# Question #832c6

Feb 2, 2017

See below.

#### Explanation:

Regarding the regular triangular prism with edge length $l$

a) The vector representing the edge $\left[{B}_{1} {A}_{1}\right]$ is ${\vec{v}}_{1} = \left(0 , 1 , 0\right)$
and the vector representing the edge $\left[B P\right]$ is ${\vec{v}}_{2} = \left(- \sin \left(\frac{\pi}{3}\right) , \cos \left(\frac{\pi}{3}\right) , \frac{1}{2}\right)$ Now making the dot product $\left\langle{\vec{v}}_{1} , {\vec{v}}_{2}\right\rangle = \cos \left(\frac{\pi}{3}\right) \ne 0$ so $\left[{B}_{1} {A}_{1}\right]$ and $\left[B P\right]$ are not perpendicular.

b) The vector associated to the edge $\left[A {A}_{1}\right]$ is ${\vec{v}}_{3} = \left(0 , 0 , 1\right)$
and the vector associated to $\left[B P\right]$ is ${\vec{v}}_{2} = \left(- \sin \left(\frac{\pi}{3}\right) , \cos \left(\frac{\pi}{3}\right) , \frac{1}{2}\right)$ so their dot product is $\left\langle{\vec{v}}_{3} , {\vec{v}}_{2}\right\rangle = \frac{1}{2}$. We know that $\left\langle{\vec{v}}_{3} , {\vec{v}}_{2}\right\rangle = \left\lVert {\vec{v}}_{3} \right\rVert \left\lVert {\vec{v}}_{2} \right\rVert \cos \left(\alpha\right)$ so $\cos \left(\alpha\right) = \frac{\frac{1}{2}}{\left\lVert {\vec{v}}_{3} \right\rVert \left\lVert {\vec{v}}_{2} \right\rVert} = \frac{\frac{1}{2}}{1 \cdot \frac{\sqrt{5}}{2}} = \frac{\sqrt{5}}{5} \ne \frac{\sqrt{2}}{2}$ so the angle between $\left[A {A}_{1}\right]$ and $\left[B P\right]$ is not $\frac{\pi}{4}$