Question

Find the component of vector $\overrightarrow{A}=(2\hat{i}+3\hat{j})$ along the direction $(\hat{i}-\hat{j})$

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

Answer 1

The correct option is A $-\frac{1}{2}(\hat{i}-\hat{j})$

Given vector $2\hat{i}+3\hat{j}$

To find the component of A along $\hat{i}-\hat{j}$ we have to find unit vector along $\hat{i}-\hat{j}$

Let the unit vector by $\hat{a}$

$\Rightarrow \hat{a}=\frac{(\hat{i}-\hat{j})}{|\hat{i}-\hat{j}|}$

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

component of A a long $\hat{i}-\hat{j}$

$=(\overrightarrow{A}.\hat{a})\hat{a}$

$=[(2\hat{i}+3\hat{j}).\frac{1}{\sqrt{2}}(\hat{i}-\hat{j})].\frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$

$=\left(\frac{1}{\sqrt{2}}(2-3)\right)\frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$

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

$=\frac{-1}{2}\hat{i}+\frac{1}{2}\hat{j}$

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

Hence, the answer is $-\frac{1}{2}(\hat{i}-\hat{j}).$