The distance between the line bar{r} = 3hat{i | vedclass

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(A) The equation of the line is $\bar{r} = \bar{a} + \lambda \bar{b}$,where $\bar{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\bar{b} = \hat{i} - 2\hat{j

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$\frac{1}{\sqrt{6}}$ units

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(A) The equation of the line is $\bar{r} = \bar{a} + \lambda \bar{b}$,where $\bar{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\bar{b} = \hat{i} - 2\hat{j}$.The equation of the plane is $\bar{r} \cdot \bar{n} = d$,where $\bar{n} = 2\hat{i} + \hat{j} + \hat{k}$ and $d = 4$.First,check if the line is parallel to the plane by calculating $\bar{b} \cdot \bar{n} = (1)(2) + (-2)(1) + (0)(1) = 2 - 2 + 0 = 0$.Since $\bar{b} \cdot \bar{n} = 0$,the line is parallel to the plane.The distance $D$ between a parallel line and a plane is given by $D = \frac{|\bar{a} \cdot \bar{n} - d|}{|\bar{n}|}$.Calculate $\bar{a} \cdot \bar{n} = (3)(2) + (-2)(1) + (1)(1) = 6 - 2 + 1 = 5$.Calculate $|\bar{n}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.Thus,$D = \frac{|5 - 4|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$ units.

The distance between the line $\bar{r} = 3\hat{i} - 2\hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and the plane $\bar{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 4$ is

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