Question

Find the equation of the plane passing through the points

(-1,1,-4) and is perpendicular to each of the planes
-2x+y+z+2=0;x+y-3z+1=0​

Answer 1

Extract the normal vectors from the given planes:

`-2x+y+z+2=0\implies\vec n_1=(-2,1,1)`

`x+y-3z+1=0\implies\vec n_2=(1,1,-3)`

(which are unique up to their signs, meaning either
`\vec n_1` or
`-\vec n_1` are valid choices for the normal vector)

The third plane must be perpendicular to both these given planes, which means it would be parallel to both
`\vec n_1` and
`\vec n_2`, which in turn means its own normal vector
`\vec n_3` should be perpendicular to both
`\vec n_1` and
`\vec n_2`.

Enter the cross product:

`\vec n_3=\vec n_1*\vec n_2=(-4,-5,-3)`

or (4, 5, 3), which also works.

The given plane passes through (-1, 1, 4), so its equation is

`(x+1,y-1,z-4)\cdot\vec n_3=0`

Simplify:

`(x+1,y-1,z-4)\cdot(4,5,3)=0`

`4(x+1)+5(y-1)+3(z-4)=0`

`\boxed{4x+5y+3z=13}`