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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.
Solution
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
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