Question
Consider the parallelepiped with side $$\overrightarrow a = 3\hat i + 2\hat j + \hat k,\,\overrightarrow b = \hat i + \hat j + 2\hat k$$ and $$\overrightarrow c = \hat i + 3\hat j + 3\hat k$$ then the angle between $$\overrightarrow a $$ and the plane containing the face determined by $$\overrightarrow b $$ and $$\overrightarrow c $$ is :
A.
$${\sin ^{ - 1}}\frac{1}{3}$$
B.
$${\cos ^{ - 1}}\frac{9}{{14}}$$
C.
$${\sin ^{ - 1}}\frac{9}{{14}}$$
D.
$${\sin ^{ - 1}}\frac{2}{3}$$
Answer :
$${\sin ^{ - 1}}\frac{9}{{14}}$$
Solution :
\[\overrightarrow b \times \overrightarrow c = \left| \begin{array}{l}
\hat i\,\,\,\hat j\,\,\,\hat k\\
1\,\,\,\,1\,\,\,\,2\\
1\,\,\,\,3\,\,\,3
\end{array} \right| = - 3\hat i - \hat j + 2\hat k\]
If $$\theta $$ is the angle between $$\overrightarrow a $$ and the plane containing $$\overrightarrow b $$ and $$\overrightarrow c ,$$ then
$$\eqalign{
& \cos \left( {{{90}^ \circ } - \theta } \right) = \left| {\frac{{\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b \times \overrightarrow c } \right|}}} \right| \cr
& \Rightarrow \cos \left( {{{90}^ \circ } - \theta } \right) = \frac{1}{{\sqrt {14} }}.\frac{1}{{\sqrt {14} }}\left| {\left( { - 9 - 2 + 2} \right)} \right| \cr
& \Rightarrow \cos \left( {{{90}^ \circ } - \theta } \right) = \frac{9}{{14}} \cr
& \Rightarrow \sin \,\theta = \frac{9}{{14}} \cr
& \Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{9}{{14}}} \right) \cr} $$