The shortest distance between the line r = 2h | vedclass

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

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

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

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

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