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If \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are three unit vectors such that \(\mathbf{a}+\mathbf{b}+\mathbf{c}=0\), then \(\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot\mathbf{c}+\mathbf{c}\cdot\mathbf{a}\) is equal to

1. \(-1\)
2. 3
3. 0
4. \(-\frac{3}{2}\)

Correct answer: \(-\frac{3}{2}\)

Solution

From \((\mathbf{a}+\mathbf{b}+\mathbf{c})^2=0\), we get \(1+1+1+2(\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot\mathbf{c}+\mathbf{c}\cdot\mathbf{a})=0\). Therefore the sum of pairwise dot products is \(-\frac{3}{2}\).

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