Distance between the parallel lines frac{x}{3 | vedclass

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(A) The vector equations of the given lines are $\bar{r}=\hat{j}+\lambda(3 \hat{i}-2 \hat{j}+\hat{k})$ and $\bar{r}=-4 \hat{i}+3 \hat{j}-2 \hat{k}+\mu

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$\sqrt{\frac{6}{7}}$ units

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(A) The vector equations of the given lines are $\bar{r}=\hat{j}+\lambda(3 \hat{i}-2 \hat{j}+\hat{k})$ and $\bar{r}=-4 \hat{i}+3 \hat{j}-2 \hat{k}+\mu(3 \hat{i}-2 \hat{j}+\hat{k})$.The distance between the parallel lines $\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\mu \bar{b}$ is given by $d=\left|\frac{(\bar{a}_2-\bar{a}_1) 	imes \bar{b}}{|\bar{b}|}ight|$.Here,$\bar{a}_1=\hat{j}$,$\bar{a}_2=-4 \hat{i}+3 \hat{j}-2 \hat{k}$,and $\bar{b}=3 \hat{i}-2 \hat{j}+\hat{k}$.Therefore,$\bar{a}_2-\bar{a}_1=-4 \hat{i}+2 \hat{j}-2 \hat{k}$.The cross product $(\bar{a}_2-\bar{a}_1) 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & 2 & -2 \ 3 & -2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(-4+6) + \hat{k}(8-6) = -2 \hat{i}-2 \hat{j}+2 \hat{k}$.The magnitude $|\bar{b}| = \sqrt{3^2+(-2)^2+1^2} = \sqrt{9+4+1} = \sqrt{14}$.Thus,$d = \frac{|-2 \hat{i}-2 \hat{j}+2 \hat{k}|}{\sqrt{14}} = \frac{\sqrt{(-2)^2+(-2)^2+2^2}}{\sqrt{14}} = \sqrt{\frac{4+4+4}{14}} = \sqrt{\frac{12}{14}} = \sqrt{\frac{6}{7}}$ units.

Distance between the parallel lines $\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}$ and $\frac{x+4}{3}=\frac{y-3}{-2}=\frac{z+2}{1}$ is

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