Question

The angles of a triangle, two of whose sides are represented by the vectors $\sqrt{3}(\overrightarrow{a}\times \overrightarrow{b})$ and $\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{a}$ where $\overrightarrow{b}$ is a non-zero vector and $\overrightarrow{a}$ is a unit vector are

• A. ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right);{\tan }^{-1}\left(\frac{1}{2}\right);{\tan }^{-1}\left(\frac{\sqrt{3}+2}{1-2\sqrt{3}}\right)$
• B. ${\tan }^{-1}\left(\sqrt{3}\right);{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right);{\cot }^{-1}\left(0\right)$
• C. ${\tan }^{-1}\left(\sqrt{3}\right);{\tan }^{-1}\left(2\right);{\tan }^{-1}\left(\frac{\sqrt{3}+2}{2\sqrt{3}-1}\right)$
• D. ${\tan }^{-1}\left(\sqrt{3}\right);{\tan }^{-1}\left(\sqrt{2}\right);{\tan }^{-1}\left(\frac{\sqrt{2}+3}{3\sqrt{2}-1}\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(\sqrt{3}\right);{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right);{\cot }^{-1}\left(0\right)$

$\overline{x}=3(\overline{a}\times \overline{b})$ and & $\overline{y}=\overline{b}-(\overline{a}.\overline{b})\overline{a}$
So, $\overline{x}.\overline{y}=0$

So one angle is ${90}^o.$

Also,$|\overline{x}|=b\sin \theta$, where $\theta$ is angle between $\overline{a}$ and $\overline{b}$

$|\overline{y}|=b\cos \theta$

Thus,$\frac{|\overline{x}|}{|\overline{y}|}=\sqrt{3}=tan\alpha$
$\therefore \alpha ={60}^o$ and $\beta ={30}^o.$