Question

Consider the parallelopiped with sides $\overrightarrow{a}=3\hat{i}+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{c}=\hat{i}+3\hat{j}+3\hat{k}$, then 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 1

The correct option is C ${\sin }^{-1}\frac{9}{14}$

$\overline{b}\times \overline{c}:\left|\begin{array}{ccc}\stackrel{\wedge }{i}& \stackrel{\wedge }{j}& \stackrel{\wedge }{k}\\ 1& 1& 2\\ 1& 3& 3\end{array}\right|$

$⟹\stackrel{\wedge }{i}(3-6)-\stackrel{\wedge }{j}(3-2)+\stackrel{\wedge }{k}(3-1)$

$⟹-3\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+2\stackrel{\wedge }{k}$

$If\theta istheanglebetween\overline{a}andtheplane$

$containing\overline{b}and\overline{c},then:$

$\cos (90-\theta )=\left|\frac{\overline{a}.(\overline{b}\times \overline{c})}{\left|\overline{a}\right|.|\overline{b}\times \overline{c}|}\right|$

$\because (90-\theta )willbetheanglewiththenormal$

$totheplane.$

$⟹\overline{a}.(\overline{b}\times \overline{c})=(3\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+\stackrel{\wedge }{k}).(-3\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+2\stackrel{\wedge }{k})$

$=-9-2+2$

$=-9$

$\left|\overline{a}\right|=\sqrt{3^2+2^2+1}=\sqrt{14}$

$|\overline{b}\times \overline{c}|=|-3\stackrel{\wedge }{i}-\stackrel{\wedge }{j}+2\stackrel{\wedge }{k}|$

$=\sqrt{3^2+1^2+4}$

$=\sqrt{9+1+4}=\sqrt{14}$

$⟹\left|\overline{a}\right|.|\overline{b}\times \overline{c}|={\left(\sqrt{14}\right)}^2=14$

$\left|\frac{\overline{a}.(\overline{b}\times \overline{c})}{\left|\overline{a}\right|.|\overline{b}\times \overline{c}|}\right|=\frac{9}{14}$

$⟹\cos (90-\theta )=\frac{9}{14}$

$⟹sin\theta =\frac{9}{14}$

$⟹\theta ={\sin }^{-1}\left(\frac{9}{14}\right)$