If $\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\ve
Question:

If A+B=AB\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\vec{B}\right|, then the angle between A\vec{A} and B\vec{B} will be

Updated On: Jul 5, 2022
  • 30?30^?
  • 45?45^?
  • 60?60^?
  • 90?90^?
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The Correct Option is D

Solution and Explanation

Let θ\theta be the angle between the vectors A\vec{A} and B\vec{B}. Then A+B=A2+B2+2ABcosθ\left|\vec{A}+\vec{B}\right|=\sqrt{A^{2}+B^{2}+2AB\,cos\,\theta} AB=A2+B22ABcosθ\left|\vec{A}-\vec{B}\right|=\sqrt{A^{2}+B^{2}-2AB\,cos\,\theta} According to given problem A+B=AB\left|\vec{A}+\vec{B}\right|=\left|\vec{A}-\vec{B}\right| A2+B2+2ABcosθ\therefore\sqrt{A^{2}+B^{2}+2AB\,cos\,\theta} =A2+B22ABcosθ=\sqrt{A^{2}+B^{2}-2AB\,cos\,\theta} Squaring both sides, we get A2+B2+2ABcosθ=A2+B22ABcosθA^{2}+B^{2}+2ABcos\theta=A^{2}+B^{2}-2ABcos\theta 4ABcosθ=0\therefore 4ABcos\theta=0 As A0A\ne0, B0B\ne0 cosθ=0\therefore cos\theta=0 or θ=90\theta=90^{\circ}
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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration