Question

The acute angle that the vector $2\hat{i}-2\hat{j}+\hat{k}$ makes with the plane contained by the two vectors $2\hat{i}+3\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+2\hat{k}$ is given by

• A. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. ${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• C. ${\tan }^{-1}\left(\sqrt{2}\right)$
• D. ${\cot }^{-1}\left(\sqrt{2}\right)$

Answer 1

The correct option is D ${\cot }^{-1}\left(\sqrt{2}\right)$

Let $\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}$, then the normal plane vector of the plane is

$\overrightarrow{n}=\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 1& -1& 2\end{array}\right|=5\hat{i}-5\hat{j}-5\hat{k}$

$\therefore$ unit vector perpendicular to the plane $\overrightarrow{a}$ and $\overrightarrow{b}$ is

$\overrightarrow{n}=\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})$

If $\theta$ is the required angle between the plane and $\overrightarrow{c}=2\hat{i}-2\hat{j}+\hat{k}$, then $(\frac{\pi }{2}-\theta )$ is the angle between normal vector and $\overrightarrow{c}.$

Thus $\cos (\frac{\pi }{2}-\theta )=\frac{2\hat{i}-2\hat{j}+\hat{k}}{3}.\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})=\frac{1}{\sqrt{3}}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$\therefore \theta ={\cot }^{-1}\left(\sqrt{2}\right)$