Question

If vectors $\overrightarrow{b}=(\tan \alpha ,-1,2\sqrt{\sin \alpha /2})$ and $\overrightarrow{c}=(\tan \alpha ,\tan \alpha ,-\frac{3}{\sqrt{\sin \alpha /2}})$ are orthogonal and vector $\overrightarrow{\alpha }=(1,3,\sin 2\alpha )$ makes an obtuse angle with the z-axis, then the value of $\alpha$ is?

• A. $\alpha =(4n+1)\pi +{\tan }^{-1}2$
• B. $\alpha =(4n+1)\pi -{\tan }^{-1}2$
• C. $\alpha =(4n+2)\pi +{\tan }^{-1}2$
• D. $\alpha =(4n+2)\pi -{\tan }^{-1}2$

Answer 1

The correct option is B $\alpha =(4n+1)\pi -{\tan }^{-1}2$

Given,

$\overrightarrow{b}=(\tan 2,-1,2\sqrt{\sin 2/2}$

$\overrightarrow{c}=(\tan \alpha ,\tan \alpha ,-\frac{3}{\sqrt{\sin 2/2}}$

Given, $\overrightarrow{b}$ and $\overrightarrow{c}$ are orthogonal.

Then,

$\overrightarrow{b}=\tan \alpha t+-\hat{ȷ}+2\sqrt{\sin \alpha /2}\hat{k}$

$\overrightarrow{c}=\tan 2\hat{ı}+\tan 2\hat{ȷ}-\frac{3}{\sqrt{\sin 2/2}}$

$\overrightarrow{b}\cdot \overrightarrow{c}=0[\because$ They are orthogonal $]$

$\overrightarrow{b}\cdot \overrightarrow{c}={\tan }^2\alpha -\tan \alpha +2\sqrt{\sin \alpha /2}x-\frac{3}{\sqrt{\sin \alpha /}2}$

$0={\tan }^2\alpha -\tan \alpha -6$

$\Rightarrow {\tan }^2\alpha -3\tan \alpha +2\tan 2-6$

$\Rightarrow \tan 2(\tan 2-3)+2(\tan 2-3)$

$\Rightarrow (\tan 2-3)(\tan 2+2)$

$\tan \alpha =3;\tan \alpha =-2$

$2={\tan }^{-1}{3}^{\prime }$ or $\alpha ={\tan }^{-1}(-2)$

$\because \sin 2\alpha =\frac{2\tan 2}{1+{\tan }^2\alpha }$

$for\tan \alpha =3\sin 2\alpha =\frac{2x3}{1+9}=\frac{3}{5}>0$

$\mathrm{for}\tan =-2,\sin 2\alpha =\frac{-4}{5}<0$

$\text{AQ}\overrightarrow{a}=\hat{r}+3\hat{ȷ}+\sin 22\hat{k}$

$\text{makes an obtuse angle with}2-\mathrm{axis}(0,0,1)$

$\overrightarrow{d}=\hat{k}$

$\overrightarrow{a}\cdot \overrightarrow{d}=|\overrightarrow{a}||\overrightarrow{a}|\cos \theta -.$

$\cos \theta \text{will be negative for obtuse angle.}$

$\therefore \overline{\theta }\cdot 2<0$

$\sin 22<0$

$\text{ie,}\sin 2\alpha ;\tan 2=-2$

$\therefore \text{option}B:\alpha =(1n+1)\pi -{\tan }^{-1}2$