Question

Let $\overrightarrow{u}=\hat{i}+\hat{j},\overrightarrow{v}=\hat{i}-\hat{j}and\overrightarrow{w}=\hat{i}+2\hat{j}+3\hat{k}$. If $\hat{n}$ is a unit vector such that $\overrightarrow{u}.\hat{n}=0and\overrightarrow{v}.\hat{n}=0$ then $\left|\overrightarrow{w}.\hat{n}\right|$ is equal to

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

Answer: (C)

3

$\because \stackrel{\wedge }{n}$ is perpendicular to both

$\overline{u}and\overline{v}⟹\stackrel{\wedge }{n}=\frac{\overline{u}\times \overline{v}}{|\overline{u}\times \overline{v}|}⟹\overline{u}\times \overline{v}=\left|\begin{array}{ccc}\stackrel{\wedge }{i}& \stackrel{\wedge }{j}& \stackrel{\wedge }{k}\\ 1& 1& 0\\ 1& -1& 0\end{array}\right|⟹\stackrel{\wedge }{i}\left(0\right)-\stackrel{\wedge }{j}\left(0\right)+\stackrel{\wedge }{k}(-1-1)⟹-2\stackrel{\wedge }{k}\therefore \stackrel{\wedge }{n}=\frac{-2\stackrel{\wedge }{k}}{2}=-\stackrel{\wedge }{k}\overline{w}.\stackrel{\wedge }{n}=(\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+3\stackrel{\wedge }{k}).(-\stackrel{\wedge }{k})=-3⟹|\overline{w}.\stackrel{\wedge }{n}|=3$