Cross product with two orthogonal vectors being zero implying a null vector

Question

Consider two orthogonal nonzero vectors $\vec{B}=(p,q,r)$ and $\vec{C}=(u,v,w)$. How to prove algebraically that if vector $\vec{A}=(a,b,c)$ satisfies $\vec{A}\times \vec{B}=\vec{A}\times\vec{C}=\vec{0}$, then $\vec{A}=\vec{0}$?

For example, if $\vec{B}=\hat{i}$ and $\vec{C}=\hat{j}$, we have

$$\begin{bmatrix} a\\b\\c\end{bmatrix} \times \begin{bmatrix} 1\\0\\0\end{bmatrix} = \begin{bmatrix} 0\\c\\-b\end{bmatrix}$$

$$\begin{bmatrix} a\\b\\c\end{bmatrix} \times \begin{bmatrix} 0\\1\\0\end{bmatrix} = \begin{bmatrix} -c\\0\\a\end{bmatrix}$$

Thus $a=b=c=0$.

What about the general case?

Answer 1

Do a change of basis. Consider the basis $B=\{e_1,e_2,e_3\}$, with $e_1=\vec B$, $e_2=\vec C$, and $e_3=\vec B\times\vec C$. Then you know that $\vec A\times e_1=\vec A\times e_2=0$. If $A=a_1e_1+a_2e_2+a_3e_3$, then this means that $a=b=c=0$, since $\vec A\times e_1=(0,c,-b)_B$ and $A\times e_2=(-c,0,a)_B$.

Answer 2

If $\vec A\times\vec B=0$, then since $\vec B\ne0$, $\vec A=\lambda\vec B$. Similarly, $\vec A=\mu\vec C$. Since $\vec B$ and $\vec C$ are orthogonal, they are linearly independent and so the only vector that is a scalar multiple of both is the zero vector. Alternatively, from $\vec A=\lambda\vec B$ we have $\vec A\times\vec C=\lambda\vec B\times\vec C=\lambda(\vec B\times\vec C)=0$, but since $\vec B$ and $\vec C$ are nonzero and orthogonal, their cross product is also nonzero and we must have $\lambda=0$.