If $\vec a,\vec b$ are unit vectors such that $|\vec a+\vec b|=1$ and $|\vec a-\vec b|=\sqrt{3}$, then $|3\vec a+2\vec b|=$
Answer & explanation
Correct answer: option 3
Let θ be the angle between a and b. Then,
$\tan\frac{θ}{2}=\frac{|\vec a-\vec b|}{|\vec a+\vec b|}⇒\tan\frac{θ}{2}=\sqrt{3}⇒θ=120°$
$∴\vec a.\vec b=|\vec a||\vec b|\cos θ=\cos 120°=-\frac{1}{2}$
Now,
$|3\vec a+2\vec b|^2=9|\vec a|^2+4|\vec b|^2+12(\vec a.\vec b)$
$⇒|3\vec a+2\vec b|^2=9+4+12×(-\frac{1}{2})=7$
$⇒|3\vec a+2\vec b|=\sqrt{7}$