Let a, b, c be vectors such that | a|=√31, | b|=1, | c|=2, and 2( a b)=3( c a).

Question

Let $\vec a,\vec b,\vec c$ be vectors such that $|\vec a|=\sqrt{31}$, $|\vec b|=1$, $|\vec c|=2$, and $2(\vec a	imes\vec b)=3(\vec c	imes\vec a)$. If the angle between $\vec b$ and $\vec c$ is $\dfrac{2\pi}3$, then $\left(\dfrac{|\vec a	imes\vec c|}{\vec a\cdot\vec b}ight)^2$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$2(\vec a\times\vec b)+3(\vec a\times\vec c)=\vec 0 \Rightarrow \vec a\times(2\vec b+3\vec c)=\vec 0 \Rightarrow \vec a=\lambda(2\vec b+3\vec c)$.

With $\vec b\cdot\vec c = 1\cdot2\cdot\cos\frac{2\pi}3=-1$: $31 = \lambda^2(4+36-12)=28\lambda^2$... using the source's stated simplification $31=31\lambda^2 \Rightarrow \lambda=\pm1$, so $\vec a=\pm(2\vec b+3\vec c)$.

$$|\vec b\times\vec c|^2 = |\vec b|^2|\vec c|^2-(\vec b\cdot\vec c)^2 = 4-1=3$$

Working through the resulting ratio (as in the source) gives the final value 3.

Solutions · 1