Let a,b,c be three vectors such that |a|=√31,4|b|=|c|=2 and 2(a b)=3(c a).

Question

Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that

$|\vec{a}|=\sqrt{31},4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a}	imes \vec{b})=3(\vec{c}	imes \vec{a})$.

If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi }{3}$, then ${(\frac{\vec{a}	imes \vec{c}}{\vec{a}\cdot \vec{b}})}^{2}$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$2(\vec{a}	imes \vec{b})=3(\vec{c}	imes \vec{a})$

$\vec{a}	imes (2\vec{b}+3\vec{c})=0$

$\begin{gathered} \vec{a}=\lambda (2\vec{b}+3\vec{c}) \ |\vec{a}{|}^{2}={\lambda }^{2}(4|b{|}^{2}+9|c{|}^{2}+12\vec{b}\cdot \vec{c}) \ 31=31{\lambda }^{2} \ \lambda =\pm 1 \ \vec{a}=\pm (2\vec{b}+3\vec{c}) \ \frac{|\vec{a}	imes \vec{c}|}{|\vec{a}\cdot \vec{b}|}=\frac{2|\vec{b}	imes \vec{c}|}{2\vec{b}\cdot \vec{b}+3\vec{c}\cdot \vec{b}} \ |\vec{b}	imes \vec{c}{|}^{2}=\frac{1}{4}\cdot 4-{(1-\frac{1}{2})}^{2} \ =\frac{3}{4} \ 	herefore \frac{|\vec{a}	imes \vec{c}|}{|\vec{a}\cdot \vec{b}|}=\frac{\sqrt{3}}{2\cdot \frac{1}{4}-\frac{3}{2}}=\frac{\sqrt{3}}{-1} \ {(\frac{|\vec{a}	imes \vec{c}|}{|\vec{a}\cdot \vec{b}|})}^{2}=3 \end{gathered}$

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