Aspire Faculty ID #3644 · Topic: NIMCET 2012 · Just now
NIMCET 2012

If $\mathbf{a},\; \mathbf{b},\; \mathbf{c}$ are unit vectors such that $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$, then the value of $\mathbf{a}\cdot \mathbf{b} + \mathbf{b}\cdot \mathbf{c} + \mathbf{c}\cdot \mathbf{a}$ is:

Solution

Given: $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$ 
 Square both sides: 
 $(\mathbf{a} + \mathbf{b} + \mathbf{c})\cdot(\mathbf{a} + \mathbf{b} + \mathbf{c}) = 0$ 
 $\mathbf{a}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{b} + \mathbf{c}\cdot\mathbf{c} + 2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = 0$ 
 Since each of $\mathbf{a},\;\mathbf{b},\;\mathbf{c}$ is a unit vector: $\mathbf{a}\cdot\mathbf{a} = \mathbf{b}\cdot\mathbf{b} = \mathbf{c}\cdot\mathbf{c} = 1$ 
 Thus: $3 + 2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = 0$ 
 $2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = -3$
 $\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a} = -\dfrac{3}{2}$ 
 Therefore: ${-\dfrac{3}{2}}$

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