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`.

Answer 1

The vector equation of the plane passing through the point `A(veca)` and parallel to the vectors `vecb` and `vecc` is `vecr*(vecb xx vecc) = veca*(vecb xx vecc)`   ...(1)

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)`

= (–1)(–2) + 2(7) + (–5)(5)

= 2 + 14 – 25

= –9

∴ From (1), the vector equation of the required plane is `vecr * (-2hati + 7hatj + 5hatk) = -9`.