A tetrahedron has vertices at O(0,0,0), A(1,2,1), B(2,1,3) and C(-1,1,2).

Question

A tetrahedron has vertices at $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$ and $C(-1,1,2)$. Then the angle between the faces $OAB$ and $ABC$ will be:

Accepted Answer

Normal to face $OAB$: $\overrightarrow{OA}\times\overrightarrow{OB}=\begin{vmatrix}\hat i&\hat j&\hat k\1&2&1\2&1&3\end{vmatrix}=5\hat i-\hat j-3\hat k$, magnitude $\sqrt{35}$.

Normal to face $ABC$: $\overrightarrow{AB}\times\overrightarrow{AC}$ with $\overrightarrow{AB}=(1,-1,2)$, $\overrightarrow{AC}=(-2,-1,1)$:

$$\begin{vmatrix}\hat i&\hat j&\hat k\1&-1&2\-2&-1&1\end{vmatrix}=\hat i-5\hat j-3\hat k,\quad\text{magnitude }\sqrt{35}$$

$$\cos\theta = \frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35} \implies \theta=\cos^{-1}\frac{19}{35}$$

Solutions · 1