If the magnitudes of bar{a},bar{b} and bar{a} | vedclass

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(D) Given that $|\bar{a}| = 3$,$|\bar{b}| = 4$,and $|\bar{a}+\bar{b}| = 5$. We know that $|\bar{a}+\bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + 2(\bar{a}

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$5$

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(D) Given that $|\bar{a}| = 3$,$|\bar{b}| = 4$,and $|\bar{a}+\bar{b}| = 5$. We know that $|\bar{a}+\bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + 2(\bar{a} \cdot \bar{b})$. Substituting the values: $5^2 = 3^2 + 4^2 + 2(\bar{a} \cdot \bar{b})$. $25 = 9 + 16 + 2(\bar{a} \cdot \bar{b}) \implies 25 = 25 + 2(\bar{a} \cdot \bar{b}) \implies \bar{a} \cdot \bar{b} = 0$. This implies that $\bar{a}$ and $\bar{b}$ are perpendicular to each other. Now,we need to find $|\bar{a}-\bar{b}|$. $|\bar{a}-\bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 - 2(\bar{a} \cdot \bar{b})$. $|\bar{a}-\bar{b}|^2 = 3^2 + 4^2 - 2(0) = 9 + 16 = 25$. Therefore,$|\bar{a}-\bar{b}| = \sqrt{25} = 5$.

If the magnitudes of $\bar{a}$,$\bar{b}$ and $\bar{a}+\bar{b}$ are respectively $3$,$4$ and $5$,then the magnitude of $\bar{a}-\bar{b}$ is

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