Physics Vectors & Equilibrium PMDC Conceptual Practice
PMDC Verified Question 32 of 50
The magnitude of the dot product of two vectors is \(6\sqrt{3}\) and the magnitude of their cross product is 6. What is the angle between the two vectors?
A
\(0^\circ\)
B
\(30^\circ\)
C
\(45^\circ\)
D
\(60^\circ\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \(30^\circ\)
Concept:
Finding the angle between vectors using the ratio of cross and dot products.

Formula:
$$\vec{A} \cdot \vec{B} = AB \cos\theta = 6\sqrt{3}$$
$$|\vec{A} \times \vec{B}| = AB \sin\theta = 6$$

Solution:
Divide the magnitude of the cross product by the dot product:
$$\frac{AB \sin\theta}{AB \cos\theta} = \frac{6}{6\sqrt{3}}$$
$$\tan\theta = \frac{1}{\sqrt{3}}$$
$$\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = 30^\circ$$

Why other options are incorrect:
  • \(0^\circ\): The cross product would be zero.
  • \(45^\circ\): The dot product and cross product would be equal in magnitude.
  • \(60^\circ\): This would swap the two values, making the dot product 6 and the cross product \(6\sqrt{3}\).

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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