Question

Find the equation of the plane containing the lines `(x - 1)/2 = (y + 1)/-1 = z/3` and `x/2 = (y - 2)/-1 = (z + 1)/3`.

Answer 1

The required plane passes through A(1, –1, 0) and B(0, 2, –1)

The plane is also parallel to the vector `2hati - hatj + 3hatk` which is along the given lines.

Also the plane is parallel to `bar(AB) = -hati + 3hatj - hatk`

Let `barm = 2hati - hatj + 3hatk`,

`barn = -hati + 3hatj - hatk`

Hence the required plane passes through

`A(bara) = hati - hatj` and parallel to `barm` and `barn`.

∴ Its equation is

`barr*(barm xx barn) = bara*(barm xx barn)`   ....(1)

Now `barm xx barn = |(hati, hatj, hatk),(2, -1, 3),(-1, 3, -1)|`

= `hati(-8) - hatj(1) + hatk(5)`

Also `bara*(barm xx barn)` = (1)(–8) + (–1)(–1) = –7

∴ From (1) equation of plane is

`barr*(-8hati - hatj + 5hatk)` = –7

i.e. – 8x – y + 5z = –7,

i.e. 8x + y – 5z = 7