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Question
Find the vector equation of the plane passing through the point A(–1, 2, –5) and parallel to the vectors `4hati - hatj + 3hatk` and `hati + hatj - hatk`.
Solution
The vector equation of the plane passi ng through the point `A(veca)` and parallel to the vectors `vecb` and `vecc` is `vecr.(vecb xx vecc) = veca.(vecb xx vecc)`
Here, `veca = -hati + 2hatj - 5hatk`
`vecb = 4hati - hatj + 3hatk`
`vecc = hati + hatj - hatk`
`vecb xx vecc = |(hati, hatj, hatk),(4, -1, 3),(1, 1, -1)|`
= `(1 - 3)hati - (-4 - 3)hatj + (4 + 1)hatk`
= `- 2hati + 7hatj + 5hatk`
∴ `veca.(vecb xx vecc) = (-hati + 2hatj - 5hatk).(-2hati + 7hatj + 5hatk)`
= (–2)(–1) + (7)(2) + (5)(–5)
= 2 + 14 – 25
= – 9
Then vector equation of required planes `vecr.(-2hati + 7hatj + 5hatk)` = – 9
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