Question

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

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

Answer 1

Answer: (A)

Given, $|u|=1,|v|=2|w|=3$ $\therefore$ $\frac{v.u}{|v|}=\frac{w.u}{|u|}$ $\Rightarrow$ $v.u=w.u$ and $u.w=0$ Now, $|u-v+w{|}^2=u^2+v^2+w^2-2u.v$ $+2u.w-2v.w$ $=1+4+9-2v.w=14-0=14$ $\therefore$ $|u-v+w|=\sqrt{14}$