Question

Find the co-ordinates of the foot of the perpendicular drawn from the point `2hati - hatj + 5hatk` to the line `barr = (11hati - 2hatj - 8hatk) + λ(10hati - 4hatj - 11hatk).` Also find the length of the perpendicular.

Answer 1

Let M be the foot of perpendicular drawn from the point `P(2hati - hatj + 5hatk)` on the line.

`barr = (11hati - 2hatj - 8hatk) + λ(10hati - 4hatj - 11hatk)`

Let the position vector of the point M be

`(11hati - 2hatj - 8hatk) + λ(10hati - 4hatj - 11hatk)`

= `(11 + 10λ)hati + (-2 - 4λ)hatj + (-8 - 11λ)hatk`.

Then PM = Position vector of M – Position vector of P

= `[(11 + 10λ)hati + (-2 - 4λ)hatj + (-8 - 11λ)hatk] - (2hati -hatj + 5hatk)`

= `(9 + 10λ)hati + (-1 - 4λ)hatj + (-13 - 11λ)hatk`

Since PM is perpendicular to the given line which is parallel to `barb = 10hati - 4hatj - 11hatk`,

PM ⊥ `barb`

∴ PM.`barb` = 0

∴ `[(9 + 10λ)hati + (-1 - 4λ) - 11(-13 - 11λ)hatk].(10hati - 4hatj - 11hatk)` = 0

∴ 10(9 + 10λ) – 4(–1 – 4λ) – 11(13 – 11λ) = 0

∴ 90 + 100λ + 4 + 16λ + 143 + 121λ = 0

∴ 237λ + 237 = 0

∴ λ = – 1

Putting this value of λ, we get the position vector of M as `hati + 2hatj + 3hatk`.

∴ Coordinates of the foot of perpendicular M are (1, 2, 3).

Now, PM = `(hati + 2hatj + 3hatk) - (2hati - hatj + 5hatk)`

= `-hati + 3hatj - 2hatk`

∴ |PM| = `sqrt((-1)^2 + (3)^2 + (-2)^2`

= `sqrt(1 + 9 + 4)`

= `sqrt(14)`

Hence, the coordinates of the foot of perpendicular are (1, 2, 3) and length of perpendicular = `sqrt(14)` units.