If bar{a}, bar{b} and bar{c} are any three no | vedclass

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(C) Let the expression be $E = (\bar{a}+2 \bar{b}+\bar{c}) \cdot [(\bar{a}-\bar{b}) 	imes (\bar{a}-\bar{b}-\bar{c})]$.First,simplify the cross produc

Vedclass · Accepted Answer

$3\left[\bar{a} \bar{b} \bar{c}\right]$

Vedclass · Answer

(C) Let the expression be $E = (\bar{a}+2 \bar{b}+\bar{c}) \cdot [(\bar{a}-\bar{b}) 	imes (\bar{a}-\bar{b}-\bar{c})]$.First,simplify the cross product part:$(\bar{a}-\bar{b}) 	imes (\bar{a}-\bar{b}-\bar{c}) = (\bar{a} 	imes \bar{a}) - (\bar{a} 	imes \bar{b}) - (\bar{a} 	imes \bar{c}) - (\bar{b} 	imes \bar{a}) + (\bar{b} 	imes \bar{b}) + (\bar{b} 	imes \bar{c})$.Since $\bar{a} 	imes \bar{a} = 0$,$\bar{b} 	imes \bar{b} = 0$,and $\bar{b} 	imes \bar{a} = -(\bar{a} 	imes \bar{b})$,we get:$= 0 - (\bar{a} 	imes \bar{b}) - (\bar{a} 	imes \bar{c}) + (\bar{a} 	imes \bar{b}) + 0 + (\bar{b} 	imes \bar{c}) = (\bar{b} 	imes \bar{c}) - (\bar{a} 	imes \bar{c}) = (\bar{b} 	imes \bar{c}) + (\bar{c} 	imes \bar{a})$.Now,compute the dot product:$E = (\bar{a}+2 \bar{b}+\bar{c}) \cdot (\bar{b} 	imes \bar{c} + \bar{c} 	imes \bar{a})$.$= \bar{a} \cdot (\bar{b} 	imes \bar{c}) + \bar{a} \cdot (\bar{c} 	imes \bar{a}) + 2\bar{b} \cdot (\bar{b} 	imes \bar{c}) + 2\bar{b} \cdot (\bar{c} 	imes \bar{a}) + \bar{c} \cdot (\bar{b} 	imes \bar{c}) + \bar{c} \cdot (\bar{c} 	imes \bar{a})$.Using properties of scalar triple products,terms like $\bar{a} \cdot (\bar{c} 	imes \bar{a}) = 0$ and $\bar{b} \cdot (\bar{b} 	imes \bar{c}) = 0$ vanish.$E = [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 2[\bar{b} \bar{c} \bar{a}] + 0 + 0$.Since $[\bar{b} \bar{c} \bar{a}] = [\bar{a} \bar{b} \bar{c}]$,we have:$E = [\bar{a} \bar{b} \bar{c}] + 2[\bar{a} \bar{b} \bar{c}] = 3[\bar{a} \bar{b} \bar{c}]$.

If $\bar{a}, \bar{b}$ and $\bar{c}$ are any three non-zero vectors,then $(\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=$

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