Question

Consider two vectors $\overrightarrow{a}=\hat{i}+\hat{j},\overrightarrow{b}=2\hat{i}-\hat{j}$ and a third vector coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$ and also perpendicular to $\overrightarrow{a}$ with unit magnitude. Then the third vector is

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

Answer 1

Answer: (B), (C)

$-\frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$

Let third vector is $\overrightarrow{c}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by,
$\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}$
taking dot product with vector $\overrightarrow{a}$ both sides,
$\Rightarrow \overrightarrow{c}\cdot \overrightarrow{a}=x|\overrightarrow{a}{|}^2+y(\overrightarrow{a}\cdot \overrightarrow{b})\Rightarrow 0=2x+y[\because \overrightarrow{c}\perp \overrightarrow{a}]\Rightarrow y=-2x\cdots \left(i\right)$
Also, $\overrightarrow{c}=x(\hat{i}+\hat{j})+y(2\hat{i}-\hat{j})=(x+2y)\hat{i}+(x-y)\hat{j}$
$\Rightarrow (x+2y{)}^2+(x-y{)}^2=1[\because |\overrightarrow{c}|=1]$
using $\left(i\right);9x^2+9x^2=1$
$\Rightarrow x=\pm \frac{1}{3\sqrt{2}},y=\mp \frac{2}{3\sqrt{2}}$
$\therefore \overrightarrow{c}=\pm (3x\hat{i}-3x\hat{j})=\pm \frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$