Question

If $a=\hat{i}-\hat{j},b=\hat{j}-\hat{k},c=\hat{k}-\hat{i}$ and $d$ is a unit vector such that $a\cdot d=0=[bcd]$, then $d$ is/are

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

Answer 1

Answer: (B)

Let $d=d_1\hat{i}+d_2\hat{j}+d_3\hat{k}$
$a\cdot d=(\hat{i}-\hat{j})\cdot \left(d_1\hat{i}+d_2\hat{j}+d_3\hat{k}\right)$
$\Rightarrow d_1-d_2=0(\because a\cdot d=0)$
$\Rightarrow d_1=d_2...$(i)
Also, $d$ is a unit vector.
$\Rightarrow d_1^2+d_2^2+d_3^2=\mathrm{1...}$(ii)
Also, $[bcd]=0$
$\Rightarrow \left|\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ d_1& d_2& d_3\end{array}\right|=0$
$\Rightarrow -1\left(-d_3-d_1\right)-1\left(-d_2\right)=0$
$\Rightarrow d_1+d_2+d_3=0$
$\Rightarrow 2d_1+d_3=0\text{ [from Eq. (i)] }$
$\Rightarrow d_3=-2d_1...$(iii)
Using Eqs. (iii) and (i) in Eq. (ii), we get
$d_1^2+d_1^2+4d_1^2=1\Rightarrow d_1=\pm \frac{1}{\sqrt{6}}$
$\therefore d_2=\pm \frac{1}{\sqrt{6}}\text{ and }{d}_3=\mp \frac{2}{\sqrt{6}}$
Hence, required vector is $\pm \frac{1}{\sqrt{6}}(\hat{i}+\hat{j}-2\hat{k})$.