Introduction to three Dimensional Geometry – Exercise 12.2 – Class XI

  1. Find the distance between the following pairs of points:

(i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1)

(iii) (–1, 3, –4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3)

Solution:

The distance between points P(x1, y1, z1) and P(x2, y2, z2) is given by

PQ = √[(x2 – x1)2+(y2 – y1)2+(z2 – z1)2]

(i) Distance between points (2, 3, 5 ) and (4, 3, 1)

= √[(4 – 2)2+(3 – 3)2+(1 – 5)2]

= √[(2)2+(0)2+(4)2]

= √[4+16]

= 2√5

(ii) Distance between points (-3, 7, 2 ) and (2, 4, -1)

= √[(2+3)2+(4 – 7)2+(-1 – 2)2]

= √[(5)2+(-3)2+(-3)2]

= √[25+9+9]

= √43

(iii) Distance between points (2-1, 3, -4) and (1, -3, 4)

= √[(1+1)2+(-3 – 3)2+(4+4)2]

= √[(2)2+(-6)2+(8)2]

= √[4+36+64]

= √104

= 2√26

(iv) Distance between points (2, -1,3) and (-2, 1, 3)

= √[(-2 – 2)2+(1+1)2+(3 – 3)2]

= √[(-4)2+(2)2+(0)2]

= √[16+4]

= √20

= 2√5


  1. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Solution:

Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively. Points P, Q, and R are collinear if they lie on a line.

PQ = √[(1+2)2+(2 – 3)2+(3 – 5)2]

= √[(3)2+(-2)2+(-2)2]

= √(9+1+4)

= √14

QR = √[(7 – 1)2+(0 – 3)2+(-1 – 5)2]

= √[(6)2+(-2)2+(-4)2]

= √(36+4+16)

= √56

= 2√14

PR = √[(7 +2)2+(0 – 3)2+(-1 – 5)2]

= √[(9)2+(-3)2+(-6)2]

= √(81+9+36)

= √126

= 3√14

Here, PQ + QR = PR

Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.


  1. Verify the following:

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

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

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

Solution:

(i) Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C respectively.

AB = √[(7 – 1)2+(0 – 3)2+(- 6 + 10)2]

= √[(1)2+(-1)2+(4)2]

= √(1+1+16)

= √18

= 3√2

BC = √[(4 – 1)2+(9 – 6)2+(- 6 + 6)2]

= √[(3)2+(3)2]

= √(9+9)

= √18

= 3√2

CA = √[(0 – 4)2+(7 – 9)2+(- 10 + 62]

= √[(-4)2+(-2)2+(-4)2]

= √(16+4+16)

= √36

= 6

Here AB = BC ≠ CA

Thus, the given points are the vertices of an isosceles triangle.

 

(i) Let points (0, 7, 10), (-1, 6, 6), and (-4, 9, 6) be denoted by A, B, and C respectively.

AB = √[(-1 – 0)2+(6 – 7)2+(6 + 10)2]

= √[(-1)2+(-1)2+(-4)2]

= √(1+1+16)

= √18

= 3√2

BC = √[(-4 + 1)2+(9 – 6)2+(6 – 6)2]

= √[(-3)2+(3)2+(0)2]

= √(9+9)

= √18

= 3√2

CA = √[(0 + 4)2+(7 – 9)2+(10 – 62]

= √[(-4)2+(-2)2+(4)2]

= √(16+4+16)

= √36

= 6

Here AB2 + BC2 = (3√2)2+(3√2)2 = 18 + 18 = 36 = AC2

Thus, the given points are the vertices of an isosceles triangle.

Hence, the given points are the vertices of a right-angled triangle.

 

(iii) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D respectively.

AB = √[(1+1)2+(-2-2)2+(5-1)2]

= √[(2)2+(-4)2+(4)2]

= √(4+16+16)

= √36

= 6

BC = √[(4 – 1)2+(-7 +2)2+(8 – 5)2]

= √(9+25+9)

= √43

CD = √[(2 – 4)2+(-3 + 7)2+(4 – 8)2]

= √(9+25+9)

= √43

Here AB = CD = 6; BC= AD = √43

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal. Thus, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.


  1. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Solution:

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

Therefore,  PA = PB

PA2 = PB2

( x – 1)2 + ( y – 2)2 + ( z – 3)2 = ( x – 3)2 + ( y – 2)2 + ( z + 1)2

x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1

–2x –4y – 6z + 14 = –6x – 4y + 2z + 14

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8z = 0

⇒ x – 2z = 0

Thus, the required equation is x – 2z = 0.


  1. Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (–4, 0, 0) is equal to 10.

Solution:

Let the coordinates of P be (x, y, z).

The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.

It is given that PA + PB = 10.

√[(x – 4)2 + y2 + z2] +√[(x + 4)2 + y2 + z2] = 10

√[(x – 4)2 + y2 + z2] = 10 – √[(x + 4)2 + y2 + z2]

Squaring on both the sides, we obtain

{√[(x – 4)2 + y2 + z2]}2= {10 – √[(x + 4)2 + y2 + z2] }2

(x – 4)2 + y2 + z2 = 102 – 2x10x(√[(x + 4)2 + y2 + z2)+ (√[(x + 4)2 + y2 + z2)2

(x – 4)2 + y2 + z2 = 100 – 20(√[x2 + 8x + 16 + y2 + z2]+ [(x + 4)2 + y2 + z2]

20(√[x2 + 8x + 16 + y2 + z2] = 100 + x2 + 8x + 16 + y2 + z2 – x2 + 8x – 16 – y2 – z2

20(√[x2 + 8x + 16 + y2 + z2]) = 100 + 16x

5(√[x2 + 8x + 16 + y2 + z2]) = (25 + 4x)

Again squaring on both the sides

{5(√[x2 + 8x + 16 + y2 + z2])}2 = {(25 + 4x)}2

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

25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 200x + 16x2

9x2 + 25y2 + 25y2 – 225 = 0

Thus the required equation is 9x2 + 25y2 + 25y2 – 225 = 0


 

 

 

 

 

 

 

 

 

 

 

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