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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
Given: - Point P(2, 4, – 1) and equation of line`(x + 5)/1 = (y + 3)/4 = (z - 6)/(-9)`
Let, Q be a point through which line passes
Thus from given equation of line coordinates of Q is Q( – 5, – 3, 6)
As we know line equation with direction ratio of given line is parallel to given line.
Hence Line is parallel to `vecb = hati + 4hatj - 9hatk`
Now, ⇒ `vec(PQ) = (-5hati - 3hatj + 6hatk) - (2hati + 4hatj - hatk)`
⇒ `vec(PQ) = (-7hati - 7hatj + 7hatk)`
Now let's find cross product of this two vectors
⇒ `|vecb xx vec(PQ)| = sqrt(1225 + 3136 + 441)`
⇒ `|vecb xx vec(PQ)| = sqrt(4802)`
The magnitude of this cross product
And magnitude of `vecb`
⇒ `|vecb| = sqrt(1 + 16 + 81)`
⇒ `|vecb| = sqrt(98)`
Thus distance of point from line is
⇒ d = `(|vecb xx vec(PQ)|)/|vecb|`
⇒ d = `sqrt(4802)/sqrt(98)`
⇒ d = 7 units.
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