Find the magnitude and direction of the vectors $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$. What are the components of a vector $\vec{A}=2\hat{i}+3\hat{j}$ along the directions of $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$?

(N/A) Let $\vec{P} = \hat{i} + \hat{j}$. The magnitude is $|\vec{P}| = \sqrt{1^2 + 1^2} = \sqrt{2}$. The direction is $\theta = \tan^{-1}(1/1) = 45^{\circ}$ with the $x$-axis.
Let $\vec{Q} = \hat{i} - \hat{j}$. The magnitude is $|\vec{Q}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. The direction is $\theta = \tan^{-1}(-1/1) = -45^{\circ}$ with the $x$-axis.
To find the component of $\vec{A} = 2\hat{i} + 3\hat{j}$ along a vector $\vec{V}$,we use the formula $\text{Component} = \left( \frac{\vec{A} \cdot \vec{V}}{|\vec{V}|} \right) \hat{V}$.
For $\vec{V}_1 = \hat{i} + \hat{j}$,$|\vec{V}_1| = \sqrt{2}$. The unit vector is $\hat{u}_1 = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
Component $= (\vec{A} \cdot \hat{u}_1) \hat{u}_1 = \left( \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} \right) \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \left( \frac{2+3}{2} \right) (\hat{i} + \hat{j}) = 2.5(\hat{i} + \hat{j})$.
For $\vec{V}_2 = \hat{i} - \hat{j}$,$|\vec{V}_2| = \sqrt{2}$. The unit vector is $\hat{u}_2 = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.
Component $= (\vec{A} \cdot \hat{u}_2) \hat{u}_2 = \left( \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} - \hat{j})}{\sqrt{2}} \right) \frac{\hat{i} - \hat{j}}{\sqrt{2}} = \left( \frac{2-3}{2} \right) (\hat{i} - \hat{j}) = -0.5(\hat{i} - \hat{j})$.