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(C) Given that the projection of $\bar{b}$ on $\bar{a}$ is equal to the projection of $\bar{c}$ on $\bar{a}$,we have $\frac{\bar{b} \cdot \bar{a}}{|\b

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$6$

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(C) Given that the projection of $\bar{b}$ on $\bar{a}$ is equal to the projection of $\bar{c}$ on $\bar{a}$,we have $\frac{\bar{b} \cdot \bar{a}}{|\bar{a}|} = \frac{\bar{c} \cdot \bar{a}}{|\bar{a}|}$.This implies $\bar{b} \cdot \bar{a} = \bar{c} \cdot \bar{a}$,or $(\bar{b} - \bar{c}) \cdot \bar{a} = 0$.Also,$\bar{b}$ is perpendicular to $\bar{c}$,so $\bar{b} \cdot \bar{c} = 0$.We need to find $|\bar{a} + \bar{b} - \bar{c}|$.Let $\bar{v} = \bar{a} + \bar{b} - \bar{c}$. Then $|\bar{v}|^2 = |\bar{a} + (\bar{b} - \bar{c})|^2 = |\bar{a}|^2 + |\bar{b} - \bar{c}|^2 + 2\bar{a} \cdot (\bar{b} - \bar{c})$.Since $(\bar{b} - \bar{c}) \cdot \bar{a} = 0$,the last term is $0$.$|\bar{b} - \bar{c}|^2 = |\bar{b}|^2 + |\bar{c}|^2 - 2\bar{b} \cdot \bar{c} = 4^2 + 4^2 - 0 = 32$.Thus,$|\bar{v}|^2 = |\bar{a}|^2 + 32 = 2^2 + 32 = 4 + 32 = 36$.Therefore,$|\bar{a} + \bar{b} - \bar{c}| = \sqrt{36} = 6$.

The vectors $\bar{a}, \bar{b}$ and $\bar{c}$ are such that $|\bar{a}|=2, |\bar{b}|=4, |\bar{c}|=4$. If the projection of $\bar{b}$ on $\bar{a}$ is equal to the projection of $\bar{c}$ on $\bar{a}$ and $\bar{b}$ is perpendicular to $\bar{c}$,then the value of $|\bar{a}+\bar{b}-\bar{c}|$ is

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