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 (integer)
14
Solution
$\because|\vec{a}+\dot{b}|^{2}=|\vec{a}|^{2}+2|b|^{2}$
<br/><br/>or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
<br/><br/>$\therefore|\vec{b}|^{2}=6$
<br/><br/>Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
<br/><br/>$75=|\vec{a}|^{2} \cdot 6-9$
<br/><br/>$\therefore|\vec{a}|^{2}=14$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Cross Product
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