Question

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

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

Answer 1

The correct option is A $\frac{3\overrightarrow{i}+2\overrightarrow{j}-2\overrightarrow{k}}{\sqrt{17}}$

Let the unit vector be $x\hat{i}+y\hat{j}+j\hat{k}$

given this is a unit vector $\sqrt{x^{{}^2}+y^2+j^2}=1$

$x^{{}^2}+y^2+j^2=1$ is equation (1)

also perpendicular to $2\hat{i}-\hat{j}+2\hat{k}$

$\overline{a}\cdot \overline{b}=0$

2x - y + 2j = 0 is equation (2)

also copalaner with

$\hat{i}+\hat{j}-\hat{k}$

$\hat{i}+\hat{j}-\hat{k}$

$\left|\begin{array}{c}xyz\\ 11-1\\ 21-1\end{array}\right|=0$

y = - j

put in equation (2)

2x + 3j = 0

$x=-\frac{3y}{j}$

put in equation (1)

$\frac{9j^2}{4}+j^2+j^2=1$

$17j^2=4$

$\begin{array}{c}{j}^2=\frac{4}{17}\\ j=\pm \frac{2}{\sqrt{17}}\end{array}$

so vector is

$\frac{3\hat{i}}{\sqrt{17}}+\frac{2\hat{j}}{\sqrt{17}}-\frac{2\hat{k}}{\sqrt{17}}$