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Question
A line passes through the point (2, – 1, 3) and is perpendicular to the lines `vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` obtain its equation.
Solution
Given lines
`vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` ...(1)
and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` ...(2)
Here `vecb_1 = 2hati - 2hatj + hatk`
and `vecb_2 = hati + 2hatj + 2hatk`
Now vector perpendicular to `vecb_1` and `vecb_2` is
`vecb = vecb_1 xx vecb_2 = |(hati, hatj, hatk),(2, -2, 1),(1, 2, 2)|`
`vecb = (-4 - 2)hati - (4 - 1)hatj + (4 + 2)hatk`
`vecb = -6hati - 3hatj + 6hatk`
Then equation of line passing through
`veca = (2, -1, 3)`
= `(2hati - hatj + 3hatk)`
and perpendicular to lines (1) and (2)
i.e., parallel to `vecb = vecb_1 xx vecb_2` is
`vecr = veca + tvecb`
`vecr = (2hati - hatj + 3hatk) + t(-6hati - 3hatj + 6hatk)`
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