Let overrightarrow{a} = 2hat{i} - 3hat{j} + 4 | vedclass

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(A) Given $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{r} 	imes \overrightarrow{b}$,we have $\overrightarrow{r} 	imes (\overright

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Given $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{r} 	imes \overrightarrow{b}$,we have $\overrightarrow{r} 	imes (\overrightarrow{a} - \overrightarrow{b}) = 0$.This implies $\overrightarrow{r}$ is parallel to $(\overrightarrow{a} - \overrightarrow{b})$.Calculate $\overrightarrow{a} - \overrightarrow{b} = (2-7)\hat{i} + (-3-1)\hat{j} + (4 - (-6))\hat{k} = -5\hat{i} - 4\hat{j} + 10\hat{k}$.So,$\overrightarrow{r} = \lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})$.Given $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,substitute $\overrightarrow{r}$:$\lambda(-5\hat{i} - 4\hat{j} + 10\hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$.$\lambda(-5 - 8 + 10) = -3 \Rightarrow -3\lambda = -3 \Rightarrow \lambda = 1$.Thus,$\overrightarrow{r} = -5\hat{i} - 4\hat{j} + 10\hat{k}$.Finally,calculate $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = (-5)(2) + (-4)(-3) + (10)(1) = -10 + 12 + 10 = 12$.

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