If |vec{a}| = 2,|vec{b}| = 3 and |2vec{a} - v | vedclass

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(C) Given $|2\vec{a} - \vec{b}| = 5$. Squaring both sides,we get $|2\vec{a} - \vec{b}|^2 = 25$. Using the property $|\vec{u} - \vec{v}|^2 = |\vec{u}|^

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$5$

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(C) Given $|2\vec{a} - \vec{b}| = 5$. Squaring both sides,we get $|2\vec{a} - \vec{b}|^2 = 25$. Using the property $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$,we have: $4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b}) = 25$. Substituting $|\vec{a}| = 2$ and $|\vec{b}| = 3$: $4(2)^2 + (3)^2 - 4(\vec{a} \cdot \vec{b}) = 25$. $16 + 9 - 4(\vec{a} \cdot \vec{b}) = 25$. $25 - 4(\vec{a} \cdot \vec{b}) = 25 \implies 4(\vec{a} \cdot \vec{b}) = 0 \implies \vec{a} \cdot \vec{b} = 0$. Now,we need to find $|2\vec{a} + \vec{b}|$. $|2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b})$. Substituting the values: $|2\vec{a} + \vec{b}|^2 = 4(4) + 9 + 4(0) = 16 + 9 = 25$. Therefore,$|2\vec{a} + \vec{b}| = \sqrt{25} = 5$.

If $|\vec{a}| = 2$,$|\vec{b}| = 3$ and $|2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}|$ equals

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