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Vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are such that a⃗+b⃗+c⃗=0\vec{a} + \vec{b} + \vec{c} = 0 and ∣a⃗∣=∣b⃗∣=∣c⃗∣|\vec{a}| = |\vec{b}| = |\vec{c}| then, the angle between any two of them, is

Solution

✅ Correct Option: 3

Squaring a⃗+b⃗+c⃗=0\vec a+\vec b+\vec c=0 with common magnitude kk gives 3k2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=03k^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0. By symmetry each dot product equals −k22-\frac{k^2}{2}, so cos⁡θ=−12\cos\theta=-\frac{1}{2} and θ=120∘\theta=120^\circ.

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