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QExample - 5: HC Verma: Concepts of Physics - Volume 1 Physics and Mathematics Example - 5 Solution: Vector A "> has a magnitude of 5 units, vector B "> has a…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often assume $ heta = arcsin(1/2)$ yields a unique answer of $30^{circ}$, ignoring that $sin( heta) = 1/2$ also holds for $ heta = 150^{circ}$. Since the magnitude of the cross product depends on $sin( heta)$, which is positive in both the first and second quadrants, both $30^{circ}$ and $150^{circ}$ are valid angles between the vectors.
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has a magnitude of 5 units, vector B has a magnitude of 6 units, and the magnitude of their cross product (A×B) is 15 units. Find the angle between A and B.
Step-by-Step Verified Solution
Solution:
Given:
- |A| = 5 units
- |B| = 6 units
- |A×B| = 15 units
To Find:
- Angle between A and B (θ)
Concept:
The magnitude of the cross product of two vectors is given by:
|A×B| = |A| |B| sin(θ)
Calculation:
- Substitute the given values into the formula:
- 15 = (5)(6) sin(θ)
- Simplify:
- 15 = 30 sin(θ)
- Solve for sin(θ):
- sin(θ) = 15 / 30 sin(θ) = 1/2
- Find the angle θ for which sin(θ) = 1/2:
- θ = arcsin(1/2)
- The angle θ can be either 30° or 150° because sin(30°) = 1/2 and sin(150°) = 1/2.
Answer:
The angle between A and B is either 30° or 150°.
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