JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 20)
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 __________.
$|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2$ 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
3
Explanation
$2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$
$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$
$$ \begin{aligned} & \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\ & |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\ & 31=31 \lambda^{2} \\\\ & \lambda=\pm 1 \\\\ & \vec{a}=\pm(2 \vec{b}+3 \vec{c}) \\\\ & \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}} \\\\ & |\vec{b} \times \vec{c}|^{2}=\frac{1}{4} \cdot 4-\left(1-\frac{1}{2}\right)^{2} \\\\ & =\frac{3}{4} \\\\ & \therefore \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{\sqrt{3}}{2 \cdot \frac{1}{4}-\frac{3}{2}}=\frac{\sqrt{3}}{-1} \\\\ & \left(\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}\right)^{2}=3 \end{aligned} $$
$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$
$$ \begin{aligned} & \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\ & |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\ & 31=31 \lambda^{2} \\\\ & \lambda=\pm 1 \\\\ & \vec{a}=\pm(2 \vec{b}+3 \vec{c}) \\\\ & \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}} \\\\ & |\vec{b} \times \vec{c}|^{2}=\frac{1}{4} \cdot 4-\left(1-\frac{1}{2}\right)^{2} \\\\ & =\frac{3}{4} \\\\ & \therefore \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{\sqrt{3}}{2 \cdot \frac{1}{4}-\frac{3}{2}}=\frac{\sqrt{3}}{-1} \\\\ & \left(\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}\right)^{2}=3 \end{aligned} $$
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