Let a, b and c be non-zero vectors such that a( b c)=13| b|| c| a.

Question

Let $\vec a,\vec b$ and $\vec c$ be non-zero vectors such that $\vec a	imes(\vec b	imes\vec c)=\dfrac13|\vec b||\vec c|\vec a$. If $	heta$ is the acute angle between the vectors $\vec b$ and $\vec c$, then $\sin	heta$ equals:

Accepted Answer

$$\vec a\times(\vec b\times\vec c) = (\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c = \frac13|\vec b||\vec c|\vec a$$

Since $\vec a,\vec b$ are non-collinear, comparing gives $\vec a\cdot\vec c=0$ and $-(\vec a\cdot\vec b)=\dfrac13|\vec b||\vec c|\cdot(\text{unit vector factor})$; working through the coplanar identity as in the analogous Q217 gives $\cos\theta=-\dfrac13$ so:

$$\sin\theta=\sqrt{1-\frac19}=\frac{2\sqrt2}3$$

Solutions · 1