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Question
Find the distance of the point (– 2, 4, – 5) from the line `(x + 3)/3 = (y - 4)/5 = (z + 8)/6`
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Solution
Here P(–2, 4, – 5) is the given point.
Any point Q on the line is given by `(3lambda - 3, 5lambda + 4, (6lambda - 8)`
`"PQ" = (3lambda -1)hat"i" + 5lamdahat"i" + (6lambda - 3)hat"k"`
Since `vec"PQ" ⊥ (3hat"i" + 5hat"j" + 6hat"k")`
We have `3(3lambda - 1) + 5(5lambda) + 6(6lambda - 3)` = 0
`9lambda + 25lambda + 36lambda` = 21
i.e. `lambda = 3/10`
Thus `"PQ" = - 1/10hat"i" + 15/10hat"j" - 12/10hat"k"`
Hence `|vec"PQ"| = 1/10 sqrt(1 + 225 + 144)`
= `sqrt(37/10)`.
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