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

The Correct Option is B

Let $x\hat{i} + y\hat{j} + z\hat{k}$ be the required unit vector. Since $\hat{a}$ is perpendicular to $\left(2\hat{i} -\hat{j}+ 2\hat{k}\right)$. $\therefore\quad2x - y + 2z = 0 \quad\quad...........\left(i\right)$ Since vector $x\hat{i} + y\hat{j} + z\hat{k}$ is coplanar with the vector $\hat{i} + \hat{j} - \hat{k} and 2\hat{i} + 2\hat{j} - \hat{k}$. $\therefore\quad x\hat{i} + y\hat{j} + z\hat{k}$ $= p\left(\hat{i} + \hat{j} - \hat{k}\right)+q \left(2\hat{i} + 2\hat{j} + \hat{k}\right)$, where p and q are some scalars. $\Rightarrow\quad x\hat{i} + y\hat{j} + z\hat{k}$ $= \left(p+2q\right)\hat{i} + \left(p+2q\right)\hat{j} - \left(p+q\right)\hat{k}$ $\Rightarrow\quad x=p + 2q, \,y=p + 2q,\,z = -p-q$ Now from equation $\left(i\right)$, $2p + 4q - p -2q - 2p - 2q = 0$ $\Rightarrow\quad- p = 0\quad\Rightarrow\quad p = 0$ $\therefore\quad x = 2q,\, y = 2q,\, z = - q$ Since vector $x\hat{i} + y\hat{j} + z\hat{k}$ is a unit vector, therefore $\left|x\hat{i} + y\hat{j} + z\hat{k}\right| = 1$ $\Rightarrow\quad\sqrt{x^{2}+y^{2}+z^{2}} = 1$ $\Rightarrow\quad x^{2} + y^{2} + z^{2} = 1$ $\Rightarrow\quad4 q^{2} + 4 q^{2} + q^{2 }= 1$ $\Rightarrow\quad9q^{2} = 1\quad\Rightarrow\quad q = \pm \frac{1}{3}$ 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 $- \frac{2}{3}\hat{i}-\frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}.$