orthogonal complement of interestion of 2 planes Question:Let $W$ be the intersection of the two planes: $x-y+z=0$ and $x+y+z=0$
Find a basis for and the dimension of the orthogonal complement, $W^{\perp}$

Answer:
The line $x+z=0$ intersects the plane, which is parameterized as $t(1, 0, -1)$
Then $W^{\perp}$ is the plane $x-z=0$

The basis for $W$ is $(1,0,1)$ I think, how do I find the basis for $W^{\perp}$ and the dimension? any help is appreciated :)
$$W = span\{(1,0,-1)^\top\}$$ the vector $(0,1,0)^\top$ is in  $W^\perp.$ to find a second vector that is orthogonal to both $(0,1,0)^\top, (1,0, -1)$ we can find the null space of $\pmatrix{1&amp;0&amp;-1\\0&amp;1&amp;0}.$ that is $(1,0,1)^\top$ is in the null space. therefore a basis for $W^\perp$ is 
$$\{(0,1,0)^\top, (1,0,1)^\top\}. $$
Find the normals to each plane. Then each is orthogonal to its respective plane, $W$ lies in both planes, so is orthogonal to both normal vectors. The planes are distinct and intersect, so the normals are not parallel. Hence they form a basis of $W^{\perp}$, since its dimension cannot be larger than $2$ for the obvious reason that $W^{\perp} \neq \mathbb{R}^3$.