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

Answer: (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
$cos\theta =\frac{5\times 1+\left(-1\right)\times \left(-5\right)+\left(-3\right)\times \left(-3\right)}{\sqrt{5^2+{\left(-1\right)}^2+{\left(-3\right)}^2}\sqrt{1^2+{\left(-5\right)}^2+\left(-3\right)}}$
$=\frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35}$
$\theta =co{s}^{-1}\left(\frac{19}{35}\right)$