Let overrightarrow{a} = hat{i} + 2hat{j} - ha | vedclass

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(C) Given $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{c} 	imes \overrightarrow{a}$,we can write $(\overrightarrow{r} - \overright

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Given $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{c} 	imes \overrightarrow{a}$,we can write $(\overrightarrow{r} - \overrightarrow{c}) 	imes \overrightarrow{a} = 0$.This implies that $\overrightarrow{r} - \overrightarrow{c}$ is parallel to $\overrightarrow{a}$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda \overrightarrow{a}$ for some scalar $\lambda$.Given $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,we substitute $\overrightarrow{r}$:$(\overrightarrow{c} + \lambda \overrightarrow{a}) \cdot \overrightarrow{b} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{b} + \lambda (\overrightarrow{a} \cdot \overrightarrow{b}) = 0$.Calculating the dot products:$\overrightarrow{c} \cdot \overrightarrow{b} = (1)(1) + (-1)(-1) + (-1)(0) = 1 + 1 = 2$.$\overrightarrow{a} \cdot \overrightarrow{b} = (1)(1) + (2)(-1) + (-1)(0) = 1 - 2 = -1$.Thus,$2 + \lambda(-1) = 0 \Rightarrow \lambda = 2$.Now,$\overrightarrow{r} \cdot \overrightarrow{a} = (\overrightarrow{c} + 2\overrightarrow{a}) \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a} + 2|\overrightarrow{a}|^2$.$\overrightarrow{c} \cdot \overrightarrow{a} = (1)(1) + (-1)(2) + (-1)(-1) = 1 - 2 + 1 = 0$.$|\overrightarrow{a}|^2 = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6$.Therefore,$\overrightarrow{r} \cdot \overrightarrow{a} = 0 + 2(6) = 12$.

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