Question

The cross product of vector \( \vec{A} \) and vector \( \vec{B} \) has a magnitude of 50 units, where vector \( \vec{A} \) has a magnitude of 10. The angle between vector \( \vec{A} \) and \( \vec{B} \) is 60 degrees. What is the magnitude of vector \( \mathbf{B} \)?

• A. \( 5/\sqrt{2} \)
• B. \( 10/\sqrt{2} \)
• C. \( 10/\sqrt{3} \)
• D. \( 5/\sqrt{3} \)

Hint:

When calculating the magnitude of a cross product, remember to use the formula \( |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \). Pay special attention to the angle between the vectors.

Answer 1

The Correct Option is C

Step 1: Formula for Cross Product Magnitude.
The magnitude of the cross product of two vectors is given by the formula: \[ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \] where:
\( |\mathbf{A}| \) is the magnitude of vector \( \mathbf{A} \),
\( |\mathbf{B}| \) is the magnitude of vector \( \mathbf{B} \),
\( \theta \) is the angle between the two vectors.
Step 2: Plugging in Known Values.
We are given:
\( |\mathbf{A}| = 10 \),
\( |\mathbf{A} \times \mathbf{B}| = 50 \),
\( \theta = 60^\circ \).
Substituting these values into the formula: \[ 50 = 10 \times |\mathbf{B}| \times \sin 60^\circ \] Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), we have: \[ 50 = 10 \times |\mathbf{B}| \times \frac{\sqrt{3}}{2} \] Step 3: Solving for \( |\mathbf{B}| \).
Now solve for \( |\mathbf{B}| \): \[ 50 = 5 \times |\mathbf{B}| \times \sqrt{3} \] \[ |\mathbf{B}| = \frac{50}{5 \times \sqrt{3}} = \frac{10}{\sqrt{3}} \] Step 4: Conclusion.
Thus, the magnitude of vector \( \mathbf{B} \) is \( \frac{10}{\sqrt{3}} \).