Question

The unit vectors orthogonal to $-i+2j+2k$ and making equal angles with x and y axis are

• A. $\pm \frac{1}{3}\left(2\overline{i}+2\overline{j}-\overline{k}\right)$
• B. $\pm \frac{1}{3}\left(\overline{i}+\overline{j}-\overline{k}\right)$
• C. $\pm \frac{1}{3}\left(2\overline{i}-2\overline{j}-\overline{k}\right)$
• D. $\pm \frac{1}{4}\left(\overline{i}+\overline{j}-\overline{k}\right)$

Answer 1

Answer: (A)

$\pm \frac{1}{3}\left(2\overline{i}+2\overline{j}-\overline{k}\right)$

Let required vector be a$\stackrel{\wedge }{i}+b\stackrel{\wedge }{j}+c\stackrel{\wedge }{k}$

$\because$ It makes equal angle with x and y-axis.

$⟹\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{b}{\sqrt{a^2+b^2+c^2}}⟹a=b$ $\because$It is a unit vector.

$⟹a^2+b^2+c^2=1-\left(i\right)$

and it is orthogonal to $-\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}+2\stackrel{\wedge }{k}$

$⟹-a+2b+2c=0\because a=b⟹-a+2a+2c=0⟹2c=-a⟹c=\frac{-a}{2}$

$Puttingb=a,c=\frac{-a}{2}in\left(i\right):⟹a^2+a^2+\frac{a^2}{4}=1⟹4a^2+4a^2+a^2=4⟹9a^2=4⟹a=\pm \frac{2}{3}\therefore a\stackrel{\wedge }{i}+b\stackrel{\wedge }{j}+c\stackrel{\wedge }{k}=a\stackrel{\wedge }{i}+a\stackrel{\wedge }{j}+\frac{-a}{2}\stackrel{\wedge }{k}=\pm \frac{1}{3}(2\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}-\stackrel{\wedge }{k})$