Question

If the measure of the angle between the vectors $\sqrt{3}\hat{i}+\hat{j}$ and a$\hat{i}+\sqrt{3}\hat{j}$ is $\frac{\pi }{3}$, then find a.

Answer 1

$\overline{x}=\sqrt{3}\cdot \hat{i}+\hat{j}=(\sqrt{3},1)$
$\overline{y}=a\hat{i}+\sqrt{3}\hat{j}=(a,\sqrt{3})$
$\therefore$ Here $\overline{x}$^$\overline{y}=\frac{\pi }{3}$
$\therefore \cos (\overline{x}$^$\overline{y})=\cos \left(\frac{\pi }{3}\right)$
$\therefore \frac{\overline{x}\cdot \overline{y}}{|\overline{x}||\overline{y}|}=\frac{1}{2}$
$\therefore \frac{(\sqrt{3},1)\cdot (a,\sqrt{3})}{\sqrt{3+1}\cdot \sqrt{a^2+3}}=\frac{1}{2}$
$\frac{\sqrt{3}a+\sqrt{3}}{2\sqrt{a^2+3}}=\frac{1}{2}$
$\therefore \sqrt{3}(a+1)=\sqrt{a^2+3}$ ..$\left(1\right)$
Now squaring on both sides.
$\therefore 3(a^2+2a+1)a^2+3$
$\therefore 2a^2+6a=0$
$2a(a+3)=0$
$\therefore a=0$ or $a=-3$
Here $a=-3$ does not satisfy result $\left(1\right)$
$\therefore a=0$.