Answer
If the vectors ${\bf{u}}$, ${\bf{v}}$, and ${\bf{w}}$ are mutually orthogonal, then
$\left( {{\bf{u}} \times {\bf{v}}} \right) \times {\bf{w}} = {\bf{0}}$ and ${\bf{u}} \times \left( {{\bf{v}} \times {\bf{w}}} \right) = {\bf{0}}$.
Work Step by Step
Suppose that vectors ${\bf{u}}$, ${\bf{v}}$, and ${\bf{w}}$ are mutually orthogonal. Let $\left\{ {{\bf{u}},{\bf{v}},{\bf{w}}} \right\}$ forms a right-handed system.
By Theorem 1, we have the vector ${\bf{u}} \times {\bf{v}} = {\bf{w}}$ orthogonal to ${\bf{u}}$ and ${\bf{v}}$. Thus,
$\left( {{\bf{u}} \times {\bf{v}}} \right) \times {\bf{w}} = {\bf{w}} \times {\bf{w}}$
But ${\bf{w}} \times {\bf{w}} = {\bf{0}}$, since the angle between the vector ${\bf{w}}$ and itself is zero. Therefore, $\left( {{\bf{u}} \times {\bf{v}}} \right) \times {\bf{w}} = {\bf{0}}$.
Similarly, by Theorem 1, we have the vector ${\bf{v}} \times {\bf{w}} = {\bf{u}}$ orthogonal to ${\bf{v}}$ and ${\bf{w}}$. Thus,
${\bf{u}} \times \left( {{\bf{v}} \times {\bf{w}}} \right) = {\bf{u}} \times {\bf{u}}$
But ${\bf{u}} \times {\bf{u}} = {\bf{0}}$, since the angle between the vector ${\bf{u}}$ and itself is zero. Therefore, ${\bf{u}} \times \left( {{\bf{v}} \times {\bf{w}}} \right) = {\bf{0}}$.
