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A tetrahedron has vertices at $\mathrm{O}(0,0,0), \mathrm{A}(1,2,1) \mathrm{B}(2,1,3)$ and $\mathrm{C}(-1,1,2)$. Then the angle between the faces $\mathrm{OAB}$ and $\mathrm{ABC}$ will be
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The correct answer is:
$cos ^{-1}\left(\frac{19}{35}\right)$
$cos ^{-1}\left(\frac{19}{35}\right)$
Vector perpendicular to the face $\mathrm{OAB}$
$=\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right|=5 \mathrm{i}-\mathrm{j}-3 \mathrm{k}$
Vector perpendicular to the face $\mathrm{ABC}$
$=\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right|=\mathrm{i}-5 \mathrm{j}-3 \mathrm{k}$
Angle between the faces $=$ Angle between their normals
$\cos \theta=\left|\frac{5+5+9}{\sqrt{35} \sqrt{35}}\right|=\frac{19}{35}$ or $\theta=\cos ^{-1}\left(\frac{19}{35}\right)$
$=\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right|=5 \mathrm{i}-\mathrm{j}-3 \mathrm{k}$
Vector perpendicular to the face $\mathrm{ABC}$
$=\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right|=\mathrm{i}-5 \mathrm{j}-3 \mathrm{k}$
Angle between the faces $=$ Angle between their normals
$\cos \theta=\left|\frac{5+5+9}{\sqrt{35} \sqrt{35}}\right|=\frac{19}{35}$ or $\theta=\cos ^{-1}\left(\frac{19}{35}\right)$
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