Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$. What is $\cos(\angle CMD)$? $\textbf{(A) } \frac14 \qquad$ $ \textbf{(B) } \frac13 \qquad$ $ \textbf{(C) } \frac25 \qquad$ $ \textbf{(D) } \frac12 \qquad$ $ \textbf{(E) } \frac{\sqrt{3}}{2}$
Solution
Place $A=(1,1,1),B=(1,-1,-1),C=(-1,1,-1),D=(-1,-1,1)$ (regular, edge $2\sqrt2$); $M=$ midpoint of $AB=(1,0,0)$.
$\vec{MC}=(-2,1,-1),\vec{MD}=(-2,-1,1)$; $\cos\angle CMD=\dfrac{\vec{MC}\cdot\vec{MD}}{|\vec{MC}||\vec{MD}|}=\dfrac{4-1-1}6=\dfrac13$. The answer is (\textbf{B}).
Cube $ABCDEFGH$ is labeled so that $ABCD$ and $EFGH$ are opposite faces, with $AE$, $BF$, $CG$, and $DH$ as edges. Point $M$ is the midpoint of $\overline{AE}$, and point $N$ is the midpoint of $\overline{FG}$. What is $\cos(\angle MBN)$?
$\textbf{(A) } \frac14 \qquad$ $ \textbf{(B) } \frac13 \qquad$ $ \textbf{(C) } \frac25 \qquad$ $ \textbf{(D) } \frac12 \qquad$ $ \textbf{(E) } \frac{\sqrt{3}}{2}$ 