Show that for every $v\in\mathbb R^6$ there is a $0\ne w\in\mathbb R^6$ such that $B(v,w)=B(w,w)=0$. 
Let $B$ be a nondegenerate symmetric bilinear form on $\mathbb R^6$ with signature $0$. Show that for every $v\in\mathbb  R^6$ there is a $0\ne w\in\mathbb R^6$ such that $B(v,w)=B(w,w)=0$.


My attempt:
Since $B$ has signature $0$ and $B$ is nondegenerate, we can choose a basis $e_1,...,e_6$ such that $$B(\textbf{x},\textbf{y})=x_1y_1+x_2y_2+x_3y_3-x_4y_4-x_5y_5-x_6y_6$$ where $\textbf{x}=\sum_{k=1}^6x_ke_k$ and $\textbf{y}=\sum_{k=1}^6 y_ke_k$. Then if we set $v=\sum_{k=1}^6v_ke_k$ then we need to show that there exists $w=\sum_{k=1}^6w_ke_k$ such that 
\begin{align}
\sum_{k=1}^3 v_kw_k-\sum_{k=4}^6v_kw_k&=0\tag 1\\
\sum_{k=1}^3 w_k^2-\sum_{k=4}^6 w_k^2&=0\tag 2
\end{align}
But I can't solve this system of equations.... So how to move on?
 A: If $v=0\in\mathbb R^6$, then clearly we are done.
Now consider the signature of $W = \text{span}(v)^\perp$, where $v\in\mathbb R^6$ and $ v\ne 0$. We know that it has dimension $d=5$ because it is defined by your equation $(1)$. From another answer, we know all the possible signatures $(a,b)$ of $W$. 
$a=0$ or $b=0$ are not possible: $B_{\mathbb{R}^6}$ has signature $(p,q)=(3,3)$, then $d-a-b \leq \min(p-a,q-b)$. If $a=0$, then $d-b \leq \min(p,q-b) = q-b \Rightarrow 5 \leq 3$.
Then consider the vector $w_++w_-$ where $w_+, w_- \in W$ and $B(w_+,w_+) = 1, B(w_-, w_-) = -1, B(w_+,w_-)=0$.
A: Hint Here's a geometric approach that requires proving a short lemma, namely that the punctured null cone $N := \{u \in \Bbb R^6 : B(u, u) = 0\}$ is (path-)connected.
Now, pick any $X \in N$: If $B(v, x) = 0$, we are done; otherwise $B(v, x)$ and $B(v, -x)$ have opposite signs. Since $N$ is (path-)connected, there is some path $y(t)$ in $N$ connecting $x, -x$. Now, consider the function $t \mapsto N(v, y(t))$.
The same approach shows, by the way, that the nondegeneracy hypothesis can be dropped, and that we can replace $\Bbb R^6$ with $\Bbb R^n$ and allow any signature $-n + 2 < s < n - 2$. In particular this argument does not apply to Lorentzian signature ($s = \pm(n - 2)$), and indeed the claim is false in that case.
