Question

The distance of the point (1, 0, 2) from the point of intersection of the line `(x - 2)/3 = (y + 1)/4 = (z - 2)/12` and the plane x - y + z = 16, is ______ 

Options

• `2sqrt14`

• 8

• `3sqrt21`

• 13

Answer 1

The distance of the point (1, 0, 2) from the point of intersection of the line `(x - 2)/3 = (y + 1)/4 = (z - 2)/12` and the plane x - y + z = 16, is 13. 

Explanation:

Let `(x - 2)/3 = (y + 1)/4 = (z - 2)/12 = lambda`

∴ the co-ordinates of any point on the line are

P ≡ (3λ + 2, 4λ - 1, 12λ + 2)

This point lies on the plane x - y + z = 16

∴ 3λ + 2 - 4λ + 1 + 12λ + 2 = 16

⇒ 11λ = 11 ⇒ λ = 1

∴ P ≡ (5, 3, 14)

∴ Let Q ≡ (1, 0, 2)

∴ distance PQ is given by

d = `sqrt((5 - 1)^2 + (3 - 0)^2 + (14 - 2)^2) = 13`

∴