JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 20)
Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3$$ and $$|\vec{a} \times \vec{b}|^{2}=75$$. Then $$|\vec{a}|^{2}$$ is equal to __________.
Answer
14
Explanation
$\because|\vec{a}+\dot{b}|^{2}=|\vec{a}|^{2}+2|b|^{2}$
or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
$\therefore|\vec{b}|^{2}=6$
Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
$$ 75=|\vec{a}|^{2} \cdot 6-9 $$
$\therefore|\vec{a}|^{2}=14$
or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
$\therefore|\vec{b}|^{2}=6$
Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
$$ 75=|\vec{a}|^{2} \cdot 6-9 $$
$\therefore|\vec{a}|^{2}=14$
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