If bar{a}, bar{b} and bar{c} are vectors such | vedclass

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(B) Given: $|\bar{a}| = 1$,$|\bar{b}| = 2$,$|\bar{c}| = 3$. Since $\bar{b} \perp \bar{c}$,we have $\bar{b} \cdot \bar{c} = 0$. The projection of $\bar

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$\sqrt{14}$

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(B) Given: $|\bar{a}| = 1$,$|\bar{b}| = 2$,$|\bar{c}| = 3$. Since $\bar{b} \perp \bar{c}$,we have $\bar{b} \cdot \bar{c} = 0$. The projection of $\bar{b}$ on $\bar{a}$ is $\frac{\bar{b} \cdot \bar{a}}{|\bar{a}|} = \bar{b} \cdot \bar{a}$ (since $|\bar{a}| = 1$). The projection of $\bar{c}$ on $\bar{a}$ is $\frac{\bar{c} \cdot \bar{a}}{|\bar{a}|} = \bar{c} \cdot \bar{a}$. Given $\bar{b} \cdot \bar{a} = \bar{c} \cdot \bar{a} = k$. Now,$|\bar{a} - \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 - 2(\bar{a} \cdot \bar{b}) + 2(\bar{a} \cdot \bar{c}) - 2(\bar{b} \cdot \bar{c})$. Substituting the values: $|\bar{a} - \bar{b} + \bar{c}|^2 = 1^2 + 2^2 + 3^2 - 2k + 2k - 2(0) = 1 + 4 + 9 = 14$. Therefore,$|\bar{a} - \bar{b} + \bar{c}| = \sqrt{14}$.

If $\bar{a}, \bar{b}$ and $\bar{c}$ are vectors such that $|\bar{a}| = |\frac{\bar{b}}{2}| = |\frac{\bar{c}}{3}| = 1$; $\bar{b}$ and $\bar{c}$ are perpendicular; and the projections of $\bar{b}$ and $\bar{c}$ on $\bar{a}$ are equal,then $|\bar{a} - \bar{b} + \bar{c}| = $

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