Q94 / 200 · NEET 2004 · Physics
If $|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}$ then the value of $|\vec{A} + \vec{B}|$ is:
- Correct answerA$\left(A^2+B^2+A B\right)^{1 / 2}$
- B$\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{1 / 2}$
- C$A+B$
- D$\left(A^2+B^2+\sqrt{3} A B\right)^{1 / 2}$
Solution
According to the question
$\begin{aligned}
& \vec{A} \times \vec{B}=\sqrt{3} \vec{A} \cdot \vec{B} \\
& A B \sin =\sqrt{3} A B \cos \theta \\
& \Rightarrow \tan \theta=\sqrt{3} \\
& \Rightarrow \theta=60^{\circ} \\
& \Rightarrow|\vec{A}+\vec{B}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2+2|A|} \overline{B \mid \cos \theta}
\end{aligned}$
Hence, the correct answer is option: 'A'.