The distance between the line r = 2i - 2j + 3 | vedclass

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(D) The line is given by $r = a + \lambda b$,where $a = 2i - 2j + 3k$ and $b = i - j + 4k$. The plane is given by $r \cdot n = d$,where $n = i + 5j +

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

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

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

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