Orthogonality of vectors in $\mathbb{R}^3$ I have to show the following: 

If $v, w \in \mathbb{R}^3 \setminus \{(0,0,0)\}$ such that the set of vectors orthogonal to both of them is a plane through the origin, then each is a scalar multiple of the other.

I've proved that if $u^\perp = \{x \in \mathbb{R}^3 : x \perp u\}=\{x \in \mathbb{R}^3 : \langle x , u \rangle=0\}$, the set of all vectors which are ortogonal to $u$, then $v^\perp=w^\perp$ [since $v^\perp \cap w^\perp$ is a plane through origin (given)
then dim$(v^\perp \cap w^\perp)$=dim$(v^\perp)$=dim$(w^\perp)$=2 and  $v^\perp \cap w^\perp \subseteq v^\perp$,$v^\perp \cap w^\perp \subseteq w^\perp$]. Will this be of any help? I can't go any further. Many thanks.
 A: Take an orthogonal basis of the plane $\{{\bf v}_1,{\bf v}_2\}$.  The set $\{{\bf v},{\bf v}_1,{\bf v}_2\}$ must be independent.
Indeed if $c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2=\bf 0$, then
$${\bf v}_1\cdot ( c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2)=0 \quad\Rightarrow \quad c_2=0 $$
and
$${\bf v}_2\cdot ( c_1{\bf v} +c_2{\bf v}_1+c_3 {\bf v}_2)=0 \quad\Rightarrow \quad c_3=0. $$
we then must have $c_1=0$.
So,  $\{{\bf v},{\bf v}_1,{\bf v}_2\}$ is an independent set and thus  a basis of $\Bbb R^3$. 
Now write $\bf u$ in terms of this basis. We have for some constants $c_1$, $c_2$, and $c_3$
$${\bf u}=c_1{\bf v}+c_2 {\bf v}_1+c_3 {\bf v}_2.$$
Now
$$0={\bf v}_1\cdot {\bf u}= c_1 {\bf v}_1\cdot{\bf v}
+c_2 {\bf v}_1\cdot {\bf v}_1+c_3 {\bf v}_1\cdot {\bf v}_2=c_2 ||{\bf v}_1||^2;$$
thus, $c_2=0$.  Similarly, taking dot products with ${\bf v}_2$, it follows that $c_3=0$. Thus ${\bf u}=c_1{\bf v} $.
A: Hint: For finite dimensional vector spaces, if you have subspaces $A,B$ such that $\text{dim}(A) = \text{dim}(B)$ and $A \subseteq B$, then $A = B$.
