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(C) Let the expression be $E = (\overrightarrow{a}+2 \overrightarrow{b}-\overrightarrow{c}) \cdot ((\overrightarrow{a}-\overrightarrow{b}) 	imes (\ov

Vedclass · Accepted Answer

$3[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$

Vedclass · Answer

(C) Let the expression be $E = (\overrightarrow{a}+2 \overrightarrow{b}-\overrightarrow{c}) \cdot ((\overrightarrow{a}-\overrightarrow{b}) 	imes (\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}))$.First, simplify the cross product term: $(\overrightarrow{a}-\overrightarrow{b}) 	imes (\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c})$.$= (\overrightarrow{a} 	imes \overrightarrow{a}) - (\overrightarrow{a} 	imes \overrightarrow{b}) - (\overrightarrow{a} 	imes \overrightarrow{c}) - (\overrightarrow{b} 	imes \overrightarrow{a}) + (\overrightarrow{b} 	imes \overrightarrow{b}) + (\overrightarrow{b} 	imes \overrightarrow{c})$.Since $\overrightarrow{a} 	imes \overrightarrow{a} = \overrightarrow{0}$ and $\overrightarrow{b} 	imes \overrightarrow{b} = \overrightarrow{0}$, and using $\overrightarrow{b} 	imes \overrightarrow{a} = -(\overrightarrow{a} 	imes \overrightarrow{b})$, we get:$= \overrightarrow{0} - (\overrightarrow{a} 	imes \overrightarrow{b}) - (\overrightarrow{a} 	imes \overrightarrow{c}) + (\overrightarrow{a} 	imes \overrightarrow{b}) + \overrightarrow{0} + (\overrightarrow{b} 	imes \overrightarrow{c}) = (\overrightarrow{b} 	imes \overrightarrow{c}) - (\overrightarrow{a} 	imes \overrightarrow{c}) = (\overrightarrow{b} 	imes \overrightarrow{c}) + (\overrightarrow{c} 	imes \overrightarrow{a})$.Now, $E = (\overrightarrow{a}+2 \overrightarrow{b}-\overrightarrow{c}) \cdot ((\overrightarrow{b} 	imes \overrightarrow{c}) + (\overrightarrow{c} 	imes \overrightarrow{a}))$.Using the definition of scalar triple product $[\overrightarrow{u} \overrightarrow{v} \overrightarrow{w}] = \overrightarrow{u} \cdot (\overrightarrow{v} 	imes \overrightarrow{w})$:$E = \overrightarrow{a} \cdot (\overrightarrow{b} 	imes \overrightarrow{c}) + \overrightarrow{a} \cdot (\overrightarrow{c} 	imes \overrightarrow{a}) + 2\overrightarrow{b} \cdot (\overrightarrow{b} 	imes \overrightarrow{c}) + 2\overrightarrow{b} \cdot (\overrightarrow{c} 	imes \overrightarrow{a}) - \overrightarrow{c} \cdot (\overrightarrow{b} 	imes \overrightarrow{c}) - \overrightarrow{c} \cdot (\overrightarrow{c} 	imes \overrightarrow{a})$.Note that any scalar triple product with repeated vectors is $0$ (e.g.,$\overrightarrow{a} \cdot (\overrightarrow{c} 	imes \overrightarrow{a}) = 0$).$E = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] + 0 + 0 + 2[\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}] - 0 - 0$.Since $[\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}] = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$, we have $E = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] + 2[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 3[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$.

$(\overrightarrow{a}+2 \overrightarrow{b}-\overrightarrow{c}) \cdot ((\overrightarrow{a}-\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}))$ is equal to

Similar Questions

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