JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 21)
Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$$ and $$|\vec{a} \times \vec{b}|=\sqrt{48}$$. Then $$(\vec{a} \cdot \vec{b})^{2}$$ is equal to ___________.
Answer
36
Explanation
$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$
$$ \begin{aligned} & \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\ & \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\ & \Rightarrow (\vec{a} \cdot \vec{b})^{2}=84-48 \\\\ &=36 \end{aligned} $$
$$ \begin{aligned} & \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\ & \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\ & \Rightarrow (\vec{a} \cdot \vec{b})^{2}=84-48 \\\\ &=36 \end{aligned} $$
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