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Let $\bar{u}, \bar{v}, \bar{w}$ be such that $|\bar{u}|=1,|\bar{v}|=2,|\bar{w}|=3$. If the projection $\bar{v}$ along $\bar{u}$ is equal to that of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$ and $\overline{\mathrm{v}}, \overline{\mathrm{w}}$ are perpendicular to each other then $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|$ equals
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The correct answer is:
$\sqrt{14}$
$\sqrt{14}$
Projection of $\bar{v}$ along $\bar{u}$ and $\bar{w}$ along $\bar{u}$ is $\frac{\bar{v} \cdot \bar{u}}{|\bar{u}|}$ and $\frac{\bar{w} \cdot \bar{u}}{|\bar{u}|}$ respectively According to question $\frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} \Rightarrow \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}$. and $\overline{\mathrm{v}} \cdot \overline{\mathrm{w}}=0$
$$
|\bar{u}-\bar{v}+\bar{w}|^2=|\bar{u}|^2+|\bar{v}|^2+|\bar{w}|^2-2 \bar{u} \cdot \bar{v}+2 \bar{u} \cdot \bar{w}-2 \bar{v} \cdot \bar{w}=14 \Rightarrow|\bar{u}-\bar{v}+\bar{w}|=\sqrt{14}
$$
$$
|\bar{u}-\bar{v}+\bar{w}|^2=|\bar{u}|^2+|\bar{v}|^2+|\bar{w}|^2-2 \bar{u} \cdot \bar{v}+2 \bar{u} \cdot \bar{w}-2 \bar{v} \cdot \bar{w}=14 \Rightarrow|\bar{u}-\bar{v}+\bar{w}|=\sqrt{14}
$$
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