Question

If \(|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})\) then the value of \(|\vec{A}+\vec{B}|\) is \(\left(\tan 60^{\circ}=\sqrt{3}, \cos 60^{\circ}=0.5\right)\)

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

Hint:

The condition fixes tan of the angle between the vectors, which gives the angle and then the resultant magnitude.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(|\vec{A}\times\vec{B}|=AB\sin\theta\) and \(\vec{A}\cdot\vec{B}=AB\cos\theta\).

Step 2: Find the angle
\(AB\sin\theta=\sqrt{3}AB\cos\theta\Rightarrow\tan\theta=\sqrt{3}\Rightarrow\theta=60^{\circ}\).

Step 3: Resultant
\(|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2AB\cos60^{\circ}}\).

Step 4: Simplify
\(=\sqrt{A^{2}+B^{2}+AB}\).

Final Answer:
The magnitude is \(\sqrt{A^{2}+B^{2}+AB}\). \[ \boxed{\text{(B) }(A^{2}+B^{2}+AB)^{\frac{1}{2}}} \]