If the distance of the point P(l, – 2, 1) from the plane x + 2y - 2z = a , where a > 0. is 5, then the foot of the perpendicular from P to the plane is

Engineering

Mathematics

Plane and Its Different Forms

Question

$(\frac{4}{3}, - \frac{4}{3}, \frac{1}{3})$

$(\frac{8}{3}, \frac{4}{3}, - \frac{7}{3})$

$(\frac{1}{3}, \frac{2}{3}, \frac{10}{3})$

$(\frac{2}{3}, - \frac{1}{3}, \frac{5}{2})$

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Perpendicular distance $= | \frac{1 - 4 - 2 - α}{3} | = 5$

$\Rightarrow | α + 5 | = 15 \Rightarrow α = 10 (as α > 5)$

foot of the perpendicular $\frac{x - 1}{1} = \frac{y - (- 2)}{2} = \frac{z - 1}{- 2} = \frac{- (1 - 4 - 2 - 10)}{9}$

$or x = \frac{8}{3}, y = \frac{4}{3} and z = \frac{- 7}{3}$