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

Let the given equation be  `(4 - x)/2 = y/6 = (1 - z)/3 = lambda`

This can be written as `(x - 4)/(-2) = y/6 = (z - 1)/(-3) = lambda`  .......(1)

∴ The coordinates of any point on the line is x = 4 − 2λ, y = 6λ, z = 1 − 3λ

Let Q(4 − 2λ, 6λ, 1 − 3λ) be the foot of perpendicular from the point P(2, 3, −8) on line  ......(1)

We know the direction ratios of any line segement PQ is given by (x₂ ​− x₁​, y₂ ​− y₁​, z₂ ​− z₁​)

The direction cosines of PQ is given by

= (−2λ + 4 − 2, 6λ − 3, −3λ + 1 + 8)

= (−2λ + 2, 6λ − 3, −3λ + 9)

Now Q is the foot of the perpendicular of the line (1)

`vec(PQ)` ​is the perpendicular to the line (1)

Hence the sum of the product of this direction ratios is 0

= (−2λ + 2)(−2) + (6λ − 3) . 6 + (−3λ + 9)(−3) = 0

⇒ 4λ − 4 + 36λ − 18 + 9λ − 27 = 0

⇒ 49λ − 49 = 0

∴ λ = 1

Substituting λ = 1 in Q we get

Q(0, 3, 6)