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Q. 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 _____

JEE MainJEE Main 2023Vector Algebra

Solution:

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