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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.
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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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