Question
माना दो सदिशों $\vec{a}$ तथा $\vec{b}$ के लिए $|\vec{a}+\vec{b}|^2=|\vec{a}|^2+2|\vec{b}|^2, \vec{a} \cdot \vec{b}=3 \quad$ तथा $|\vec{a} \times \vec{b}|^2=75$ हैं। तो $|\overrightarrow{ a }|^2$ बराबर है $............$
As $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
$|\vec{b}|^{2}=2 \vec{a} \cdot \vec{b}=6$
$|\vec{a} \times \vec{b}|^{2}=75$
$|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}=75$
$6|\vec{a}|^{2}-9=75 \Rightarrow|\vec{a}|^{2}=14$
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