Question

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

• A. $\pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}$
• B. $\pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}$
• C. $\pm \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$
• D. $\pm \hat{k}$

Answer 1

Answer: (A)

Let $\overrightarrow{d}=x\hat{i}+y\hat{j}+z\hat{k}$
where, $x^2+y^2+z^2=\mathrm{1......}$(i)
$[\because \vec{d}$ being unit vector $]$
Since, $\overrightarrow{a}\cdot \overrightarrow{d}=0$
$\Rightarrow x-y=0\Rightarrow x=y......$(ii)
Also, $[\overrightarrow{b}\overrightarrow{c}\vec{d}]=0$
$\Rightarrow \left|\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ x& y& z\end{array}\right|=0\Rightarrow x+y+z=0$
$\Rightarrow 2x+z=0$ [from Eq. (ii)] ...(iii)
From Eqs. (i), (ii) and (iii),
$x^2+x^2+4x^2=1\Rightarrow x=\pm \frac{1}{\sqrt{6}}$
$\therefore \overrightarrow{d}=\pm \frac{1}{\sqrt{6}}(\hat{i}+\hat{j}-2\hat{k})$