# Question #63619

Jun 6, 2016

${\theta}_{z} = {90}^{\circ}$

#### Explanation:

Let $\vec{v} = \left\{{v}_{x} , {v}_{y} , {v}_{z}\right\}$ and ${\hat{e}}_{x} , {\hat{e}}_{y} , {\hat{e}}_{z}$ the axis unity vectors.

We have

$\left\langle\vec{v} , {\hat{e}}_{x}\right\rangle = \left\lVert \vec{v} \right\rVert \cos \left({\theta}_{x}\right) = {v}_{x}$
$\left\langle\vec{v} , {\hat{e}}_{y}\right\rangle = \left\lVert \vec{v} \right\rVert \cos \left({\theta}_{y}\right) = {v}_{y}$
$\left\langle\vec{v} , {\hat{e}}_{z}\right\rangle = \left\lVert \vec{v} \right\rVert \cos \left({\theta}_{z}\right) = {v}_{z}$

also

${v}_{x}^{2} + {v}_{y}^{2} + {v}_{z}^{2} = {\left\lVert \vec{v} \right\rVert}^{2} = {\left\lVert \vec{v} \right\rVert}^{2} {\cos}^{2} \left({\theta}_{x}\right) + {\left\lVert \vec{v} \right\rVert}^{2} {\cos}^{2} \left({\theta}_{y}\right) + {\left\lVert \vec{v} \right\rVert}^{2} {\cos}^{2} \left({\theta}_{z}\right)$

simplifying we have

${\cos}^{2} \left({\theta}_{x}\right) + {\cos}^{2} \left({\theta}_{y}\right) + {\cos}^{2} \left({\theta}_{z}\right) = 1$

so

$\cos \left({\theta}_{z}\right) = \pm \sqrt{1 - {\cos}^{2} \left({\theta}_{x}\right) - {\cos}^{2} \left({\theta}_{y}\right)}$

having now ${\theta}_{x} = {150}^{\circ}$ and ${\theta}_{y} = {60}^{\circ}$

we have $\cos \left({\theta}_{z}\right) = 0$ so ${\theta}_{z} = \pm \frac{\pi}{2} = \pm {90}^{\circ}$