Let overrightarrow{a} = 2hat{i} - 7hat{j} + 5 | vedclass

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Question

(A) Given $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j}

Vedclass · Accepted Answer

$\frac{11}{7} \sqrt{2}$

Vedclass · Answer

(A) Given $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j} - 3\hat{k}$.From the condition $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{c} 	imes \overrightarrow{a}$,we have $(\overrightarrow{r} - \overrightarrow{c}) 	imes \overrightarrow{a} = 0$.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 $\overrightarrow{r}$:$(\overrightarrow{c} + \lambda\overrightarrow{a}) \cdot \overrightarrow{b} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{b} + \lambda(\overrightarrow{a} \cdot \overrightarrow{b}) = 0$.Calculate the dot products:$\overrightarrow{c} \cdot \overrightarrow{b} = (1)(1) + (2)(0) + (-3)(1) = 1 - 3 = -2$.$\overrightarrow{a} \cdot \overrightarrow{b} = (2)(1) + (-7)(0) + (5)(1) = 2 + 5 = 7$.Substituting these values: $-2 + 7\lambda = 0 \Rightarrow \lambda = \frac{2}{7}$.Now,$\overrightarrow{r} = \overrightarrow{c} + \frac{2}{7}\overrightarrow{a} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \frac{2}{7}(2\hat{i} - 7\hat{j} + 5\hat{k}) = (1 + \frac{4}{7})\hat{i} + (2 - 2)\hat{j} + (-3 + \frac{10}{7})\hat{k} = \frac{11}{7}\hat{i} - \frac{11}{7}\hat{k}$.Finally,$|\overrightarrow{r}| = \sqrt{(\frac{11}{7})^2 + 0^2 + (-\frac{11}{7})^2} = \sqrt{2 	imes (\frac{11}{7})^2} = \frac{11}{7}\sqrt{2}$.

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