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IF |bara|=4,|barb|=2 and the angle betwe...

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

A

64

B

16

C

`|bara|^2`

D

6

Text Solution

AI Generated Solution

The correct Answer is:
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 of the vectors and the angle between them, we can use the formula for the magnitude of the cross product. ### Step-by-Step Solution: 1. **Identify the 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. **Use the Formula for the Magnitude of the Cross Product:** The magnitude of the cross product of two vectors is given by: \[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \] 3. **Substitute the Known Values:** Substitute the values into the formula: \[ |\mathbf{a} \times \mathbf{b}| = 4 \times 2 \times \sin\left(\frac{\pi}{6}\right) \] 4. **Calculate \( \sin\left(\frac{\pi}{6}\right) \):** We know that: \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] Therefore: \[ |\mathbf{a} \times \mathbf{b}| = 4 \times 2 \times \frac{1}{2} \] 5. **Perform the Multiplication:** \[ |\mathbf{a} \times \mathbf{b}| = 4 \times 2 \times \frac{1}{2} = 4 \] 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: \[ (\mathbf{a} \times \mathbf{b})^2 = 16 \]
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