Two vectors A and B have magnitudes 2 units and 4 units respectively. Find A. B is angle between these two vectors is (a) `0^(@)` (b) `60^(@)` (c) `90^(@)` (d) `120^(@)` .

Explanation

To solve the problem of finding the dot product \( A \cdot B \) for the given vectors \( A \) and \( B \) with magnitudes of 2 units and 4 units respectively, we will use the formula for the dot product:

\[ A \cdot B = |A| |B| \cos \theta \]

where \( |A| \) and \( |B| \) are the magnitudes of the vectors \( A \) and \( B \), and \( \theta \) is the angle between them.

### Step-by-Step Solution:

1. **Identify the Magnitudes and Angles**:
- Magnitude of vector \( A \) = 2 units
- Magnitude of vector \( B \) = 4 units
- Angles to consider: \( 0^\circ, 60^\circ, 90^\circ, 120^\circ \)

2. **Calculate \( A \cdot B \) for each angle**:

**(a) For \( \theta = 0^\circ \)**:
\[ \cos(0^\circ) = 1 \]
\[ A \cdot B = 2 \times 4 \times 1 = 8 \]

**(b) For \( \theta = 60^\circ \)**:
\[ \cos(60^\circ) = \frac{1}{2} \]
\[ A \cdot B = 2 \times 4 \times \frac{1}{2} = 4 \]

**(c) For \( \theta = 90^\circ \)**:
\[ \cos(90^\circ) = 0 \]
\[ A \cdot B = 2 \times 4 \times 0 = 0 \]

**(d) For \( \theta = 120^\circ \)**:
\[ \cos(120^\circ) = -\frac{1}{2} \]
\[ A \cdot B = 2 \times 4 \times -\frac{1}{2} = -4 \]

3. **Summarize the Results**:
- For \( \theta = 0^\circ \), \( A \cdot B = 8 \)
- For \( \theta = 60^\circ \), \( A \cdot B = 4 \)
- For \( \theta = 90^\circ \), \( A \cdot B = 0 \)
- For \( \theta = 120^\circ \), \( A \cdot B = -4 \)

### Final Answer:
- \( A \cdot B \) values for the respective angles are:
- (a) \( 8 \)
- (b) \( 4 \)
- (c) \( 0 \)
- (d) \( -4 \)