If $|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}$ then the value of $|\vec{A} + \vec{B}|$ is:
If $|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}$ then the value of $|\vec{A} + \vec{B}|$ is:
- $\left(A^2+B^2+A B\right)^{1 / 2}$
- $\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{1 / 2}$
- $A+B$
- $\left(A^2+B^2+\sqrt{3} A B\right)^{1 / 2}$
Solution
According to the question
$\begin{aligned}
& \vec{A} \times \vec{B}=\sqrt{3} \vec{A} \cdot \vec{B} \\
& A B \sin \theta =\sqrt{3} A B \cos \theta \\
& \Rightarrow \tan \theta=\sqrt{3} \\
& \Rightarrow \theta=60^{\circ} \\
& \Rightarrow|\vec{A}+\vec{B}|=\sqrt{|\vec{A}|^2+|\vec{B}|^2+2|A| |B| \cos \theta}
\end{aligned}$
Asked in: NEET 2004
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