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Question
Find the foot of the perpendicular from the point (1, 2, 0) upon the plane x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane.
Solution
The equation of the line perpendicular to the plane and passing through the point (1, 2, 0) is `(x - 1)/1 = (y - 2)/(-3) = z/2`
The coordinates of the foot of the perpendicular are `(mu + 1, -3mu + 2, 2mu)` for some `mu`
These coordinates will satisfy the equation of the plane. Hence, we have `mu + 1 - 3(-3mu + 2) + 2(2mu)` = 9
⇒ `mu` = 1
The foot of the perpendicular is (2, –1, 2)
Hence, the required distance = `sqrt((1 - 2)^2 + (2 + 1)^2 + (0 - 2)^2) = sqrt(14)` units
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