The angle between `A=hat(i)+hat(j) " and " B=hat(i)-hat(j)` is

Options

A. `45^(@)`
B. `90^(@)`
C. `-45^(@)`
D. `180^(@)`

Explanation

To find the angle between the vectors \( \mathbf{A} = \hat{i} + \hat{j} \) and \( \mathbf{B} = \hat{i} - \hat{j} \), we can use the formula for the cosine of the angle between two vectors:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

### Step 1: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)

The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:

\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} + \hat{j}) \cdot (\hat{i} - \hat{j}) \]

Calculating this, we have:

\[ \mathbf{A} \cdot \mathbf{B} = \hat{i} \cdot \hat{i} + \hat{i} \cdot (-\hat{j}) + \hat{j} \cdot \hat{i} + \hat{j} \cdot (-\hat{j}) \]

Using the properties of dot products:

\[ = 1 + 0 + 0 - 1 = 0 \]

### Step 2: Calculate the magnitudes \( |\mathbf{A}| \) and \( |\mathbf{B}| \)

The magnitude of vector \( \mathbf{A} \) is:

\[ |\mathbf{A}| = \sqrt{(\hat{i})^2 + (\hat{j})^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

The magnitude of vector \( \mathbf{B} \) is:

\[ |\mathbf{B}| = \sqrt{(\hat{i})^2 + (-\hat{j})^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]

### Step 3: Substitute into the cosine formula

Now we can substitute the values into the cosine formula:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = \frac{0}{2} = 0 \]

### Step 4: Determine the angle \( \theta \)

Since \( \cos \theta = 0 \), we find that:

\[ \theta = 90^\circ \]

### Final Answer

The angle between the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( 90^\circ \).

---