If r→=x1(a→×b→)+x2(b→×a→)+x3(c→×d→) and 4[a→b→c→]=1, then x1+x2+x3 is equal to

r→=x1(a→×b→)+x2(b→×c→)+x3(c→×a→)⇒ r→⋅a→=x2[a→b→c→],r→⋅b→=x3[b→c→a→] and r→⋅c→=x1[c→a→b→]=x1[a→b→c→]⇒ x1+x2+x3=4r→⋅(a→+b→+c→)

If r→=x1(a→×b→)+x2(b→×a→)+x3(c→×d→) and 4[a→b→c→]=1, then x1+x2+x3 is equal to

r→=x1(a→×b→)+x2(b→×c→)+x3(c→×a→)⇒ r→⋅a→=x2[a→b→c→],r→⋅b→=x3[b→c→a→] and r→⋅c→=x1[c→a→b→]=x1[a→b→c→]⇒ x1+x2+x3=4r→⋅(a→+b→+c→)

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If $\vec{r} = x_{1} (\vec{a} \times \vec{b}) + x_{2} (\vec{b} \times \vec{a}) + x_{3} (\vec{c} \times \vec{d}) and 4 [\vec{a} \vec{b} \vec{c}] = 1, then x_{1} + x_{2} + x_{3}$ is equal to

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$- \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$

$4 \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$

$2 \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$

$\frac{1}{2} \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$

answer is D.

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Detailed Solution

$\begin{array}{l} \vec{r} = x_{1} (\vec{a} \times \vec{b}) + x_{2} (\vec{b} \times \vec{c}) + x_{3} (\vec{c} \times \vec{a}) \\ \Rightarrow \vec{r} \cdot \vec{a} = x_{2} [\vec{a} \vec{b} \vec{c}], \vec{r} \cdot \vec{b} = x_{3} [\vec{b} \vec{c} \vec{a}] \\ and \vec{r} \cdot \vec{c} = x_{1} [\vec{c} \vec{a} \vec{b}] = x_{1} [\vec{a} \vec{b} \vec{c}] \\ \Rightarrow x_{1} + x_{2} + x_{3} = 4 \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) \end{array}$