The distance of the point (1, 3, – 7) from the plane passing through the point (1, – 1, – 1) having normal perpendicular to both the lines and is

Engineering

Mathematics

Plane and Its Different Forms

Question

The distance of the point (1, 3, – 7) from the plane passing through the point (1, – 1, – 1) having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $= \frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ is

$\frac{10}{\sqrt{74}}$

$\frac{10}{\sqrt{83}}$

$\frac{20}{\sqrt{74}}$

$\frac{5}{\sqrt{83}}$

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Equation of plane: a (x – 1) + b (y + 1) + c (z + 1) = 0 this plane perpendicular to given two lines, then a – 2b + 3c = 0

2a – b – c = 0

$\frac{a}{5} = \frac{b}{7} = \frac{c}{3}$

then equation of plane is 5 (x – 1) + 7 (y + 1) + 3 (z + 1) = 0

5x + 7y + 3z + 5 = 0

$d = | \frac{5 + 21 - 21 + 5}{\sqrt{25 + 49 + 9}} | = \frac{10}{\sqrt{83}} .$