Cross product and sum of vectors

Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be three vectors such that

$$\mathbf{a \times b} = \mathbf{b \times c} =\mathbf{c \times a} \neq \mathbf{0}$$

Prove that $\mathbf{a+b+c}=\mathbf{0}$.

Solution

We have successively:

\begin{align*}
\mathbf{a \times b} = \mathbf{b \times c}= -\mathbf{c \times b} &\Leftrightarrow \mathbf{a\times b} + \mathbf{c \times b}=0 \\
 &\Leftrightarrow \left ( \mathbf{a} + \mathbf{c} \right ) \times \mathbf{b}=0 
\end{align*}

From the last relation we conclude that the vector $\mathbf{a}+\mathbf{c}$ and $\mathbf{b}$ are parallel. This means that there exists an $\ell$ such that $\mathbf{b}=\ell (\mathbf{a}+\mathbf{c}) \quad (*)$.

We also have that:

\begin{align*}
\mathbf{b \times c}= \mathbf{c \times a}=-\mathbf{a \times c} &\Leftrightarrow \mathbf{b \times c} + \mathbf{a \times c} =0 \\
 &\Leftrightarrow \mathbf{\left ( b+a \right )} \times \mathbf{c} =0 \\
 &\overset{(*)}{\Leftrightarrow }\left [ \ell \left ( \mathbf{a+c} \right )+\mathbf{a} \right ]\times \mathbf{c}=0 \\
 &\Leftrightarrow \left [ \left ( 1+\ell \right )\mathbf{a} + \ell \mathbf{c} \right ] \times \mathbf{c} =0  \\
 &\Leftrightarrow \left ( 1+\ell \right ) \left ( \mathbf{a \times c} \right )+\cancelto{0}{ \ell \mathbf{c \times c}} =0 \\
 &\Leftrightarrow \left ( 1+\ell \right ) \left ( \mathbf{a\times c} \right ) =0
\end{align*}

From the last relation we get that $\ell=-1$ since $\mathbf{a \times c} \neq 0$. Substituting back in $(*)$ we get that $\mathbf{a+b+c}=\mathbf{0}$, which is exactly what we wanted.

The exercise can also be found at mathematica.gr