Question

If $|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$ then the value of $|\vec{A} + \vec{B}|$ is

• A. $(A^2 + B^2 + AB)^{1/2}$
• B. $(A^2 + B^2 + \frac{AB}{\sqrt{3}})^{1/2}$
• C. $A + B$
• D. $(A^2 + B^2 + \sqrt{3}AB)^{1/2}$

Hint:

The ratios of cross-product to dot-product are common exam topics. If the ratio is $1$, $\theta = 45^\circ$; if the ratio is $\sqrt{3}$, $\theta = 60^\circ$; if the ratio is $1/\sqrt{3}$, $\theta = 30^\circ$.

Answer 1

The Correct Option is A

Step 1: Solve for the angle $\theta$: Given: $|\vec{A} \times \vec{B}| = \sqrt{3} (\vec{A} \cdot \vec{B})$ We know: $|\vec{A} \times \vec{B}| = AB \sin \theta$ $\vec{A} \cdot \vec{B} = AB \cos \theta$ Substituting these: $AB \sin \theta = \sqrt{3} AB \cos \theta$ $\frac{\sin \theta}{\cos \theta} = \sqrt{3} \implies \tan \theta = \sqrt{3}$ Therefore, $\theta = 60^\circ$.

Step 2: Find the magnitude of the resultant $|\vec{A} + \vec{B}|$: The formula for the magnitude of the sum of two vectors is: $$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$$ Substituting $\theta = 60^\circ$: $$R = \sqrt{A^2 + B^2 + 2AB \cos 60^\circ}$$ Since $\cos 60^\circ = 1/2$: $$R = \sqrt{A^2 + B^2 + 2AB(1/2)}$$ $$R = \sqrt{A^2 + B^2 + AB}$$ $$R = (A^2 + B^2 + AB)^{1/2}$$ The resultant magnitude matches Option (A).