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If $|\vec{A} + \vec{B}| = |\vec{A}| = |\vec{B}|$ then angle between $\vec{A}$ and $\vec{B}$ will be : -
A
$90^\circ$
B
$120^\circ$
C
$0^\circ$
D
$60^\circ$
Detailed Solution
$|\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\cos\theta$
Given $|\vec{A} + \vec{B}| = A = B$: $A^2 = A^2 + A^2 + 2A^2\cos\theta$
$\Rightarrow 0 = A^2 + 2A^2\cos\theta \Rightarrow \cos\theta = -\dfrac{1}{2}$
$\theta = 120^\circ$
Given $|\vec{A} + \vec{B}| = A = B$: $A^2 = A^2 + A^2 + 2A^2\cos\theta$
$\Rightarrow 0 = A^2 + 2A^2\cos\theta \Rightarrow \cos\theta = -\dfrac{1}{2}$
$\theta = 120^\circ$