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

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(C) Given $\overrightarrow{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overrightarrow{c} = 5\hat{i}

Vedclass · Accepted Answer

$339$

Vedclass · Answer

(C) Given $\overrightarrow{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overrightarrow{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$.From $\overrightarrow{r} 	imes \overrightarrow{b} = \overrightarrow{c} 	imes \overrightarrow{b}$,we have $(\overrightarrow{r} - \overrightarrow{c}) 	imes \overrightarrow{b} = 0$.This implies $\overrightarrow{r} - \overrightarrow{c} = \lambda \overrightarrow{b}$ for some scalar $\lambda$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda \overrightarrow{b}$.Given $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,we substitute $\overrightarrow{r}$:$(\overrightarrow{c} + \lambda \overrightarrow{b}) \cdot \overrightarrow{a} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{a} + \lambda(\overrightarrow{b} \cdot \overrightarrow{a}) = 0$.Calculate dot products:$\overrightarrow{c} \cdot \overrightarrow{a} = (5)(1) + (-3)(2) + (3)(3) = 5 - 6 + 9 = 8$.$\overrightarrow{b} \cdot \overrightarrow{a} = (1)(1) + (-1)(2) + (2)(3) = 1 - 2 + 6 = 5$.Thus,$8 + 5\lambda = 0 \Rightarrow \lambda = -\frac{8}{5}$.Now,$\overrightarrow{r} = \overrightarrow{c} - \frac{8}{5}\overrightarrow{b} = (5\hat{i} - 3\hat{j} + 3\hat{k}) - \frac{8}{5}(\hat{i} - \hat{j} + 2\hat{k}) = \frac{1}{5}(25\hat{i} - 15\hat{j} + 15\hat{k} - 8\hat{i} + 8\hat{j} - 16\hat{k}) = \frac{1}{5}(17\hat{i} - 7\hat{j} - \hat{k})$.$|\overrightarrow{r}|^2 = \frac{1}{25}(17^2 + (-7)^2 + (-1)^2) = \frac{1}{25}(289 + 49 + 1) = \frac{339}{25}$.Therefore,$25|\overrightarrow{r}|^2 = 339$.

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