Question

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

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

Answer 1

Answer: (C)

$\sqrt{14}$

Given, $v\cdot u=w\cdot u$ and $v\perp w$ $\Rightarrow v\cdot w=0$
Now, consider
${|u-v+w|}^2={\left|u\right|}^2+{\left|v\right|}^2+{\left|w\right|}^2-2u\cdot v-2w\cdot v+2u\cdot w$
$=1+9+4=14$
$\Rightarrow |u-v+w|=\sqrt{14}$