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. $cos{1}^{-1}\left(19/35\right)$
• B. $co{s}^{-1}\left(17/31\right)$
• C. ${30}^0$
• D. ${90}^0$

Answer 1

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

Solution - In $△OAB$ $\begin{array}{c}OA=\hat{ı}+2\hat{ȷ}+\hat{k}\\ OB=2\hat{ı}+\hat{ȷ}+3\hat{k}\end{array}$ Vector $\perp$ to $OA$ and $OB=(\overrightarrow{OA}\times \overrightarrow{OB})$ $$\begin{array}{cc}& =\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|\\ {\overrightarrow{n}}_1=5\hat{ı}-\hat{ȷ}+(-3)\hat{k}& \end{array}$$ $\sin △ABC$ $$\begin{array}{c}AB=\hat{ı}-\hat{ȷ}+2\hat{k}\\ AC=2\hat{ı}-\hat{ȷ}+\hat{k}\end{array}$$ Vector $L$ to $AB$ and $AC={\overrightarrow{n}}_2$

${\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& k\\ 1& -1& 2\\ 2& -1& 1\end{array}\right|=-\hat{ı}+5\hat{ȷ}+3\hat{k}$

Angle between ${\overrightarrow{n}}_1$ and ${\overrightarrow{n}}_2\Rightarrow \cos \theta =\left|\frac{{\overrightarrow{n}}_1\cdot {\overrightarrow{n}}_2}{\left|{\overrightarrow{n}}_1\right|\left|{\overrightarrow{n}}_2\right|}\right|$ $\cos \theta =(\frac{(5\hat{ı}-\hat{ȷ}+(3\left)\hat{k}\right)\cdot (2\hat{ı}-\hat{ȷ}+\hat{k})}{\sqrt{35}\cdot \sqrt{35}}\Rightarrow \theta ={\cos }^4\left(\frac{19}{35}\right)$

Hence (A) is the correct option.