Question

Find the distance of a point (2, 4, –1) from the line `(x + 5)/1 = (y + 3)/4 = (z - 6)/(-9)`

Answer 1

The given equation of line is `(x + 5)/1 = (y + 3)/4 = (z - 6)/(-9) = lambda` and any point P(2, 4, – 1)

Let Q be any point on the given line

∴ Coordinates of Q are x = λ – 5, y = 4λ – 3, z = – 9λ + 6

D’ratios of PQ are λ – 5 – 2, 4λ – 3 – 4, – 9λ + 6 + 1

i.e., λ – 7, 4λ – 7, – 9λ + 7

And the d’ratios of the line are 1, 4, – 9

If PQ ⊥ line then

1(λ – 7) + 4(4λ – 7) – 9(– 9λ + 7) = 0

λ – 7 + 16λ – 28 + 81λ – 63 = 0

⇒ 98λ – 98 = 0

∴ λ = 1

So, the coordinates of Q are 1 – 5, 4 × 1 – 3, – 9 × 1 + 6

i.e., – 4, 1, – 3

∴ PQ = `sqrt((-4 - 2)^2 + (1 - 4)^2 + (-3 + 1)^2)`

= `sqrt((-6)^2 + (-3)^2 + (-2)^2)`

= `sqrt(36 + 9 + 4)`

= `sqrt(49)`

= 7

Hence, the required distance is 7 units.