IF `|bara|=4,|barb|=2` and the angle between `bara and barb` is `pi/6` then `(bara times barb)^2=`

Options

A. 64
B. 16
C. `|bara|^2`
D. 6

Answer

To solve the problem, we need to find the square of the magnitude of the cross product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \). Given the magnitudes and the angle between them, we can use the formula for the magnitude of the cross product.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Magnitude of vector \( \mathbf{a} \): \( |\mathbf{a}| = 4 \)
- Magnitude of vector \( \mathbf{b} \): \( |\mathbf{b}| = 2 \)
- Angle between \( \mathbf{a} \) and \( \mathbf{b} \): \( \theta = \frac{\pi}{6} \)

2. **Formula for Cross Product Magnitude**:
The magnitude of the cross product of two vectors is given by:
\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \]

3. **Substituting Values**:
Substitute the known values into the formula:
\[ |\mathbf{a} \times \mathbf{b}| = 4 \cdot 2 \cdot \sin\left(\frac{\pi}{6}\right) \]

4. **Calculate \( \sin\left(\frac{\pi}{6}\right) \)**:
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

5. **Calculate the Magnitude of the Cross Product**:
Now substitute \( \sin\left(\frac{\pi}{6}\right) \) into the equation:
\[ |\mathbf{a} \times \mathbf{b}| = 4 \cdot 2 \cdot \frac{1}{2} = 4 \]

6. **Square the Magnitude of the Cross Product**:
Now we need to find \( |\mathbf{a} \times \mathbf{b}|^2 \):
\[ |\mathbf{a} \times \mathbf{b}|^2 = 4^2 = 16 \]

### Final Answer:
\[ (\mathbf{a} \times \mathbf{b})^2 = 16 \]