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, we will calculate the dot product of two vectors A and B for the given angles. The formula for the dot product of two vectors A and B is:

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

where:
- \( |A| \) is the magnitude of vector A,
- \( |B| \) is the magnitude of vector B,
- \( \theta \) is the angle between the two vectors.

Given:
- Magnitude of vector A, \( |A| = 2 \) units
- Magnitude of vector B, \( |B| = 4 \) units

Now, we will calculate the dot product for each angle.

### Step 1: Calculate \( A \cdot B \) for \( \theta = 0^\circ \)

Using the formula:
\[ A \cdot B = |A| |B| \cos(0^\circ) \]
\[ A \cdot B = 2 \times 4 \times 1 \]
\[ A \cdot B = 8 \text{ units} \]

### Step 2: Calculate \( A \cdot B \) for \( \theta = 60^\circ \)

Using the formula:
\[ A \cdot B = |A| |B| \cos(60^\circ) \]
\[ A \cdot B = 2 \times 4 \times \frac{1}{2} \]
\[ A \cdot B = 4 \text{ units} \]

### Step 3: Calculate \( A \cdot B \) for \( \theta = 90^\circ \)

Using the formula:
\[ A \cdot B = |A| |B| \cos(90^\circ) \]
\[ A \cdot B = 2 \times 4 \times 0 \]
\[ A \cdot B = 0 \text{ units} \]

### Step 4: Calculate \( A \cdot B \) for \( \theta = 120^\circ \)

Using the formula:
\[ A \cdot B = |A| |B| \cos(120^\circ) \]
\[ A \cdot B = 2 \times 4 \times \cos(120^\circ) \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ A \cdot B = 2 \times 4 \times -\frac{1}{2} \]
\[ A \cdot B = -4 \text{ units} \]

### Summary of Results:
- For \( \theta = 0^\circ \), \( A \cdot B = 8 \) units
- For \( \theta = 60^\circ \), \( A \cdot B = 4 \) units
- For \( \theta = 90^\circ \), \( A \cdot B = 0 \) units
- For \( \theta = 120^\circ \), \( A \cdot B = -4 \) units