Vector Algebra Question 52
Question: Consider the parallelepiped with side $ \vec{a}=3\hat{i}+2\hat{j}+\hat{k},\vec{b}=\hat{i}+\hat{j}+2\hat{k} $ and $ \vec{c}=\hat{i}+3\hat{j}+3\hat{k} $ then the angle between $ \vec{a} $ and the plane containing the face determined by $ \vec{b} $ and $ \vec{c} $ is
Options:
A) $ \sin {{,}^{-1}}\frac{1}{3} $
B) $ \cos {{,}^{-1}}\frac{1}{14} $
C) $ sin{{,}^{-1}}\frac{9}{14} $
D) $ sin{{,}^{-1}}\frac{2}{3} $
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Answer:
Correct Answer: C
Solution:
- [c] $ \vec{b}\times \vec{c}= \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 1 & 2 \\ 1 & 3 & 3 \\ \end{vmatrix} =-3\hat{i}-\hat{j}+2\hat{k} $
If $ \theta $ is the angle between $ \overset{\to }{\mathop{a}}, $ and the plane containing $ \overset{\to }{\mathop{b}}, $ and $ \overset{\to }{\mathop{c}}, $ , then $ \cos (90{}^\circ -\theta )=| \frac{\overset{\to }{\mathop{a}},.(\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},)}{|\overset{\to }{\mathop{a}},||\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},|} | $ $ =\frac{1}{\sqrt{14}}.\frac{1}{\sqrt{14}}|(-9-2+2)|=\frac{9}{14} $
$ \Rightarrow \sin \theta =\frac{9}{14}\Rightarrow \theta ={{\sin }^{-1}}( \frac{9}{14} ). $
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