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Question
A line l passes through point (– 1, 3, – 2) and is perpendicular to both the lines `x/1 = y/2 = z/3` and `(x + 2)/-3 = (y - 1)/2 = (z + 1)/5`. Find the vector equation of the line l. Hence, obtain its distance from the origin.
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Solution
Line `vecl` is passing through the point (– 1, 3, – 2) and the lines are
`x/1 = y/2 = z/3` and `(x + 2)/-3 = (y - 1)/2 = (z + 1)/5`
Direction cosines of first line is
`veca = (hati + 2hatj + 3hatk)`
Direction cosines of second line is
`vecb = (-3hati + 2hatj + 5hatk)`
∴ `veca xx vecb = (hati + 2hatj + 3hatk) xx (-3hati + 2hatj + 5hatk)`
= `|(hati, hatj, hatk),(1, 2, 3),(-3, 2, 5)|`
= `hati(10 - 6) - hatj(5 + 9) + hatk(2 + 6)`
`vecA = 4hati - 14hatj + 8hatk`
Position vector of the given line is
`vecB = -hati + 3hatj - 2hatk`
Vector equation of line is,
`vecr = vecB + λvecA`
= `(-hati + 3hatj + 2hatk) + λ(4hati - 14hatj + 0hatk)`
or `vecr = (-hati + 3hatj - 2hatk) + λ(2hati - 7hatj + 4hatk)`
Distance of the line ‘l’ passing through the point (–1, 3, –2) from the origin is
= `sqrt((-1 - 0)^2 + (3 - 0)^2 + (-2 - 0)^2`
= `sqrt((-1)^2 + 3^2 + 2^2)`
= `sqrt(1 + 9 + 4)`
= `sqrt(14)` units.
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