Question

A terahedron 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 PQR is :

• A. ${\cos }^{-1}\left(\frac{9}{35}\right)$
• B. ${\cos }^{-1}\left(\frac{19}{35}\right)$
• C. ${\cos }^{-1}\left(\frac{17}{31}\right)$
• D. ${\cos }^{-1}\left(\frac{7}{31}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{19}{35}\right)$

Vector $\perp$ to face OPQ is $h_1$

$=OP\times OQ=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|$

$=5i-\hat{j}-3k$

Vector $\perp$ to face PQR is $h_2$

$PQ\times PR=\left|\begin{array}{ccc}i& j& k\\ 1& -1& 2\\ -2& -1& 1\end{array}\right|$

$=1-5j-3k$

Angle between faces is the angle between their normal

$\cos \theta =\frac{n_1\cdot n_2}{|n_1||n_2|}$

$=\frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35}$

$\theta ={\cos }^{-1}\left(\frac{19}{35}\right)$.