. A tetrahedron has. vertices at 0(0, 0, 0), A(l, 2, 1) B(2, 1, 3) and C(-1, 1, 2). Then, the angle between the faces 0AB and ABC will be

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$90 °$

$\cos^{- 1} (\frac{19}{35})$

$\cos^{- 1} (\frac{17}{31})$

$30 °$

answer is B.

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Detailed Solution

$Let {\vec{n}}_{1}$ and ${\vec{n}}_{2}$ be the vectors normal to the faces OAB and ABC. Then

${\vec{n}}_{1} = \vec{O A} \times \vec{O B} = |\begin{array}{l} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array}| = 5 \hat{i} - \hat{j} - 3 \hat{k}$

and ${\vec{n}}_{2} = \vec{A B} \times \vec{A C} = |\begin{array}{r} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 2 \\ - 2 & - 1 & 1 \end{array}| = \hat{i} - 5 \hat{j} - 3 \hat{k}$

If 0 is the angle between the faces OAB and ABC, then

$\cos θ = \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{|{\vec{n}}_{1}| |{\vec{n}}_{2}|} \Rightarrow \cos θ = \frac{5 + 5 + 9}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \frac{19}{35}$