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, $x^{2}+y^{2}+z^{2}=1 ......$(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 \begin{vmatrix} 0 & 1 & -1 \\-1 & 0 & 1 \\x & y & z\end{vmatrix}=0 \Rightarrow x+y+z=0 $
$\Rightarrow 2x + z = 0 $ [from Eq. (ii)] ...(iii)
From Eqs. (i), (ii) and (iii),
$x^{2}+x^{2}+4 x^{2} =1 \Rightarrow x=\pm \frac{1}{\sqrt{6}}$
$ \therefore \overrightarrow{ d } =\pm \frac{1}{\sqrt{6}}(\hat{ i }+\hat{ j }-2 \hat{ k })$