Question

Find the components of $\overrightarrow{a}=2\hat{i}+3j$ along the directions of vectors $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$:

Answer 1

Let the given vector be

$\overrightarrow{a}=2\hat{i}+3\hat{j}$

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 be $\hat{a}$

$\hat{a}=\frac{\hat{i}+\hat{j}}{|\hat{i}+\hat{j}}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

Component of $\overrightarrow{a}$ along $\hat{i}+\hat{j}$

$=(\overrightarrow{a}\cdot \hat{a})\hat{a}$

$=\left[\right(2\hat{i}+3\hat{j})\cdot \frac{1}{\sqrt{2}}(\hat{i}+\hat{j}\left)\right]\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

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

$=\frac{5}{\sqrt{2}}\left[\frac{1}{\sqrt{2}}\right(\hat{i}+\hat{j}\left)\right]$

$=\frac{5}{2}\hat{i}+\frac{5}{\sqrt{2}}\hat{j}$

Component of $\overrightarrow{a}$ along $\hat{i}-\hat{j}$

$=(\overrightarrow{a}\cdot \hat{a})\hat{a}$

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

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

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

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