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• RD Sharma Class 11 Solutions for Maths

# 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:             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:  = 3 = 3 = 3 = 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: ⇒ x2 + y2 + z2 = x2 − 4x + 4 + y2 +z2

⇒ 4x = 4

⇒ x = 1

Similarly, OP2 = PB2 ⇒ x2 + y2 + z2 = x2 + y2 − 6y +9 +z2

⇒ 6y = 9

⇒ y = 3/2

Also, OP2 = PC2 ⇒ 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, , we have: 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, we have: ⇒ 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:        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:   ⇒ AB2 = 586  ⇒ BC2 = 4571  ⇒ CA2 = 2709

Since, AB2 + BC2 ≠ AC2

AB2 + AC2 ≠ BC2

BC2 + AC2 ≠ AB2

ABC is not a right triangle.

### (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:      = 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:      = 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:         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:         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: ⇒ 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:             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,  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, ⇒ -6x + 9 – 8y + 16 + 10z + 25 = 4x + 4 – 2y +1 – 8z +16

⇒ 10x + 6y – 18z -29 = 0.

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