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Question
Find the distance of a point (2, 4, –1) from the line `(x + 5)/1 = (y + 3)/4 = (z - 6)/(-9)`
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Solution
The given equation of line is `(x + 5)/1 = (y + 3)/4 = (z - 6)/(-9) = lambda` and any point P(2, 4, – 1)
Let Q be any point on the given line
∴ Coordinates of Q are x = λ – 5, y = 4λ – 3, z = – 9λ + 6
D’ratios of PQ are λ – 5 – 2, 4λ – 3 – 4, – 9λ + 6 + 1
i.e., λ – 7, 4λ – 7, – 9λ + 7
And the d’ratios of the line are 1, 4, – 9
If PQ ⊥ line then
1(λ – 7) + 4(4λ – 7) – 9(– 9λ + 7) = 0
λ – 7 + 16λ – 28 + 81λ – 63 = 0
⇒ 98λ – 98 = 0
∴ λ = 1
So, the coordinates of Q are 1 – 5, 4 × 1 – 3, – 9 × 1 + 6
i.e., – 4, 1, – 3
∴ PQ = `sqrt((-4 - 2)^2 + (1 - 4)^2 + (-3 + 1)^2)`
= `sqrt((-6)^2 + (-3)^2 + (-2)^2)`
= `sqrt(36 + 9 + 4)`
= `sqrt(49)`
= 7
Hence, the required distance is 7 units.
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