The angle between ` vec A= hat i- + hat j` and ` vec B= hat i- hat j` is.

Options

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

Explanation

To find the angle between the vectors \(\vec{A} = \hat{i} + \hat{j}\) and \(\vec{B} = \hat{i} - \hat{j}\), we can follow these steps:

### Step 1: Write down the vectors
We have:
\[ \vec{A} = \hat{i} + \hat{j} \]
\[ \vec{B} = \hat{i} - \hat{j} \]

### Step 2: Calculate the dot product of the vectors
The dot product \(\vec{A} \cdot \vec{B}\) is given by:
\[ \vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) = 1 - 1 = 0 \]

### Step 3: Calculate the magnitudes of the vectors
The magnitude of \(\vec{A}\) is:
\[ |\vec{A}| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

The magnitude of \(\vec{B}\) is:
\[ |\vec{B}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Use the dot product to find the cosine of the angle
The cosine of the angle \(\theta\) between the vectors is given by the formula:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
Substituting the values we calculated:
\[ \cos \theta = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = \frac{0}{2} = 0 \]

### Step 5: Calculate the angle \(\theta\)
Since \(\cos \theta = 0\), we find:
\[ \theta = \cos^{-1}(0) = 90^\circ \]

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

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