The distance between the lines frac{x}{2}=fra | vedclass

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(NONE) The given lines are parallel because their direction ratios are the same: $(2, -1, 2)$.Let the lines be $L_1: \vec{r} = 0\hat{i} + 0\hat{j} + 0

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$\sqrt{2}$ units

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(NONE) The given lines are parallel because their direction ratios are the same: $(2, -1, 2)$.Let the lines be $L_1: \vec{r} = 0\hat{i} + 0\hat{j} + 0\hat{k} + \lambda(2\hat{i} - \hat{j} + 2\hat{k})$ and $L_2: \vec{r} = 1\hat{i} + 1\hat{j} + 2\hat{k} + \mu(2\hat{i} - \hat{j} + 2\hat{k})$.Here,$\vec{a_1} = (0, 0, 0)$,$\vec{a_2} = (1, 1, 2)$,and $\vec{b} = (2, -1, 2)$.The distance $d$ between two parallel lines is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) 	imes \vec{b}|}{|\vec{b}|}$.$\vec{a_2} - \vec{a_1} = (1-0)\hat{i} + (1-0)\hat{j} + (2-0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$.$(\vec{a_2} - \vec{a_1}) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 2 \ 2 & -1 & 2 \end{vmatrix} = \hat{i}(2 - (-2)) - \hat{j}(2 - 4) + \hat{k}(-1 - 2) = 4\hat{i} + 2\hat{j} - 3\hat{k}$.Magnitude $|(\vec{a_2} - \vec{a_1}) 	imes \vec{b}| = \sqrt{4^2 + 2^2 + (-3)^2} = \sqrt{16 + 4 + 9} = \sqrt{29}$.Magnitude $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.Thus,$d = \frac{\sqrt{29}}{3}$ units. Note: The provided options do not match the calculated result. Based on standard evaluation,the correct distance is $\frac{\sqrt{29}}{3}$.

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

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