Let vec{a} and vec{b} be two vectors such tha | vedclass

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(A) Given $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$ and $\vec{a} \cdot \vec{b}=3$. Expanding the left side,we have $|\vec{a}|^{2}+|\vec{b}|

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) Given $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$ and $\vec{a} \cdot \vec{b}=3$. Expanding the left side,we have $|\vec{a}|^{2}+|\vec{b}|^{2}+2(\vec{a} \cdot \vec{b})=|\vec{a}|^{2}+2|\vec{b}|^{2}$. Subtracting $|\vec{a}|^{2}$ from both sides,we get $|\vec{b}|^{2}=2(\vec{a} \cdot \vec{b})$. Substituting $\vec{a} \cdot \vec{b}=3$,we find $|\vec{b}|^{2}=2(3)=6$. Using the Lagrange identity $|\vec{a} 	imes \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}=75$. Substituting the known values,$|\vec{a}|^{2}(6)-(3)^{2}=75$. $6|\vec{a}|^{2}-9=75$. $6|\vec{a}|^{2}=84$. $|\vec{a}|^{2}=14$.

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:

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