Given $\angle (\vec{u} , \vec{v}) = 30^\circ$, $\Vert \vec{w} \Vert = 4$ and $\vec{w}$ is $\perp$ to both find $\vec{u} \cdot \vec{v} \times \vec{w}$

Question

I am given the following problem:

Knowing that the angle between the unit vectors $\angle (\vec{u} , \vec{v}) = 30^\circ$, $\Vert \vec{w} \Vert = 4$ and that $\vec{w}$ is orthogonal to both of them, evaluate $\vec{u} \cdot \vec{v} \times \vec{w}$

I am not sure where to begin this exercise. I know that

$$ \vec{u} \cdot \vec{w} = \vec{v} \cdot \vec{w} = 0 $$

but that won't take me far.

Answer 1

Hint: Note that $$ u \cdot (v \times w) = w \cdot (u \times v) $$ That is, we can cyclically permute the vectors in a triple scalar product.

Answer 2

So it is understood that you are looking for the value of $s$ in $$ s = \vec u \cdot \left( {\vec v \times \vec w} \right)$$

First note that $s$ is a scalar, that is a real number (the volume - with sign - of the paralleliped defined by the three vectors). Second, since ${\vec v}$ and ${\vec w}$ are orthogonal to each other, their cross product will be a vector orthogonal to both, whose modulus is $$\vec t = \vec v \times \vec w\quad \Rightarrow \vec v \bot \vec w \Rightarrow \quad \left| {\vec t} \right| = \left| {\vec v} \right|\left| {\vec w} \right| = 4\left| {\vec v} \right|$$

Since ${\vec u}$ is also orthogonal to ${\vec w}$, then ${\vec t}$ will lay on the same plane as ${\vec u}$ and ${\vec v}$

Then since $\angle (\vec{u} , \vec{v}) = 30^\circ$, the $\angle (\vec{u} , \vec{t})$ will be either $60^\circ$ or $120^\circ$ (don't know if you indicated oriented angle or absolute).

Thus $$ s = \vec u \cdot \left( {\vec v \times \vec w} \right) = 4\left| {\vec u} \right|\left| {\vec v} \right|\cos \left( {30^\circ \pm 90^\circ} \right) $$