Vectors \( \vec{a},\vec{b},\vec{c} \) are such that \( \vec{a}+\vec{b}+\vec{c}=0 \) and \( |\vec{a}|=|\vec{b}|=|\vec{c}| \) then, the angle between any two of them, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 120°
Let each vector have length k and let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). Since \( \vec{a}+\vec{b}=-\vec{c} \),
\( \displaystyle |\vec{a}+\vec{b}|^2=|\vec{c}|^2 \)
\( \displaystyle k^2+k^2+2k^2\cos\theta=k^2 \)
\( \displaystyle \cos\theta=-\frac{1}{2} \)
\( \displaystyle \theta=120^\circ \)
By symmetry the same holds for every pair.
Why the other options are incorrect
Option (1) 60° — Incorrect. This is the angle of the equilateral triangle formed when the vectors are placed head to tail; the angle between the vectors themselves (drawn from a common point) is 180° − 60° = 120°.
Option (2) 90° — Incorrect. \( \cos\theta=-\frac{1}{2} \), not 0.
Option (4) 150° — Incorrect. \( \cos 150^\circ=-\frac{\sqrt{3}}{2} \), not \( -\frac{1}{2} \).