Question

Which of the following is an equation for the plane through the point (1, 1, −1) and parallel

to the plane x − y + z = 3? Show all your work.

Answer 1

The given plane,
`x-y+z=3`, has normal vector
`\vec n = \langle1,-1,1\rangle`. Any plane parallel to this one has the same normal vector.

Let
`(x,y,z)` be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is

`\langle x,y,z\rangle - \langle 1,1,-1\rangle = \langle x-1, y-1, z+1 \rangle`

and this is perpendicular to
`\vec n`.

So the equation of the plane is

`\langle x-1, y-1, z+1 \rangle \cdot \vec n = 0 \implies (x-1) - (y-1) + (z+1) = 0 \implies \boxed{x - y + z = -1}`

or equivalently,

`\boxed{-x + y - z = 1}`