Question

Let $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ be such that $|\overrightarrow{u}|=1,|\overrightarrow{v}|=2,|\overrightarrow{w}|=3$. If the projection of $\overrightarrow{v}$ along $\overrightarrow{u}$ is equal to that of $\overrightarrow{w}$ along $\overrightarrow{u}$ and $\overrightarrow{v},\overrightarrow{w}$ are perpendicular to each other, then $|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}|$ equals

• A. $2$
• B. $\sqrt{7}$
• C. $\sqrt{14}$
• D. $14$

Answer 1

Answer: (C)

$\sqrt{14}$

Projection of $\overline{v}$ along $\overline{u}=\frac{\overline{v}\cdot \overline{u}}{|\overline{u}|}$ -----(1)
Projection of $\overline{w}$ along $\overline{u}=\frac{\overline{w}\cdot \overline{u}}{|\overline{u}|}$ -----(2)
$\overline{v}\cdot \overline{w}=0$
$|\overline{u}-\overline{v}+\overline{w}|=\sqrt{|\overline{u}{|}^2+|\overline{v}{|}^2+|\overline{w}{|}^2-2\overline{u}\cdot \overline{v}-2\overline{w}\cdot \overline{v}+2\overline{u}\cdot \overline{w}}$
$=\sqrt{1+4+9}$
$=\sqrt{14}$