Find the distance between the line vec{r} = ( | vedclass

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(B) The line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$,where $\vec{a} = 2\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 4\ha

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$\frac{10}{3\sqrt{3}}$

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(B) The line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$,where $\vec{a} = 2\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 4\hat{k}$.The plane is given by $\vec{r} \cdot \vec{n} = d$,where $\vec{n} = \hat{i} + 5\hat{j} + \hat{k}$ and $d = 5$.First,check if the line is parallel to the plane by calculating $\vec{b} \cdot \vec{n} = (1)(1) + (-1)(5) + (4)(1) = 1 - 5 + 4 = 0$.Since $\vec{b} \cdot \vec{n} = 0$,the line is parallel to the plane.The distance $D$ between a parallel line and a plane is given by $D = \frac{|\vec{a} \cdot \vec{n} - d|}{||\vec{n}||}$.Calculate $\vec{a} \cdot \vec{n} = (2)(1) + (-2)(5) + (3)(1) = 2 - 10 + 3 = -5$.Calculate $||\vec{n}|| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}$.Thus,$D = \frac{|-5 - 5|}{3\sqrt{3}} = \frac{|-10|}{3\sqrt{3}} = \frac{10}{3\sqrt{3}}$.

Find the distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$.

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