Question

Find the equation of the plane through the intersection of the planes `vecr.(hati + 3hatj - hatk) = 9` and `vecr.(2hati - hatj + hatj) = 3` and passing through the origin.

Answer 1

The equation of I plane is `vecr.(hati + 3hatj -hatk) = 9`

i.e `(xhati + yhatj + zhatk).(hati + 3hatk - hatk) = 9`

x + 3y - z = 9

x + 3y - z - 9 = 0    ....(i)

Equation of II plane is `vecr.(2hati - hatj + hatk) = 3`

i.e `(xhati + yhatj +  zhatk).(2hati -hatj + hatk) = 3`

i.e 2x - y + z = 3

i.e 2x - y + z - 3  = 0 ....(ii)

Now, equation of a plane passing through intersection of given planes is

`(x + 3y - z - 9) + lambda(2x - y + z - 3)  = 0`

`(1+2lambda) x + (3-lambda)y + (-1+lambda)z - 9-3lambda = 0`

Since plane is passing through the origin (0, 0, 0)

`-9-3lambda = 0`

`-3lambda = 9`

`:. lambda = -3`