Question

Two vectors \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \). The angle between the two vectors is:

• A. \( 30^\circ \)
• B. \( 60^\circ \)
• C. \( 45^\circ \)
• D. \( 90^\circ \)

Hint:

To find the angle between vectors using cross and dot products, equate \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \) and solve for \( \theta \) using trigonometric identities.

Answer 1

The Correct Option is C

We are given that \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \), and we need to find the angle \( \theta \) between the vectors \( \vec{a} \) and \( \vec{b} \). - The magnitude of the cross product \( |\vec{a} \times \vec{b}| \) is given by: \[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \] - The dot product \( \vec{a} \cdot \vec{b} \) is given by: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \] According to the problem: \[ |\vec{a}| |\vec{b}| \sin \theta = |\vec{a}| |\vec{b}| \cos \theta \] Assuming \( |\vec{a}| \) and \( |\vec{b}| \) are non-zero (as vectors have magnitude), we can divide both sides by \( |\vec{a}| |\vec{b}| \) (provided \( |\vec{a}| |\vec{b}| \neq 0 \)):
\[ \sin \theta = \cos \theta \] Divide both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \), i.e., \( \theta \neq 90^\circ \)):
\[ \frac{\sin \theta}{\cos \theta} = 1 \] \[ \tan \theta = 1 \] The angle \( \theta \) for which \( \tan \theta = 1 \) is \( \theta = 45^\circ \) (in the first quadrant, where \( \theta \) is between \( 0^\circ \) and \( 90^\circ \)).
Verification
- If \( \theta = 45^\circ \), then \( \sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2} \).
- \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin 45^\circ = |\vec{a}| |\vec{b}| \cdot \frac{\sqrt{2}}{2} \).
- \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 45^\circ = |\vec{a}| |\vec{b}| \cdot \frac{\sqrt{2}}{2} \).
- Thus, \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \), which satisfies the given condition.
Other angles like \( 90^\circ \) (where \( \sin 90^\circ = 1 \) and \( \cos 90^\circ = 0 \)) would make \( \vec{a} \cdot \vec{b} = 0 \), but \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \), which does not hold unless \( \vec{a} \) or \( \vec{b} \) is zero, contradicting the problem's intent. Hence, \( \theta = 45^\circ \) is correct.
\[ \boxed{45^\circ} \]