The distance between the parallel lines frac{ | vedclass

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(B) The given lines are $L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ and $L_2: \vec{r} = (0\hat{i} + 0\ha

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$\frac{5 \sqrt{5}}{3}$ units

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(B) The given lines are $L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ and $L_2: \vec{r} = (0\hat{i} + 0\hat{j} + 0\hat{k}) + \mu(2\hat{i} - 2\hat{j} + \hat{k})$.Here,$\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{a_2} = 0$,and $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$.The distance $d$ between two parallel lines is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) 	imes \vec{b}|}{|\vec{b}|}$.First,$\vec{a_2} - \vec{a_1} = -\hat{i} - 2\hat{j} - 3\hat{k}$.Now,$(\vec{a_2} - \vec{a_1}) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -2 & -3 \ 2 & -2 & 1 \end{vmatrix} = \hat{i}(-2 - 6) - \hat{j}(-1 + 6) + \hat{k}(2 + 4) = -8\hat{i} - 5\hat{j} + 6\hat{k}$.The magnitude is $|(\vec{a_2} - \vec{a_1}) 	imes \vec{b}| = \sqrt{(-8)^2 + (-5)^2 + 6^2} = \sqrt{64 + 25 + 36} = \sqrt{125} = 5\sqrt{5}$.The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Thus,the distance $d = \frac{5\sqrt{5}}{3}$ units.

The distance between the parallel lines $\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1}$ and $\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$ is

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