Let a, b satisfy |2 a+3 b|=|3 a+ b|, angle between them 60^.

Question

Let $\vec a,\vec b$ satisfy $|2\vec a+3\vec b|=|3\vec a+\vec b|$, angle between them $60^{\circ}$. If $\dfrac18\vec a$ is a unit vector, $|\vec b|$ is:

Accepted Answer

With $\vec a\cdot\vec b=8|\vec b|\cos60^{\circ}=4|\vec b|$: $5(64)-6(4|\vec b|)-8|\vec b|^2=0 \implies 320-24|\vec b|-8|\vec b|^2=0$
$$|\vec b|^2+3|\vec b|-40=0 \implies (|\vec b|+8)(|\vec b|-5)=0 \implies |\vec b|=5$$

Solutions · 1