Question

Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{31},4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a}\times \vec{b})=3(\vec{c}\times \vec{a})$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi }{3}$, then ${\left(\frac{\vec{a}\times \vec{c}}{\vec{a}\cdot \vec{b}}\right)}^2$ is equal to _____

Answer 1

Answer: 3

$2(\vec{a}\times \vec{b})=3(\vec{c}\times \vec{a})$
$\vec{a}\times (2\vec{b}+3\vec{c})=0$
$\vec{a}=\lambda (2\vec{b}+3\vec{c})$
$|\vec{a}{|}^2={\lambda }^2|2\vec{b}+3\vec{c}{|}^2$
$|\vec{a}{|}^2={\lambda }^2\left(4|\vec{b}{|}^2+9|\vec{c}{|}^2+12\vec{b}\cdot \vec{c}\right)$
$31=31{\lambda }^2\Rightarrow \lambda =\pm 1$
$\vec{a}=\pm (2\vec{b}+3\vec{c})$
$\frac{|\vec{a}\times \vec{c}|}{|\vec{a}\cdot \vec{b}|}=\frac{2|\vec{b}\times \vec{c}|}{2\vec{b}\cdot \vec{b}+3\vec{c}\cdot \vec{b}}$
$|\vec{b}\times \vec{c}{|}^2=|\vec{b}{|}^2|\vec{c}{|}^2-(\vec{b}\cdot \vec{c}{)}^2=\frac{3}{4}$
$\frac{|\vec{a}\times \vec{c}|}{|\vec{a}\cdot \vec{b}|}=\frac{2\times \frac{\sqrt{3}}{2}}{2\cdot \frac{1}{4}-\frac{3}{2}}=-\sqrt{3}$
${\left(\frac{\vec{a}\times \vec{c}}{\vec{a}\cdot \vec{b}}\right)}^2=3$