Obtain the angle between A+B and A-B if `A = 2hati +3hatj and B = hati - 2hatj.`

Explanation

To find the angle between the vectors \( \mathbf{A} + \mathbf{B} \) and \( \mathbf{A} - \mathbf{B} \), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \mathbf{A} = 2\hat{i} + 3\hat{j} \]
\[ \mathbf{B} = \hat{i} - 2\hat{j} \]

### Step 2: Calculate \( \mathbf{A} + \mathbf{B} \)
We add the vectors:
\[ \mathbf{A} + \mathbf{B} = (2\hat{i} + 3\hat{j}) + (\hat{i} - 2\hat{j}) \]
\[ = (2 + 1)\hat{i} + (3 - 2)\hat{j} \]
\[ = 3\hat{i} + 1\hat{j} \]

### Step 3: Calculate \( \mathbf{A} - \mathbf{B} \)
We subtract the vectors:
\[ \mathbf{A} - \mathbf{B} = (2\hat{i} + 3\hat{j}) - (\hat{i} - 2\hat{j}) \]
\[ = (2 - 1)\hat{i} + (3 + 2)\hat{j} \]
\[ = 1\hat{i} + 5\hat{j} \]

### Step 4: Define \( \mathbf{P} \) and \( \mathbf{Q} \)
Let:
\[ \mathbf{P} = \mathbf{A} + \mathbf{B} = 3\hat{i} + 1\hat{j} \]
\[ \mathbf{Q} = \mathbf{A} - \mathbf{B} = 1\hat{i} + 5\hat{j} \]

### Step 5: Calculate the dot product \( \mathbf{P} \cdot \mathbf{Q} \)
The dot product is given by:
\[ \mathbf{P} \cdot \mathbf{Q} = (3\hat{i} + 1\hat{j}) \cdot (1\hat{i} + 5\hat{j}) \]
\[ = 3 \cdot 1 + 1 \cdot 5 \]
\[ = 3 + 5 = 8 \]

### Step 6: Calculate the magnitudes of \( \mathbf{P} \) and \( \mathbf{Q} \)
Magnitude of \( \mathbf{P} \):
\[ |\mathbf{P}| = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

Magnitude of \( \mathbf{Q} \):
\[ |\mathbf{Q}| = \sqrt{(1)^2 + (5)^2} = \sqrt{1 + 25} = \sqrt{26} \]

### Step 7: Use the cosine formula to find the angle
The cosine of the angle \( \theta \) between the vectors is given by:
\[ \cos \theta = \frac{\mathbf{P} \cdot \mathbf{Q}}{|\mathbf{P}| |\mathbf{Q}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{8}{\sqrt{10} \cdot \sqrt{26}} = \frac{8}{\sqrt{260}} = \frac{8}{\sqrt{10 \cdot 26}} \]

### Step 8: Calculate \( \theta \)
To find \( \theta \):
\[ \theta = \cos^{-1} \left( \frac{8}{\sqrt{260}} \right) \]

### Final Answer
Thus, the angle \( \theta \) between \( \mathbf{A} + \mathbf{B} \) and \( \mathbf{A} - \mathbf{B} \) can be calculated using the above expression.

---