A) \[{{\cos }^{-1}}\left( \frac{19}{35} \right)\]
B) \[{{\cos }^{-1}}\left( \frac{17}{31} \right)\]
C) \[30{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: A
Solution :
Angle between two plane faces is equal to the angle between the normals \[{{\mathbf{n}}_{1}}\] and \[{{\mathbf{n}}_{2}}\] to the planes. \[{{\mathbf{n}}_{1}}\] the normal of face OAB is given by \[\overrightarrow{OA}\times \overrightarrow{OB}=\left| \,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \\ \end{matrix}\, \right|=5\mathbf{i}-\mathbf{j}-3\mathbf{k}\] .....(i) \[{{\mathbf{n}}_{2}}\] the normal of face ABC is given by \[\overrightarrow{AB}\times \overrightarrow{AC}\] \[2-1,\,\,1-2,\,\,3-1\] and \[-1-1,\,\,1-2,\,\,2-1\]i.e., \[1,\,-1,\,\,2\] and \[-2,\,\,-1,\,\,1.\] \[\therefore \,\,\,{{\mathbf{n}}_{2}}=\left| \,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \\ \end{matrix}\, \right|=\mathbf{i}-5\mathbf{j}-3\mathbf{k}\] ?..(ii) If q be the angle between \[{{\mathbf{n}}_{1}}\] and \[{{\mathbf{n}}_{2}}\], then \[\cos \theta =\frac{{{\mathbf{n}}_{1}}\,.\,{{\mathbf{n}}_{2}}}{|\,\,{{\mathbf{n}}_{1}}\,\,|\,\,.\,\,|\,\,{{\mathbf{n}}_{2}}\,\,|}=\frac{5\,+5+9}{\sqrt{35}\,.\,\sqrt{35}}=\frac{19}{35}\] \[\Rightarrow \theta ={{\cos }^{-1}}\left( \frac{19}{35} \right)\].
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