Question

Find the foot of perpendicular from the point (2, 3, –8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.

Answer 1

Given that, `(4 - x)/2 = y/6 = (1 - z)/3` is the equation of line

⇒ `(x - 4)/(-2) = y/6 = (z - 1)/(-3) = lambda`

∴ Coordinates of any point Q on the line are x = – 2λ + 4, y = 6λ and z = – 3λ + 1 and the given point is P(2, 3, – 8)

Direction ratios of PQ are – 2λ + 4 – 2, 6λ – 3, – 3λ +1 + 8

i.e., – 2λ + 2, 6λ – 3, – 3λ + 9

And the D’ratios of the given line are – 2, 6, – 3.

If PQ ⊥ line

Then – 2(– 2λ + 2) + 6(6λ – 3) – 3(– 3λ + 9) = 0

⇒ 4λ – 4 + 36λ – 18 + 9λ – 27 = 0

⇒ 49λ – 49 = 0

⇒ λ = 1

∴ The foot of the perpendicular is – 2(1) + 4, 6(1), – 3(1) + 1

i.e., 2, 6, – 2

Now, distance PQ = `sqrt((2 - 2)^2 + (3 - 6)^2 + (-8 + 2)^2)`

= `sqrt(9 + 36)`

= `sqrt(45)`

= `3sqrt(5)`

Hence, the required coordinates of the foot of perpendicular are 2, 6, – 2 and the required distance is `3sqrt(5)` units.