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:

(i) (2, 3, 5) and (4, 3,1)

Let P be (2, 3, 5) and Q be (4, 3, 1)

By using the 3d distance formula,

Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

So here,

x₁ = 2, y₁ = 3, z₁ = 5

x₂ = 4, y₂ = 3, z₂ = 1

PQ = √(4 - 2)² + (3 - 3)² + (1- 5)²

= √2² + 0² +(- 4)²

= √4 + 16

= √20

= 2√5

Therefore, the required distance is 2√5 units.

(ii) (- 3, 7, 2) and (2, 4, - 1)

Let P be (- 3, 7, 2) and Q be (2, 4, - 1)

By using the formula,

Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

So here,

x₁ = - 3, y₁ = 7, z₁ = 2

x₂ = 2, y₂ = 4, z₂ = - 1

PQ = √(2 - (- 3)² + (4 - 7)² + (- 1 - 2)²

= √5² + (- 3)² +(- 3)²

= √25 + 9 + 9

= √43

Therefore, the required distance is √43 units.

(iii) (- 1, 3, - 4) and (1, - 3, 4)

Let P be (- 1, 3, - 4) and Q be (1, - 3, 4)

By using the formula,

Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

So here,

x₁ = - 1, y₁ = 3, z₁ = - 4

x₂ = 1, y₂ = - 3, z₂ = 4

PQ = √(1 - (- 1)² + (- 3 - 3)² + (4 - (- 4))²

= √2² + (- 6)² +(8)²

= √4 + 36 + 64

= √104

= 2√26

Therefore, the required distance is 2√26 units.

(iv) (2, - 1, 3) and (- 2, 1, 3)

Let P be (2, - 1, 3) and Q be (- 2, 1, 3)

By using the formula,

Distance = √(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

So here,

x₁ = 2, y₁ = - 1, z₁ = 3

x₂ = - 2, y₂ = 1, z₂ = 3

PQ = √(- 2 - 2)² + (1 - (- 1))² + (3 - 3)²

= √(- 4)² + (2)² +(0)²

= √16 + 4

= √20

= 2√5

Therefore, the required distance is 2√5 units

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NCERT Solutions Class 11 Maths Chapter 12 Exercise 12.2 Question 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)

Summary:

(i) the required distance is 2√5 units
(ii) the required distance is √43 units.
(iii) the required distance is 2√26 units.
(iv)the required distance is 2√5 units.