What is the distance between the line r = (2i | vedclass

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(D) The distance $d$ between a line $r = a + \lambda b$ and a plane $r \cdot n = p$ is given by the formula: $d = \left| \frac{p - a \cdot n}{|n|} \ri

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

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(D) The distance $d$ between a line $r = a + \lambda b$ and a plane $r \cdot n = p$ is given by the formula: $d = \left| \frac{p - a \cdot n}{|n|} \right|$.Given $a = 2i - 2j + 3k$,$b = i - j + 4k$,$n = i + 5j + k$,and $p = 5$.First,calculate the dot product $a \cdot n = (2)(1) + (-2)(5) + (3)(1) = 2 - 10 + 3 = -5$.Next,calculate the magnitude $|n| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}$.Substitute these values into the formula:$d = \left| \frac{5 - (-5)}{3\sqrt{3}} \right| = \left| \frac{10}{3\sqrt{3}} \right| = \frac{10}{3\sqrt{3}}$.

What is the distance between the line $r = (2i - 2j + 3k) + \lambda (i - j + 4k)$ and the plane $r \cdot (i + 5j + k) = 5$?

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