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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
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.
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