Question

A unit vector which is perpendicular to the vector $2\hat{i}-\hat{j}+2\hat{k}$ and is coplanar with the vectors $\hat{i}+\hat{j}-\hat{k}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is

• A. $\frac{2\hat{j}-\hat{k}}{\sqrt{5}}$
• B. $\frac{3\hat{i}+2\hat{j}-2\hat{k}}{\sqrt{17}}$
• C. $\frac{3\hat{i}+2\hat{j}+2\hat{k}}{\sqrt{17}}$
• D. $\frac{3\hat{i}+2\hat{j}-2\hat{k}}{3}$

Answer 1

Answer: (B)

Let $x\hat{i}+y\hat{j}+z\hat{k}$ be the required unit vector.
Since $\hat{a}$ is perpendicular to $\left(2\hat{i}-\hat{j}+2\hat{k}\right)$.
$\therefore 2x-y+2z=0...........\left(i\right)$
Since vector $x\hat{i}+y\hat{j}+z\hat{k}$ is coplanar with the vector $\hat{i}+\hat{j}-\hat{k}and2\hat{i}+2\hat{j}-\hat{k}$.
$\therefore x\hat{i}+y\hat{j}+z\hat{k}$
$=p\left(\hat{i}+\hat{j}-\hat{k}\right)+q\left(2\hat{i}+2\hat{j}+\hat{k}\right)$,
where p and q are some scalars.
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}$
$=\left(p+2q\right)\hat{i}+\left(p+2q\right)\hat{j}-\left(p+q\right)\hat{k}$
$\Rightarrow x=p+2q,y=p+2q,z=-p-q$
Now from equation $\left(i\right)$,
$2p+4q-p-2q-2p-2q=0$
$\Rightarrow -p=0\Rightarrow p=0$
$\therefore x=2q,y=2q,z=-q$
Since vector $x\hat{i}+y\hat{j}+z\hat{k}$ is a unit vector, therefore
$\left|x\hat{i}+y\hat{j}+z\hat{k}\right|=1$
$\Rightarrow \sqrt{x^2+y^2+z^2}=1$
$\Rightarrow x^2+y^2+z^2=1$
$\Rightarrow 4q^2+4q^2+q^2=1$
$\Rightarrow 9q^2=1\Rightarrow q=\pm \frac{1}{3}$
When $q=\frac{1}{3}$, then $x=\frac{2}{3},y=\frac{2}{3}$,
$z=-\frac{1}{3}$
When $q=-\frac{1}{3}$, then $x=-\frac{2}{3},y=-\frac{2}{3}$,
$z=\frac{1}{3}$
Here required unit vector is $\frac{2}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}$
or $-\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}.$