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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.
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Solution
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.
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