Question

A tetrahedron has vertices $P(1,2,1),Q(2,1,3),R(-1,1,2)$ and $O(0,0,0).$ The angle between the faces $OPQ$ and $OQR$ is:

• A. ${\cos }^{-1}\left(\frac{7}{\sqrt{2065}}\right)$
• B. ${\cos }^{-1}\left(\frac{9}{\sqrt{2065}}\right)$
• C. ${\cos }^{-1}\left(\frac{7}{\sqrt{2265}}\right)$
• D. ${\cos }^{-1}\left(\frac{11}{\sqrt{2265}}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{7}{\sqrt{2065}}\right)$

Let $\overrightarrow{x_1}$ and $\overrightarrow{x_2}$ are the normal vectors to the faces $OPQ$ and $OQR$
The vertices of tetrahedron are $P(1,2,1),Q(2,1,3),R(-1,1,2)$ and $O(0,0,0).$

$\Rightarrow \overrightarrow{x_1}=\overrightarrow{OP}\times \overrightarrow{OQ}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|$
$=\hat{i}(6-1)-\hat{j}(3-2)+\hat{k}(1-4)$
$\Rightarrow \overrightarrow{x_1}=5\hat{i}-\hat{j}-3k$
Similarly,
$\overrightarrow{x_2}=\overrightarrow{\text{OQ}}\times \overrightarrow{\text{OR}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 3\\ -1& 1& 2\end{array}\right|$
$=\hat{i}(2-3)-\hat{j}(4+3)+\hat{k}(2+1)$
$\Rightarrow \overrightarrow{x_2}=-\hat{i}-7\hat{j}+3\hat{k}$
Angle between the faces $OPQ$ and $OQR$
$\cos \theta =\frac{\overrightarrow{x_1}.\overrightarrow{x_2}}{|\overrightarrow{x_1}|.|\overrightarrow{x_2}|}=\frac{|-5+7-9|}{\sqrt{35}\sqrt{59}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{7}{\sqrt{2065}}\right)$