Find angle between A and B where, (a) `A = 2hati and B =- 6 hati` (b) `A = 6hatj` and `B =- 2hatk` (c) `A = (2hati = 3hatj)` and `B = 4 hatk` (d) `A = 4hati` and `B = (-3hati + 3hatj)` .

Explanation

To find the angle between two vectors A and B, we can use the formula:

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

where \( A \cdot B \) is the dot product of the vectors, and \( |A| \) and \( |B| \) are the magnitudes of the vectors A and B, respectively.

### (a) \( A = 2\hat{i} \) and \( B = -6\hat{i} \)

1. **Calculate the dot product \( A \cdot B \)**:
\[ A \cdot B = (2\hat{i}) \cdot (-6\hat{i}) = 2 \times -6 \times (\hat{i} \cdot \hat{i}) = -12 \]

2. **Calculate the magnitudes of A and B**:
\[ |A| = \sqrt{(2)^2} = 2 \]
\[ |B| = \sqrt{(-6)^2} = 6 \]

3. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{-12}{2 \times 6} = \frac{-12}{12} = -1 \]

4. **Find the angle \( \theta \)**:
\[ \theta = \cos^{-1}(-1) = 180^\circ \]

### (b) \( A = 6\hat{j} \) and \( B = -2\hat{k} \)

1. **Calculate the dot product \( A \cdot B \)**:
\[ A \cdot B = (6\hat{j}) \cdot (-2\hat{k}) = 6 \times -2 \times (\hat{j} \cdot \hat{k}) = 0 \]

2. **Calculate the magnitudes of A and B**:
\[ |A| = \sqrt{(6)^2} = 6 \]
\[ |B| = \sqrt{(-2)^2} = 2 \]

3. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{0}{6 \times 2} = 0 \]

4. **Find the angle \( \theta \)**:
\[ \theta = \cos^{-1}(0) = 90^\circ \]

### (c) \( A = 2\hat{i} - 3\hat{j} \) and \( B = 4\hat{k} \)

1. **Calculate the dot product \( A \cdot B \)**:
\[ A \cdot B = (2\hat{i} - 3\hat{j}) \cdot (4\hat{k}) = 2 \times 4 \times (\hat{i} \cdot \hat{k}) - 3 \times 4 \times (\hat{j} \cdot \hat{k}) = 0 \]

2. **Calculate the magnitudes of A and B**:
\[ |A| = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]
\[ |B| = \sqrt{(4)^2} = 4 \]

3. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{0}{\sqrt{13} \times 4} = 0 \]

4. **Find the angle \( \theta \)**:
\[ \theta = \cos^{-1}(0) = 90^\circ \]

### (d) \( A = 4\hat{i} \) and \( B = -3\hat{i} + 3\hat{j} \)

1. **Calculate the dot product \( A \cdot B \)**:
\[ A \cdot B = (4\hat{i}) \cdot (-3\hat{i} + 3\hat{j}) = 4 \times -3 \times (\hat{i} \cdot \hat{i}) + 4 \times 3 \times (\hat{i} \cdot \hat{j}) = -12 + 0 = -12 \]

2. **Calculate the magnitudes of A and B**:
\[ |A| = \sqrt{(4)^2} = 4 \]
\[ |B| = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]

3. **Substitute into the cosine formula**:
\[ \cos \theta = \frac{-12}{4 \times 3\sqrt{2}} = \frac{-12}{12\sqrt{2}} = \frac{-1}{\sqrt{2}} \]

4. **Find the angle \( \theta \)**:
\[ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = 135^\circ \]

### Summary of Angles
- (a) \( 180^\circ \)
- (b) \( 90^\circ \)
- (c) \( 90^\circ \)
- (d) \( 135^\circ \)