JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Three Dimensional Geometry

The angle between two faces is the angle between their normals, and each normal is the cross product of two edge vectors of that face. For face $OPQ$, $\overrightarrow{OP}=(1,2,1)$ and $\overrightarrow{OQ}=(2,1,3)$ give $\overrightarrow{n_1}=\overrightarrow{OP}\times \overrightarrow{OQ}=(5,-1,-3)$. For face $PQR$, $\overrightarrow{PQ}=(1,-1,2)$ and $\overrightarrow{PR}=(-2,-1,1)$ give $\overrightarrow{n_2}=(1,-5,-3)$. Then $\cos \theta =\frac{|\overrightarrow{n_1}\cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}=\frac{|5+5+9|}{\sqrt{35}\sqrt{35}}=\frac{19}{35}$, so the angle between the faces is ${\cos }^{-1}\frac{19}{35}$.