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

• A. $\cos^{-1}\left(\frac{19}{35}\right)$
• B. $\cos^{-1}\left(\frac{17}{31}\right)$
• C. 30$^\circ$
• D. 90$^\circ$

Answer 1

The Correct Option is A

Perpendicular to face $OAB$ $\overrightarrow{OA} \times\overrightarrow{OB}$ $=\left(\hat{i} +2\hat{j}+\hat{k}\right)\times \left(2 \hat{i}+\hat{j}+3\hat{k}\right)$ $=\left(5 \hat{i}-\hat{j}-3\hat{k}\right)$ Vector perpendicular to face $ABC$

$\overrightarrow{AB} \times\overrightarrow{AC}$ $=\left(\hat{i}-\hat{j}+2 \hat{k}\right)\times\left(-2 \hat{i}-\hat{j}+\hat{k}\right)$ $=\hat{i}-5 \hat{j}-3 \hat{k}$ Since angle between face equals angle between their normals. Therefore $cos\, \theta=\frac{5\times1+\left(-1\right)\times\left(-5\right)+\left(-3\right)\times\left(-3\right)}{\sqrt{5^{2}+\left(-1\right)^{2}+\left(-3\right)^{2}}\sqrt{1^{2}+\left(-5\right)^{2}+\left(-3\right)}}$ $=\frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35}$ $\theta=cos^{-1}\left(\frac{19}{35}\right)$