# How do you find the range of the scalar triple product of the vectors <cos alpha, cos beta, 0>, <0, cos beta, cos gamma> and <cos alpha, 0, cos gamma>?

Aug 19, 2016

$\left[- 2 , 2\right]$

#### Explanation:

The triple scalar product of vectors $\vec{u} , \vec{v} , \vec{w} \in {\mathbb{R}}^{3}$ is given by $\vec{u} \cdot \left(\vec{v} \times \vec{w}\right)$. A useful fact is that this can also be calculated as the determinant of a $3 \times 3$ matrix with the given vectors as the rows or columns. That is,

$\vec{u} \cdot \left(\vec{v} \times \vec{w}\right) = \det \left(\begin{matrix}{u}_{1} & {u}_{2} & {u}_{3} \\ {v}_{1} & {v}_{2} & {v}_{3} \\ {w}_{1} & {w}_{2} & {w}_{3}\end{matrix}\right)$

We will use that to calculate the given scalar triple product.

Let
$\vec{u} = < \cos \left(\alpha\right) , \cos \left(\beta\right) , 0 >$
$\vec{v} = < 0 , \cos \left(\beta\right) , \cos \left(\gamma\right) >$
$\vec{w} = < \cos \left(\alpha\right) , 0 , \cos \left(\gamma\right) >$

Then we have:

$\vec{u} \cdot \left(\vec{v} \times \vec{w}\right) = \det \left(\begin{matrix}{u}_{1} & {u}_{2} & {u}_{3} \\ {v}_{1} & {v}_{2} & {v}_{3} \\ {w}_{1} & {w}_{2} & {w}_{3}\end{matrix}\right)$

$= \det \left(\begin{matrix}\cos \left(\alpha\right) & \cos \left(\beta\right) & 0 \\ 0 & \cos \left(\beta\right) & \cos \left(\gamma\right) \\ \cos \left(\alpha\right) & 0 & \cos \left(\gamma\right)\end{matrix}\right)$

$= \cos \left(\alpha\right) \left(\cos \left(\beta\right) \cos \left(\gamma\right) - 0 \cos \left(\gamma\right)\right)$

$- \cos \left(\beta\right) \left(0 \cos \left(\gamma\right) - \cos \left(\alpha\right) \cos \left(\gamma\right)\right)$

$+ 0 \left(0 \cdot 0 - \cos \left(\alpha\right) \cos \left(\beta\right)\right)$

$= 2 \cos \left(\alpha\right) \cos \left(\beta\right) \cos \left(\gamma\right)$

Thus, it suffices to find the range of $2 \cos \left(\alpha\right) \cos \left(\beta\right) \cos \left(\gamma\right)$.

With no restrictions on $\alpha , \beta , \gamma$ beyond being in $\mathbb{R}$, we have $\cos \left(\alpha\right) , \cos \left(\beta\right) , \cos \left(\gamma\right) \in \left[- 1 , 1\right]$.
Thus their product will also have the range $\cos \left(\alpha\right) \cos \left(\beta\right) \cos \left(\gamma\right) \in \left[- 1 , 1\right]$.
Multiplying by $2$ gives our desired range:
$2 \cos \left(\alpha\right) \cos \left(\beta\right) \cos \left(\gamma\right) \in \left[- 2 , 2\right]$

Thus the range of the scalar triple product of $< \cos \left(\alpha\right) , \cos \left(\beta\right) , 0 > , < 0 , \cos \left(\beta\right) , \cos \left(\gamma\right) > ,$ and $< \cos \left(\alpha\right) , 0 , \cos \left(\gamma\right) >$, is $\left[- 2 , 2\right]$