If `vec(A)=2hat(i)+hat(j)` and `vec(B)=hat(i)-hat(j)`, sketch vector graphically and find the component of `vec(A)` along `vec(B)` and perpendicular to `vec(B)`

Explanation

To solve the problem step by step, we will first sketch the vectors graphically, then calculate the components of vector A along vector B and perpendicular to vector B.

### Step 1: Sketch the Vectors Graphically
1. **Draw the Coordinate Axes**: Draw the x-axis (horizontal) and y-axis (vertical).
2. **Plot Vector A**: Vector A is given as \( \vec{A} = 2\hat{i} + \hat{j} \). This means it has a component of 2 along the x-axis and 1 along the y-axis. Plot the point (2, 1) and draw an arrow from the origin (0, 0) to this point.
3. **Plot Vector B**: Vector B is given as \( \vec{B} = \hat{i} - \hat{j} \). This means it has a component of 1 along the x-axis and -1 along the y-axis. Plot the point (1, -1) and draw an arrow from the origin to this point.

### Step 2: Find the Magnitudes of Vectors A and B
1. **Magnitude of Vector A**:
\[ |\vec{A}| = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

2. **Magnitude of Vector B**:
\[ |\vec{B}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 3: Find the Unit Vectors
1. **Unit Vector A**:
\[ \hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{2\hat{i} + \hat{j}}{\sqrt{5}} \]

2. **Unit Vector B**:
\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} \]

### Step 4: Find the Component of A Along B
The component of vector A along vector B can be calculated using the dot product:
\[ \text{Component of } \vec{A} \text{ along } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} \]

1. **Calculate the Dot Product**:
\[ \vec{A} \cdot \vec{B} = (2\hat{i} + \hat{j}) \cdot (\hat{i} - \hat{j}) = 2 \cdot 1 + 1 \cdot (-1) = 2 - 1 = 1 \]

2. **Calculate the Component**:
\[ \text{Component of } \vec{A} \text{ along } \vec{B} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \hat{B} \]

### Step 5: Find the Component of A Perpendicular to B
The component of vector A perpendicular to vector B can be found by subtracting the component along B from vector A:
\[ \text{Component of } \vec{A} \text{ perpendicular to } \vec{B} = \vec{A} - \text{Component of } \vec{A} \text{ along } \vec{B} \]

1. **Calculate the Component Along B**:
\[ \text{Component along } \vec{B} = \frac{1}{\sqrt{2}} \cdot \hat{B} = \frac{1}{\sqrt{2}} \left( \hat{i} - \hat{j} \right) \]

2. **Subtract from A**:
\[ \text{Component perpendicular} = (2\hat{i} + \hat{j}) - \left( \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \right) \]

To simplify:
\[ = \left( 2 - \frac{1}{\sqrt{2}} \right) \hat{i} + \left( 1 + \frac{1}{\sqrt{2}} \right) \hat{j} \]

### Final Result
- The component of vector A along vector B is \( \frac{1}{\sqrt{2}} \hat{B} \).
- The component of vector A perpendicular to vector B is \( \left( 2 - \frac{1}{\sqrt{2}} \right) \hat{i} + \left( 1 + \frac{1}{\sqrt{2}} \right) \hat{j} \).