Question

If \(|\vec{A}\times \vec{B}|=\sqrt{3}(\vec{A}\cdot \vec{B})\) then the value of \(|\vec{A}+\vec{B}|\) is (\(\tan 60^{\circ }=\sqrt{3}\), \(\cos 60^{\circ }=\frac{1}{2}\))

• A. \(\left[A^{2}+B^{2}+\frac{AB}{\sqrt{3}}\right]^{\frac{1}{2}}\)
• B. \(A+B\)
• C. \((A^{2}+B^{2}+\sqrt{3}AB)^{\frac{1}{2}}\)
• D. \((A^{2}+B^{2}+AB)^{\frac{1}{2}}\)

Hint:

Divide the cross product magnitude by the dot product to get the angle between the vectors, then use the resultant formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The magnitude of the cross product is \(AB\sin\theta\) and the dot product is \(AB\cos\theta\). Their ratio gives \(\tan\theta\).

Step 2: Find the Angle
Given \(AB\sin\theta=\sqrt{3}\,AB\cos\theta\), so \(\tan\theta=\sqrt{3}\).
Hence \(\theta=60^{\circ}\).

Step 3: Resultant Magnitude
\(|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2AB\cos\theta}\).
With \(\cos 60^{\circ}=\frac{1}{2}\), this becomes \(\sqrt{A^{2}+B^{2}+AB}\).

Step 4: Compare the options
This matches \((A^{2}+B^{2}+AB)^{\frac{1}{2}}\).

Final Answer:
The magnitude of the resultant is the square root of A squared plus B squared plus AB. \[ \boxed{\text{(D) }(A^{2}+B^{2}+AB)^{\frac{1}{2}}} \]