Question

For some non zero vector $\overrightarrow{V},$ if the sum of $\overrightarrow{V}$ and the vector obtained from $\overrightarrow{V}$ by rotating it by an angle $2\alpha$ equals to the vector obtained from $\overline{V}$ by rotating it by $\alpha$ then the value of $\alpha ,$ is where $n$ is an integer.

• A. $2n\pi \pm \frac{\pi }{3}$
• B. $n\pi \pm \frac{\pi }{3}$
• C. $2n\pi \pm \frac{2\pi }{3}$
• D. $n\pi \pm \frac{2\pi }{3}$

Answer 1

The correct option is A $2n\pi \pm \frac{\pi }{3}$

Let us take $\overrightarrow{V}=x\hat{i}+y\hat{j}$
The vector obtained by rotating the vector $\overrightarrow{V}$ by an angle $\alpha$ in the anticlockwise direction is $(x\cos \alpha -y\sin \alpha )\hat{i}+(y\cos \alpha +x\sin \alpha )\hat{j}$

According to the condition in the question, we have $x\hat{i}+y\hat{j}+(x\cos 2\alpha -y\sin 2\alpha )\hat{i}+(y\cos 2\alpha +x\sin 2\alpha )\hat{j}$

$=(x\cos \alpha -y\sin \alpha )\hat{i}+(y\cos \alpha +x\sin \alpha )\hat{j}$
$\Rightarrow x+x\cos 2\alpha -y\sin 2\alpha =x\cos \alpha -y\sin \alpha$ and
$y+y\cos 2\alpha +x\sin 2\alpha =y\cos \alpha +x\sin \alpha$

Solving we get $x(1-2\sin \frac{\alpha }{2}\sin \frac{3\alpha }{2})=y\left(\sin \frac{\alpha }{2}\cos \frac{3\alpha }{2}\right)$ and
$-x\left(2\sin \frac{\alpha }{2}\cos \frac{3\alpha }{2}\right)=y(1-2\sin \frac{\alpha }{2}\sin \frac{3\alpha }{2})$

Dividing the two equations to get rid of the variables, we get
$-4{\left(\sin \frac{\alpha }{2}\right)}^2{\left(\cos \frac{3\alpha }{2}\right)}^2=1+4{\left(\sin \frac{\alpha }{2}\right)}^2{\left(\sin \frac{3\alpha }{2}\right)}^2-4\sin \frac{\alpha }{2}\sin \frac{3\alpha }{2}$

Solving ahead, we get a quadratic equation $\Rightarrow {\cos }^2\alpha -\cos \alpha +0.25=0$
$\Rightarrow \cos \alpha =0.5$

Hence option $A$.