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(C) Given that $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar vectors,the scalar triple product $[\overline{a} \overline{b} \overline{c}]

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given that $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar vectors,the scalar triple product $[\overline{a} \overline{b} \overline{c}] 
eq 0$. We know that $\overline{a} \cdot \overline{p} = \overline{a} \cdot \frac{\overline{b} 	imes \overline{c}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{a} \overline{b} \overline{c}]}{[\overline{a} \overline{b} \overline{c}]} = 1$. Similarly,$\overline{b} \cdot \overline{q} = \overline{b} \cdot \frac{\overline{c} 	imes \overline{a}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{b} \overline{c} \overline{a}]}{[\overline{a} \overline{b} \overline{c}]} = 1$. And $\overline{c} \cdot \overline{r} = \overline{c} \cdot \frac{\overline{a} 	imes \overline{b}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{c} \overline{a} \overline{b}]}{[\overline{a} \overline{b} \overline{c}]} = 1$. Substituting these values into the expression: $2 \overline{a} \cdot \overline{p} + \overline{b} \cdot \overline{q} + \overline{c} \cdot \overline{r} = 2(1) + 1 + 1 = 4$.

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