Question

A vector $\overrightarrow{r}$ is equally inclined with the vectors $\overrightarrow{a}=cos\theta \hat{i}+sin\theta \hat{j},\overrightarrow{b}=-sin\theta \hat{i}+cos\theta \hat{j}$ and $\overrightarrow{c}=\hat{k},$ then the angle between $\overrightarrow{r}$ and $\overrightarrow{a}$ is

• A. $co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• B. $co{s}^{-1}\left(\frac{1}{3}\right)$
• C. $co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. $co{s}^{-1}\frac{1}{2}$

Answer 1

Answer: (C)

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors and mutually perpendicular vectors and $\overrightarrow{r}$ is equally inclined to all $3$ vectors
$\Rightarrow \overrightarrow{r}=\lambda \left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$
$\Rightarrow \overrightarrow{r}=\lambda \left(\left(cos\theta -sin\theta \right)\hat{i}+\left(sin\theta +cos\theta \right)\hat{j}+\hat{k}\right)$
Angle between $\overrightarrow{r}$ and $\overrightarrow{a}$ is
$cos\alpha =\frac{\overrightarrow{r}\cdot \overrightarrow{a}}{|\overrightarrow{r}||\overrightarrow{a}|}=\frac{\lambda \left({\left(cos\right)}^2\theta -sin\theta cos\theta +{\left(sin\right)}^2\theta +sin\theta cos\theta \right)}{1\times \sqrt{3}\lambda }$
$cos\alpha =\frac{1}{\sqrt{3}}\Rightarrow \alpha =co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$