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:

  1. (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, $\tan \theta=\sqrt{3}, \quad \therefore \theta=60^{\circ}$

$\therefore|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos 60^{\circ}}$

$-=\sqrt{A^{2}+B^{2}+A B}$