The resultant of two equal vectors is equal in magnitude to each of them when the angle between the vectors is 120∘120^\circ 120∘.

Key idea in simple words

Take two vectors A⃗\vec AA and B⃗\vec BB of the same magnitude AAA, with an angle θ\theta θ between them.

The magnitude of their resultant R⃗\vec RR is

R=A2+A2+2A2cos⁡θ=2A2(1+cos⁡θ).R=\sqrt{A^2+A^2+2A^2\cos\theta}=\sqrt{2A^2(1+\cos\theta)}.R=A2+A2+2A2cosθ​=2A2(1+cosθ)​.

You want the resultant to be equal to either of the vectors, so set R=AR=AR=A.

A=2A2(1+cos⁡θ).A=\sqrt{2A^2(1+\cos\theta)}.A=2A2(1+cosθ)​.

Squaring both sides and simplifying (assuming A≠0A\neq 0A=0) gives

1=2(1+cos⁡θ)⇒1=2+2cos⁡θ⇒−1=2cos⁡θ⇒cos⁡θ=−12.1=2(1+\cos\theta)\Rightarrow 1=2+2\cos\theta \Rightarrow -1=2\cos\theta \Rightarrow \cos\theta =-\frac12.1=2(1+cosθ)⇒1=2+2cosθ⇒−1=2cosθ⇒cosθ=−21​.

The angle whose cosine is −12-\tfrac12 −21​ is 120∘120^\circ 120∘.

So, when the angle between two equal vectors is 120∘120^\circ 120∘, their resultant has the same magnitude as each vector.

Visual mini-story (to imagine it)

  • Picture two equal arrows starting from the same point.
  • Open them so that the angle between them is 120∘120^\circ 120∘.
  • If you now draw the diagonal of the parallelogram formed by these two arrows, that diagonal (the resultant) will have exactly the same length as each arrow.

So the condition is:

When two equal vectors have an angle of 12 0∘120^\circ 120∘ between them, the resultant is equal in magnitude to each of the vectors.

TL;DR: For two equal vectors, the resultant equals either one of them (in magnitude) only if the angle between them is 120∘120^\circ 120∘.

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