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If a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} are three unit vectors such that a⃗+b⃗+c⃗=0\vec{a} + \vec{b} + \vec{c} = 0, then the value of a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} is

Solution

✅ Correct Option: 2

Square the magnitude: ∣a⃗+b⃗+c⃗∣2=0|\vec{a}+\vec{b}+\vec{c}|^2 = 0.

Expand: ∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0.

Since each is a unit vector: 1+1+1+2S=01+1+1+2S = 0, so S=−32S = -\frac{3}{2}.

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