Question

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

• A. $\pm \left(\frac{\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}}{\sqrt{6}}\right)$
• B. $\pm \left(\frac{\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}}{\sqrt{3}}\right)$
• C. $\pm \left(\frac{\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}}{\sqrt{3}}\right)$
• D. $\pm \overrightarrow{k}$

Answer 1

The correct option is A $\pm \left(\frac{\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}}{\sqrt{6}}\right)$

Let $\overrightarrow{d}=x\hat{i}+y\hat{j}+z\hat{k}$
$\overrightarrow{a}\cdot \overrightarrow{d}=0\Rightarrow x=y\cdots \left(i\right)$
$\left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}\right]=\left|\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ x& y& z\end{array}\right|=-(-z-x)-(-y)$
$\Rightarrow x+y+z=0$
Using $\left(i\right),$ we get
$z=-2x\cdots \left(ii\right)$
Since $\overrightarrow{d}$ is unit vector
$\therefore x^2+y^2+z^2=1\cdots \left(iii\right)$
Using $\left(i\right),\left(ii\right)$ and $\left(iii\right),$ we get
$x=\pm \frac{1}{\sqrt{6}}$
$\therefore \overrightarrow{d}=\pm \left(\frac{\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}}{\sqrt{6}}\right)$