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 _____

Hint:

In vector problems involving cross and dot products, ensure to calculate the magnitude of the cross product using the formula \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \), and the dot product as \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).

Answer 1

Correct Answer: 3

\(2(\vec{a}\times\vec{b})=3(\vec{c}\times\vec{a})\)
\(\vec{a}\times(2\vec{b}+3\vec{c})=0\)
\(\vec{a}=\lambda (2\vec{b}+3\vec{c})\)
\(|\vec{a}|^{2}=\lambda ^{2}|2\vec{b}+3\vec{c}|^{2}\)
\(|\vec{a}|^{2}=\lambda ^{2}(4|\vec{b}|^{2}+9|\vec{c}|^{2}+12\vec{b}.\, \vec{c})\)
\(31=31\lambda ^{2}\Rightarrow \lambda =\pm 1\)
\(\vec{a}=\pm (2\vec{b}+3\vec{c})\)
\( \frac{|\vec{a}\times\vec{c}|}{|\vec{a}.\vec{b}|}=\frac{2|\vec{b}\times\vec{c}|}{2\vec{b}.\vec{b}+3\vec{c}.\vec{b}}\)
\(|\vec{b}\times\vec{c}|=|\vec{b}|^{2}|\vec{c}|^{2}-(\vec{b}.\vec{c})^{2}=\frac{3}{4}\)
\(\frac{ |\vec{a}\times\vec{c}|}{|\vec{a}.\vec{b}|}=\frac{2\times\frac{\sqrt{3}}{2}}{2.\frac{1}{4}-\frac{3}{2}}=-\sqrt{3}\)
\((\frac{\vec{a}\times\vec{c}}{\vec{a}.\vec{b}})^{2}=3\)
So, the correct answer is 3.