Statement (A): If |vec{a}| = 2, |vec{b}| = 3, | vedclass

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(A) Given $|2\vec{a} - \vec{b}| = 5$.Squaring both sides: $|2\vec{a} - \vec{b}|^2 = 25$.$4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b}) = 25$.S

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$A$ and $R$ are both true and $R$ is the correct explanation of $A$.

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(A) Given $|2\vec{a} - \vec{b}| = 5$.Squaring both sides: $|2\vec{a} - \vec{b}|^2 = 25$.$4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b}) = 25$.Substituting $|\vec{a}| = 2$ and $|\vec{b}| = 3$: $4(4) + 9 - 4(\vec{a} \cdot \vec{b}) = 25$.$16 + 9 - 4(\vec{a} \cdot \vec{b}) = 25 \implies 25 - 4(\vec{a} \cdot \vec{b}) = 25 \implies \vec{a} \cdot \vec{b} = 0$.Now,$|2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 16 + 9 + 0 = 25$.Thus,$|2\vec{a} + \vec{b}| = 5$. So,Statement $A$ is true.For Reason $R$,$|\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$ implies $|\vec{p} - \vec{q}|^2 = |\vec{p} + \vec{q}|^2$.$|\vec{p}|^2 + |\vec{q}|^2 - 2(\vec{p} \cdot \vec{q}) = |\vec{p}|^2 + |\vec{q}|^2 + 2(\vec{p} \cdot \vec{q}) \implies 4(\vec{p} \cdot \vec{q}) = 0 \implies \vec{p} \cdot \vec{q} = 0$.This means $\vec{p} \perp \vec{q}$. Since the reason correctly explains the condition for the equality of magnitudes of sum and difference vectors,$R$ is true and is the correct explanation of $A$.

Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.
Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$

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Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$