA = \sqrt{(1 − 0)^2 + (1 −1)^2 + (0 −1)^2}

= \sqrt{2}

OA = \sqrt{(0 − 0)^2 + (0 −1)^2 + (0 −1)^2}

= \sqrt{2}

OB = \sqrt{(0 − 1)^2 + (0 − 0)^2 + (0 −1)^2}

= \sqrt{2}

OC = \sqrt{(0 − 1)^2 + (0 −1)^2 + (0 −0)^2}

= \sqrt{2}

Clearly, OA = OB = OC = AB = BC = CA.

Hence O, A, B and C represent a regular tetrahedron.

Question 14. Show that the points (3,2,2), (-1,1,3), (0,5,6), (2,1,2) lie on a sphere whose centre is (1,3,4). Also find its radius.

Solution:

Given: The points A(3,2,2), B(-1,1,3), C(0,5,6), D(2,1,2) and E(1,3,4)

We know that the distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

EA = \sqrt{(1 − 3)^2 + (3 − 2)^2 + (4 − 2)^2}

= 3

EB = \sqrt{(1 + 1)^2 + (3 − 1)^2 + (4 − 3)^2}

= 3

EC = \sqrt{(1 − 0)^2 + (3 − 5)^2 + (4 − 6)^2}

= 3

ED = \sqrt{(1 − 2)^2 + (3 − 1)^2 + (4 − 2)^2}

= 3

Since EA = EB = EC = ED, the points lie on a sphere with centre E.

Radius of the sphere = 3 units.

Question 15. Find the coordinates of the point which is equidistant from the four points O(0,0,0), A(2,0,0), B(0,3,0) and C(0,0,8).

Solution:

Given: Points O(0,0,0), A(2,0,0), B(0,3,0) and C(0,0,8).

Let the required point be P(x,y,z).

We are given that OP = PA ⇒ OP² = PA²

Using formula, we have:

\sqrt{(x − 0)^2 + (y − 0)^2 + (z − 0)^2} = \sqrt{(x − 2)^2 + (y − 0)^2 + (z − 0)^2}

⇒ x² + y² + z² = x² − 4x + 4 + y² +z²

⇒ 4x = 4

⇒ x = 1

Similarly, OP² = PB²

⇒\sqrt{(x − 0)^2 + (y − 0)^2 + (z − 0)^2} = \sqrt{(x − 0)^2 + (y − 3)^2 + (z − 0)^2}

⇒ x² + y² + z² = x² + y² − 6y +9 +z²

⇒ 6y = 9

⇒ y = 3/2

Also, OP² = PC²

⇒\sqrt{(x − 0)^2 + (y − 0)^2 + (z − 0)^2} = \sqrt{(x − 0)^2 + (y − 0)^2 + (z − 8)^2}

⇒ x² + y² + z² = x² + y² +z² − 16z + 64

⇒ 16z = 64

⇒ z = 4

Hence the point is P[1, 3/2, 4].

Question 16. If A(-2,2,3) and B(13,-3,13) are two points, find the locus of a point P which moves in a way such that 3PA = 2PB.

Solution:

Given: A(-2,2,3) and B(13,-3,13)

Let P = (x, y, z) be the required point.

We are given 3PA = 2PB

Using the formula,\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2} , we have:

3\sqrt{(x + 2)^2+(y − 2)^2+(z − 3)^2} = 2\sqrt{(x − 13)^2+(y + 3)^2+(z − 13)^2}

Squaring both sides, we have;

9(x² + 4x +4 + y² + 4 − 4y + z² + 9 − 6z) = 4(x² + 169 − 26x + y² +9 + 6y + z² + 169 − 26z)

⇒ 5(x² + y² +z²) + 140x − 60y + 50z − 1235 = 0.

Question 17. Find the locus of P if PA² + PB² = 2k², where A and B are the points (3,4,5) and (-1,3,-7).

Solution:

Given: A(3,4,5) and B(-1,3,-7)

Let P(x, y, z) be the required point.

PA² + PB² = 2k². Using the formula,\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}, we have:

\sqrt{(x − 3)^2 + (y − 4)^2 + (z − 5)^2} + \sqrt{(x+1)^2 + (y − 3)^2 + (z+7)^2} = 2k^2

⇒ 2x² + 2y² + 2z² − 4x −14y + 4z + 109 − 2k² = 0

⇒ 2(x² + y² +z²) − 4x −14y + 4z + 109 − 2k2 = 0.

Question 18. Show that the points A(a, b, c), B(b, c, a) and C(c, a, b) are vertices of an equilateral triangle.

Solution:

Given: points A(a, b, c), B(b, c, a) and C(c, a, b)

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(a − b)^2 + (b − c)^2 + (c − a)^2}

= \sqrt{a^2 + b^2 - 2ab + b^2 + c^2 - 2bc + c^2 + a^2 - 2ac}

= \sqrt{2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac}

BC = \sqrt{(b − c)^2 + (c − a)^2 + (a − b)^2}

= \sqrt{2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac}

CA = \sqrt{(a − c)^2 + (b − a)^2 + (c − b)^2}

= \sqrt{2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac}

Since AB = BC = CA

ABC is an equilateral triangle.

Question 19. Are points A(3,6,9), B(10,20,30), and C(25,41,5) the vertices of a right-angled triangle?

Solution:

Given: A(3,6,9), B(10,20,30) and C(25,41,5)

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(3 − 10)^2 + (6 − 20)^2 + (9 − 30)^2}

= \sqrt{49 + 196 + 441}

⇒ AB² = 586

BC = \sqrt{(10 − 25)^2 + (20 + 41)^2 + (30 − 5)^2}

= \sqrt{225 + 3721 + 625}

⇒ BC² = 4571

CA = \sqrt{(3 − 25)^2 + (6 + 41)^2 + (9 − 5)^2}

= \sqrt{484 + 2209 + 16}

⇒ CA² = 2709

Since, AB² + BC² ≠ AC²

AB² + AC² ≠ BC²

BC² + AC² ≠ AB²

ABC is not a right triangle.

Question 20. Verify that:

(i) (0,7,-10), (1,6,-6) and (4,9,-6) are the vertices of an isosceles triangle.

Solution:

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

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(0 − 1)^2 + (7 − 6)^2 + (-10 + 6)^2}

= \sqrt{1 + 1 + 16} = 3\sqrt{2}

BC = \sqrt{(1 − 4)^2 + (6 − 9)^2 + (6 − 6)^2}

= \sqrt{9 + 9} = 3\sqrt{2}

CA = \sqrt{(0 − 4)^2 + (7 − 9)^2 + (-10 + 6)^2}

= 6

Since AB = BC, ABC is an isosceles triangle.

(ii) (0,7,-10), (-1,6,6) and (4,9,-6) are the vertices of a right-angled triangle.

Solution:

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

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(0 + 1)^2 + (7 − 6)^2 + (10 - 6)^2}

= \sqrt{1 + 1 + 16} = 3\sqrt{2}

BC = \sqrt{(-1 + 4)^2 + (6 − 9)^2 + (6 − 6)^2}

= \sqrt{9 + 9} = 3\sqrt{2}

CA = \sqrt{(-4 − 0)^2 + (9 − 7)^2 + (6 + 10)^2}

= 6

Since AB² + BC² = AC², ABC is a right triangle.

(iii) (-1,2,1), (1,-2,5), (4,-7,8) and (2,-3,4) are the vertices of a parallelogram.

Solution:

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

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(-1 - 1)^2 + (2 + 2)^2 + (1 - 5)^2}

= \sqrt{36} = 6

BC = \sqrt{(1 − 4)^2 + (-2 + 7)^2 + (5− 8)^2}

= \sqrt{43}

CD = \sqrt{(4 − 2)^2 + (-7 + 3)^2 + (8 − 4)^2}

= \sqrt{36} = 6

DA = \sqrt{(2 + 1)^2 + (-3 − 2)^2 + (4− 1)^2}

= \sqrt{43}

Since the opposite sides are equal, ABCD is a parallelogram.

(iv) (5,-1,1), (7,-4,7), (1,-6,10) and (-1,-3,4) are vertices of a rhombus.

Solution:

Given: A(5,-1,1), B(7,-4,7), C(1,-6,10) and D(-1,-3,4)

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(7 − 5)^2 + (-4 + 1)^2 + (7 − 1)^2}

= \sqrt{49} = 7

BC = \sqrt{(1 − 7)^2 + (-6 + 4)^2 + (10 − 7)^2}

= \sqrt{49} = 7

CD = \sqrt{(-1 − 1)^2 + (-3 + 6)^2 + (4 − 10)^2}

= \sqrt{49} = 7

AD = \sqrt{(-1 − 5)^2 + (-3 + 1)^2 + (4 − 1)^2}

= \sqrt{49} = 7

Since AB = BC = CA = AD

ABCD is a rhombus.

Question 21. Find the locus of the points which are equidistant from the points (1,2,3) and (3,2,-1).

Solution:

Let P(x, y, z) be the point equidistant from the points A(1,2,3) and B(3,2,-1).

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

⇒ AP = BP or AP² = BP²

⇒ (x − 1)² + (y − 2)² + (z − 3)² = (x − 3)² + (y − 2)² + (z + 1)²

⇒ 4x − 8z = 14 − 14

⇒ x − 2z = 0.

Question 22. Show that the points A(1,2,3), B(-1.-2,-1), C(2,3,2) and D(7,4,6) are the vertices of a parallelogram ABCD.

Solution:

Using formula, distance between two points (a, b, c) and (d, e, f) is given as follows:

\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}

AB = \sqrt{(1 + 1)^2 + (2 + 2)^2 + (3 + 1)^2}

= \sqrt{36} = 6

BC = \sqrt{(-1 − 2)^2 + (-2 − 3)^2 + (-1− 2)^2}

= \sqrt{43}

CD = \sqrt{(2 − 4)^2 + (3 − 7)^2 + (2 − 6)^2}

= \sqrt{36} = 6

DA = \sqrt{(4 − 1)^2 + (7 − 2)^2 + (6− 3)^2}

= \sqrt{43}

AC = \sqrt{(1 -2)^2 + (2 - 3)^2 + (3 - 2)^2}

= \sqrt{3}

BD = \sqrt{(-1− 4)^2 + (-2 − 7)^2 + (-1 − 6)^2}

= \sqrt{155}

Since the opposite sides are equal, ABCD is a parallelogram.

Question 23: Find the locus of the point, the sum of whose distances from the points A(4,0,0) and B(-4,0,0) is equal to 10.

Solution:

Let P(x, y, z) be the required locus.

Given: PA + PB = 10. Using distance formula,

\sqrt{(x − 4)^2+(y − 0)^2+(z − 0)^2} + \sqrt{(x + 4)^2+(y − 0)^2+(z − 0)^2} = 10

⇒ 4x + 25 = 5\sqrt{x^2 + y^2 + z^2 + 8x + 16}

Squaring both sides, we get

16x² + 625 + 200x = 25(x² + y² + z² + 8x + 16)

⇒ 9x² + 25y² + 25z² - 225 = 0

Question 24. Find the equation of the set of points P such that its distances from the points A(3,4,-5) and B(-2,1,4) are equal.

Solution:

Given: A(3,4,-5) and B(-2,1,4)

Let P(x, y, z) be the required point, It is given that PA = PB.

Hence, PA² = PB²

Using distance formula, we have,

⇒ \sqrt{(x − 3)^2+(y − 4)^2+(z + 5)^2} = \sqrt{(x + 2)^2+(y − 1)^2+(z − 4)^2}

⇒ -6x + 9 - 8y + 16 + 10z + 25 = 4x + 4 - 2y +1 - 8z +16

⇒ 10x + 6y - 18z -29 = 0

Summary

Exercise 28.2 Set 2 deals with the distance formula and section formula in 3D coordinate geometry. Students will learn to find the distance between two points, coordinates of the centroid of a triangle, and coordinates of a point which divides a line segment in a given ratio. They will also learn to apply the distance formula and section formula to solve problems in 3D geometry. This exercise helps students understand the concepts of distance and division of line segments in 3D space, which is crucial for understanding advanced topics in calculus, physics, and engineering.