Question

If $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k},\, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, $ and $\overrightarrow{c}$ is a unit vector perpendicular to the vector $\overrightarrow{a}$ and coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$ , then a unit vector $\overrightarrow{d}$ perpendicular, to both $\overrightarrow{a}$ and $\overrightarrow{c}$ is

• A. $\frac{1}{\sqrt6}(2\hat{i}-\hat{j}+\hat{k})$
• B. $\frac{1}{\sqrt2}(\hat{i}+\hat{j})$
• C. $\frac{1}{\sqrt2}(\hat{j}+\hat{k})$
• D. $\frac{1}{\sqrt2}(\hat{i}+\hat{k})$

Answer 1

Answer: (C)

Let $\vec{c}=x\,\vec{a}+y\,\vec{b}$
$=x\left(\hat{i}+\hat{j}-\hat{k}\right)+y\left(\hat{i}-\hat{j}+\hat{k}\right)$
$=\left(x+y\right)\hat{i}+\left(x-y\right)\hat{j}-\left(x-y\right)\hat{k}$
Since $\vec{c}$ is perpendicular to $\vec{a}$
$\therefore x + y + x- y + x - y = 0$
$\Rightarrow 3x=y$
Since $\left|\vec{c}\right|=1$
$\therefore \left(x+y\right)^{2}+\left(x-y\right)^{2}+\left(x-y\right)^{2}=1$
$\therefore \left(4x\right)^{2}+\left(-2x\right)^{2}+\left(-2x\right)^{2}=1$ or $24x^{2}=1$
$\therefore x=\frac{1}{2\sqrt{6}}$
$\therefore \vec{c}=4x\,\hat{i}-2x\,\hat{j}+2x\,\hat{k}$
$=\frac{2}{\sqrt{6}}\,\hat{i}-\frac{1}{\sqrt{6}}\,\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$
i.e., $\vec{c}=\frac{1}{\sqrt{6}}\left(2\,\hat{i}-\hat{j}+\hat{k}\right)$
Let $\vec{d}=d_{1}\,\hat{i}+d_{2}\,\hat{j}+d_{3}\,\hat{k}$
Since $\vec{d}$ is $\bot$ to a $\vec{a}$ and $\vec{c}$
$\therefore \vec{a}\cdot\vec{a}=0$
$\Rightarrow d_{1}+d_{2}-d_{3}=0$
and $\vec{d}\cdot\vec{c}=0$
$\Rightarrow 2d_{1}-d_{2}+d_{3}=0$
$\Rightarrow 3d_{1}=0$
$\Rightarrow d_{1}=0$
$\therefore d_{2}=d_{3}$
Since $\left|\vec{d}\right|=1$
$\therefore d^{2}_{1}+d^{2}_{2}+d^{2}_{3}=1$
$\therefore d_{2}=d_{3}=\frac{1}{\sqrt{2}}$.
Hence $\vec{d}=\frac{1}{\sqrt{2}} \left(\hat{j}+\hat{k}\right)$