Question

Consider the parallopiped with sides $\overline{a}=3i+2j+k,\overline{b}=\overline{i}+2\overline{k}$ and $\overline{c}=\overline{i}+3\overline{j}+3\overline{k}$, the angle between $\overline{a}$ and the plane containing the face determined by $\overline{b}$ and $\overline{c}$ is

• A. ${\sin }^{-1}\left(\frac{-17}{\sqrt{14}.\sqrt{46}}\right)$
• B. ${\cos }^{-1}\frac{1}{3}$
• C. ${\sin }^{-1}\frac{9}{14}$
• D. ${\sin }^{-1}\frac{2}{3}$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{-17}{\sqrt{14}.\sqrt{46}}\right)$

$\overline{b}\times \overline{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 2\\ 1& 3& 3\end{array}\right|$

$\Rightarrow \hat{i}(-6)-\hat{j}(3-2)+\hat{k}\left(3\right)$

$\Rightarrow \overline{b}\times \overline{c}=-6\hat{i}-\hat{j}+3\hat{k}$

Angle between $\overline{a}$ and normal to plane containing $\overline{b}$ and $\overline{c}$ is:

$\Rightarrow cos\theta =\frac{\overline{a}.(\overline{b}\times \overline{c})}{|\overline{a}|.|\overline{b}\times \overline{c}|}$

$=\frac{(3\hat{i}+2\hat{j}+\hat{k}).(-6\hat{i}-\hat{j}+3\hat{k})}{\left(\sqrt{3^2+2^2+1}\right)\sqrt{6^2+1+3^2}}$

$=\frac{-18-2+3}{\sqrt{14}.\sqrt{46}}$

$cos\theta =\frac{-17}{\sqrt{14}.\sqrt{46}}$

$\because$ Normal and the plane must make ${90}^{\circ }$

$\Rightarrow$ Let angle between plane and normal $=\alpha =90-\theta$

$\Rightarrow \theta =90-\alpha$

$\Rightarrow cos(90-\alpha )=\frac{-17}{\sqrt{14}.\sqrt{46}}$

$\Rightarrow sin\left(\alpha \right)=\frac{-17}{\sqrt{14}.\sqrt{46}}$

$\alpha =si{n}^{-1}\left(\frac{-17}{\sqrt{14}.\sqrt{46}}\right)$