Let a= i+ j+ k, c= j- k, and b satisfy a b= c, a b=3.

Question

Let $\vec a=\hat i+\hat j+\hat k$, $\vec c=\hat j-\hat k$, and $\vec b$ satisfy $\vec a	imes\vec b=\vec c$, $\vec a\cdot\vec b=3$. Then $|\vec b|$ equals:

Accepted Answer

$|\vec a|=\sqrt3$, $|\vec c|=\sqrt2$. From $\vec a\times\vec b=\vec c$: $|\vec a||\vec b|\sin\theta=\sqrt2$, and $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=3$, so $\tan\theta=\dfrac{\sqrt2}3$, giving $\sin\theta=\dfrac{\sqrt2}{\sqrt{11}}$.

$$\sqrt3|\vec b|\cdot\frac{\sqrt2}{\sqrt{11}} = \sqrt2 \implies |\vec b| = \frac{\sqrt{11}}{\sqrt3} = \sqrt{\frac{11}3}$$

Solutions · 1