For a Lie algebra $L$ with a quadratic form, is there always a bivector representation $\rho:L\to Cl(V,q)$ making $L$ isomorphic to vectors in $V$? For an $n$-dimensional Lie algebra $L$ with a quadratic form $Q$, is there always a bivector representation $\rho:L\to Cl(V,q)$ and a single vector $v\in V$ such that $\{\rho(x)\times v\mid x\in L\}$ is an $n$-dimensional subspace of $V$, and $Q(x)=q\big(\rho(x)\times v\big)$ ?
This is a sequel to my previous question. The answer there gives a matrix representation $\rho$ and a vector $v$ such that $x\mapsto\rho(x)v$ is a linear isomorphism. Now I'm asking for an also quadratic isomorphism.
Any matrix Lie algebra is isomorphic to a bivector Lie algebra, as shown in my answer here. (I don't know about fields of characteristic $2$; I'm focusing on $\mathbb R$. And everything is finite-dimensional.)
The cross product of a $k$-vector $A=\langle A\rangle_k\in Cl(V,q)$ with a bivector $B=\langle B\rangle_2$ is defined as the commutator, and it preserves grade:
$$A\times B=\frac{AB-BA}{2}=\langle AB\rangle_k$$
so $\rho(x)\times v=\langle\rho(x)\times v\rangle_1\in V$.
I'd accept a weaker answer, ignoring $Q$ and just requiring $q$ to have a certain signature on the $n$-dimensional subspace. As a special case, we could just require $q$ to be positive-definite on the subspace.
 A: $L$ has a $Q$-orthogonal basis $\{x_1,x_2,\cdots,x_n\}$. (This means for $i\neq j$, $Q(x_i+x_j)=Q(x_i)+Q(x_j)$).
Using the idea from YCor's answer, if for each $x_i$ we can find a representation $\rho_i$ and a vector $v_i$ such that $\rho_i(x_i)\times v_i\neq0$ and $q\big(\rho_i(x_i)\times v_i\big)=Q(x_i)$, then the orthogonal direct sum $\rho=(\rho_1\oplus\rho_2\oplus\cdots\oplus\rho_n)$, $v=(v_1+v_2+\cdots+v_n)$ will be a solution. Conversely, if we have a solution $\rho$, $v$, then for each individual $x_i$ we have $\rho(x_i)\times v\neq0$ and $q\big(\rho(x_i)\times v\big)=Q(x_i)$. So this question is equivalent to

For any $x\neq0$ in a Lie algebra $L$ with a quadratic form $Q$, is there always a bivector representation $\rho:L\to Cl(V,q)$ and a vector $v\in V$ such that $\rho(x)\times v\neq0$, and $Q(x)=q\big(\rho(x)\times v\big)$ ?

Consider the case $Q(x)=0$. For any matrix representation with $\rho(x)\neq0$ (such as that given by Ado's theorem), there must be some $v$ such that $\rho(x)v\neq0$. Converting this to a bivector representation $\rho':L\to Cl(V\oplus V^*)$, we have a solution, because $q=0$ on the subspace $V$.
Now consider $Q(x)\neq0$, and let $\rho(x)=B$. Using some Clifford algebra identities, if we've found a solution, then
$$0\neq q(B\times v)=(B\times v)\,\lrcorner\,(B\times v)$$
$$=(B\times v)\,\lrcorner\,(B\,\llcorner\,v)$$
$$=\big((B\times v)\,\lrcorner\,B\big)\,\llcorner\,v$$
$$=\big((B\times v)\times B\big)\,\llcorner\,v$$
$$=\big(-B\times(B\times v)\big)\,\llcorner\,v,$$
so  the linear transformation $(B\,\times)^2\neq0$. (Note that $B\,\times$ is another representation of $x$, similar to an adjoint representation; Jacobi's identity gives $[A\,\times,B\,\times]=(A\times B)\times$ .) Conversely, suppose we have a matrix representation $\rho(x)=M$ with $M^2\neq0$. Then there must be $v\in V$ such that $0\neq M^2v\in V$, which itself implies that there must be $\phi\in V^*$ such that $0\neq\phi(M^2v)=(\phi M)(Mv)$. Converting to the bivector representation,
$$0\neq(\phi\times B)\cdot(B\times v)$$
$$=-(B\times\phi)\cdot(B\times v)$$
$$=-\frac12\Big((B\times\phi)\cdot(B\times\phi)+(B\times\phi)\cdot(B\times v)+(B\times v)\cdot(B\times\phi)+(B\times v)\cdot(B\times v)\Big)$$
$$=-\frac12\Big(B\times(\phi+v)\Big)\cdot\Big(B\times(\phi+v)\Big)$$
$$=-\frac12q\Big(B\times(v+\phi)\Big),$$
and we can scale $\phi$ by $Q(x)$ to get a solution. So this question is equivalent to
For any $x\neq0$ in a Lie algebra, is there always a matrix representation $\rho$ such that $\rho(x)^2\neq0$?
