Question

Let $\overline{a}=\hat{i}-\hat{j},\overline{b}=\hat{j}-\hat{k}and\overline{c}=\hat{k}-\hat{i}$ is a unit vector such that $\overline{a}\cdot \overline{d}=0$ and $\left[\overline{b}\overrightarrow{c}\overline{d}\right]=0$ then $\overline{d}$ is

• A. $\pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{6}}$
• B. $\pm \frac{\hat{i}+\hat{j}-\widehat{2k}}{\sqrt{6}}$
• C. $\pm \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$
• D. $\frac{\hat{i}-\hat{j}+\widehat{2k}}{\sqrt{6}}$

Answer 1

Answer: (B)

$\pm \frac{\hat{i}+\hat{j}-\widehat{2k}}{\sqrt{6}}$

Let, $\overline{d}=x\hat{i}+y\hat{j}+z\hat{k}$

$\therefore \overrightarrow{a}\cdot \overline{d}=0givesx-y=0$............(i)

and $\left[\overline{b}\overrightarrow{c}\overline{d}\right]=0$ gives X + Y + Z = 0............(ii)

$\therefore x=yandz=-2x$

and $x^2+y^2+z^2=1$......(iii)

$\therefore x=\pm \frac{1}{\sqrt{6}}$; $y=\pm \frac{1}{\sqrt{6}}$; $z=\mp \frac{2}{\sqrt{6}}$

$\therefore \overline{d}=\pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}$