In $R^2$, if the magnitude of the projection vector of the vector $α\hat i+β\hat j$ on $\sqrt{3}\hat i+\hat j$ is $\sqrt{3}$ and if $α = 2+ \sqrt{3}β$, then possible value(s) of $|α|$, is (are)

1, 2

Let $\vec a = α\hat i+β\hat j$ and $\vec b =\sqrt{3}\hat i+\hat j$. It is given that the magnitude of projection of $\vec a$ on $\vec b$ is $\sqrt{3}$.

$∴\left|\frac{\vec a.\vec b}{|\vec a|}\right|=\sqrt{3}$

$⇒\left|\frac{\sqrt{3}α+β}{2}\right|=\sqrt{3}⇒\sqrt{3}α+β=±2\sqrt{3}$

$⇒\sqrt{3}(2+\sqrt{3}β)+β=±2\sqrt{3}$   $[∵α=2+\sqrt{3}β]$

$⇒2\sqrt{3}+4β=±2\sqrt{3}$

$⇒β=0$ or, $β=-\sqrt{3}$

Substituting these values in $α = 2\sqrt{3}β$, we obtain $α = 2$ or $α = 1$.