Motion in a Plane - Result Question 9
11. If $|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}$ then the value of $|\vec{A}+\vec{B}|$ is
[2004]
(a) $(A^{2}+B^{2}+\sqrt{3} A B)^{1 / 2}$
(b) $(A^{2}+B^{2}+A B)^{1 / 2}$
(c) $(A^{2}+B^{2}+\frac{A B}{\sqrt{3}})^{1 / 2}$
(d) $A+B$
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Answer:
Correct Answer: 11. (b)
Solution:
- (b) $|\vec{A} \times \vec{B}|=A B \sin \theta$
$\vec{A} \cdot \vec{B}=A B \cos \theta$
$|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B} \Rightarrow A B \sin \theta=\sqrt{ } 3 A B \cos \theta$
or, $\tan \theta=\sqrt{3}, \quad \therefore \theta=60^{\circ}$
$\therefore|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos 60^{\circ}}$
$-=\sqrt{A^{2}+B^{2}+A B}$
