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. $\frac{1}{\sqrt{6}}(2\hat{i}-\hat{j}+\hat{k})$
• B. $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
• C. $\frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$
• D. $\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})$

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 $\perp$ 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 $\left|\vec{d}\right|=1$
$\therefore d_1^2+d_2^2+d_3^2=1$
$\therefore d_2=d_3=\frac{1}{\sqrt{2}}$.
Hence $\vec{d}=\frac{1}{\sqrt{2}}\left(\hat{j}+\hat{k}\right)$