Let $\overline{\mathrm{u}}, \overline{\mathrm{v}}$ and $\overline{\mathrm{w}}$ be the vectors such that…
Let $\overline{\mathrm{u}}, \overline{\mathrm{v}}$ and $\overline{\mathrm{w}}$ be the vectors such that $|\overline{\mathrm{u}}|=1$; $|\overline{\mathrm{v}}|=2 ;|\overline{\mathrm{w}}|=3$. If the projection of $\overline{\mathrm{v}}$ along $\overline{\mathrm{u}}$ is equal to that of $\bar{w}$ along $\bar{u}$ and $\bar{v}, \bar{w}$ are perpendicular to each other, then $|\bar{u}-\bar{v}+\bar{w}|$ is equal to
$2$
$\sqrt 7$
$\sqrt {14}$
$14$
Solution
Projection of $\overline{\mathrm{v}}$ along $\overline{\mathrm{u}}=$ Projection of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$
$\begin{aligned}
& \Rightarrow \frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overline{\mathrm{u}}|} \\
& \Rightarrow \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}... (i)
\end{aligned}$
Also, $\bar{v}$ and $\bar{w}$ are perpendicular to each other.
$\therefore \quad \overline{\mathrm{v}} \cdot \overline{\mathrm{w}}=0... (ii)$
Now, $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|^2=|\overline{\mathrm{u}}|^2+|\overline{\mathrm{v}}|^2+|\overline{\mathrm{w}}|^2-2(\overline{\mathrm{u}} \cdot \overline{\mathrm{v}})$
$-2(\overline{\mathrm{v}} \cdot \overline{\mathrm{w}})+2(\overline{\mathrm{u}} \cdot \overline{\mathrm{w}})$
$\Rightarrow|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|^2=1+4+9$
$\Rightarrow|\bar{u}-\bar{v}+\bar{w}|=\sqrt{14}$