overline{u}, overline{v}, overline{w} are thr | vedclass

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(C) Given: $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. According to the condition,the projection of $\overline{v}$ along $\overline{u}$ is

Vedclass · Accepted Answer

$\sqrt{14}$

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(C) Given: $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. According to the condition,the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$. $\frac{\overline{v} \cdot \overline{u}}{|\overline{u}|} = \frac{\overline{w} \cdot \overline{u}}{|\overline{u}|}$ $\implies \overline{v} \cdot \overline{u} = \overline{w} \cdot \overline{u}$ $\implies (\overline{w} - \overline{v}) \cdot \overline{u} = 0$. Now,consider $|\overline{u} - \overline{v} + \overline{w}|^2 = |\overline{u} + (\overline{w} - \overline{v})|^2$. $= |\overline{u}|^2 + |\overline{w} - \overline{v}|^2 + 2\overline{u} \cdot (\overline{w} - \overline{v})$. Since $(\overline{w} - \overline{v}) \cdot \overline{u} = 0$,the last term is $0$. $= |\overline{u}|^2 + |\overline{w}|^2 + |\overline{v}|^2 - 2(\overline{w} \cdot \overline{v})$. Since $\overline{v}$ and $\overline{w}$ are perpendicular,$\overline{w} \cdot \overline{v} = 0$. $= (1)^2 + (3)^2 + (2)^2 - 0 = 1 + 9 + 4 = 14$. Therefore,$|\overline{u} - \overline{v} + \overline{w}| = \sqrt{14}$.

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