2004 AIPMT Motion in a PlaneScalar and Vector Products Medium Question 3 of 3 in Scalar and Vector Products

If $|\vec{A} \times \vec{B}| = \sqrt{3}\,\vec{A}\cdot\vec{B}$ then the value of $|\vec{A} + \vec{B}|$ is:-

A $\left(A^2 + B^2 + \dfrac{AB}{\sqrt{3}}\right)^{1/2}$
B $A + B$
C $(A^2 + B^2 + \sqrt{3}AB)^{1/2}$
D $(A^2 + B^2 + AB)^{1/2}$

The answer is D.

Solution

Given $|\vec{A} \times \vec{B}| = \sqrt{3}\,\vec{A}\cdot\vec{B}$
$\Rightarrow AB\sin\theta = \sqrt{3}AB\cos\theta$
$\Rightarrow \tan\theta = \sqrt{3}$
$\theta = 60^\circ$
Therefore $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}$
$= \sqrt{A^2 + B^2 + 2AB\cos 60^\circ} = \sqrt{A^2 + B^2 + 2AB \times \dfrac{1}{2}}$
$= (A^2 + B^2 + AB)^{1/2}$

Concept Angle between vectors from cross and dot products

Scalar and Vector Products in past papers

3 questions from this chapter have appeared across 3 exam years.