$\overline{\mathrm{u}}, \overline{\mathrm{v}}, \overline{\mathrm{w}}$ are three vectors such that…
$\overline{\mathrm{u}}, \overline{\mathrm{v}}, \overline{\mathrm{w}}$ are three vectors such that $|\overline{\mathrm{u}}|=1$, $|\vec{v}|=2,|\bar{w}|=3$. If the projection of $\bar{v}$ along $\bar{u}$ is equal to projection of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$ and $\overline{\mathrm{v}}, \overline{\mathrm{w}}$ are perpendicular to each other, then $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|=$
4
$\sqrt{7}$
$\sqrt{14}$
2
Solution
$|\overline{\mathrm{u}}|=1,|\overline{\mathrm{v}}|=2,|\overline{\mathrm{w}}|=3$
According to the given condition,
(Projection of $\bar{v}$ along $\bar{u}$)$
=(\text { Projection of } \bar{w} \text { along } \bar{u})$
$\begin{array}{ll}
\therefore & \frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} \\
\therefore & \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}} \\
\therefore & (\overline{\mathrm{w}}-\overline{\mathrm{v}}) \cdot \overline{\mathrm{u}}=0
\end{array}$
Now consider, $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|=\sqrt{|\overline{\mathrm{u}}+\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2}$
$\begin{aligned}
& =\sqrt{|\overline{\mathrm{u}}|^2+|\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2+2 \overline{\mathrm{u}} \cdot(\overline{\mathrm{w}}-\overline{\mathrm{v}})} \\
& =\sqrt{(1)^2+|\overline{\mathrm{w}}-\overline{\mathrm{v}}|^2+0} \\
& =\sqrt{1+|\overline{\mathrm{w}}|^2+|\overline{\mathrm{v}}|^2-2(\overline{\mathrm{w}} \cdot \overline{\mathrm{v}})} \\
& =\sqrt{1+9+4+0}
\end{aligned}$
$\ldots[\because \overline{\mathrm{w}}$ and $\overline{\mathrm{v}}$ are perpendicular $]$
$=\sqrt{14}$