If bar{x}=frac{bar{b} times bar{c}}{[bar{a} b | vedclass

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(A) Given $\bar{x}=\frac{\bar{b} 	imes \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{y}=\frac{\bar{c} 	imes \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \bar{z}

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$3$

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(A) Given $\bar{x}=\frac{\bar{b} 	imes \bar{c}}{[\bar{a} \bar{b} \bar{c}]}, \bar{y}=\frac{\bar{c} 	imes \bar{a}}{[\bar{a} \bar{b} \bar{c}]}, \bar{z}=\frac{\bar{a} 	imes \bar{b}}{[\bar{a} \bar{b} \bar{c}]}$.We need to evaluate $S = \bar{x} \cdot(\bar{a}+\bar{b})+\bar{y} \cdot(\bar{b}+\bar{c})+\bar{z} \cdot(\bar{c}+\bar{a})$.Substituting the values of $\bar{x}, \bar{y}, \bar{z}$:$S = \frac{(\bar{b} 	imes \bar{c}) \cdot (\bar{a}+\bar{b})}{[\bar{a} \bar{b} \bar{c}]} + \frac{(\bar{c} 	imes \bar{a}) \cdot (\bar{b}+\bar{c})}{[\bar{a} \bar{b} \bar{c}]} + \frac{(\bar{a} 	imes \bar{b}) \cdot (\bar{c}+\bar{a})}{[\bar{a} \bar{b} \bar{c}]}$.Using the property of scalar triple product $[\bar{a} \bar{b} \bar{c}] = (\bar{a} 	imes \bar{b}) \cdot \bar{c}$ and the fact that the scalar triple product is zero if any two vectors are identical:$S = \frac{[\bar{b} \bar{c} \bar{a}] + [\bar{b} \bar{c} \bar{b}] + [\bar{c} \bar{a} \bar{b}] + [\bar{c} \bar{a} \bar{c}] + [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{a}]}{[\bar{a} \bar{b} \bar{c}]}$.Since $[\bar{b} \bar{c} \bar{b}] = 0, [\bar{c} \bar{a} \bar{c}] = 0, [\bar{a} \bar{b} \bar{a}] = 0$ and $[\bar{b} \bar{c} \bar{a}] = [\bar{c} \bar{a} \bar{b}] = [\bar{a} \bar{b} \bar{c}]$,we get:$S = \frac{[\bar{a} \bar{b} \bar{c}] + 0 + [\bar{a} \bar{b} \bar{c}] + 0 + [\bar{a} \bar{b} \bar{c}] + 0}{[\bar{a} \bar{b} \bar{c}]} = \frac{3[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 3$.

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