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Question
Find the co-ordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2.
Solution
Let us take the point P(4, 3, 2) and Q(x1, y1, z1)
⇒ (x1 – 4, y1 – 3, z1 – 2)
Plane x + 2y + 3z = 2 ........(1)
Cartesian equation of PQ, `(x_1 - 4)/1 = (y_1 - 3)/2 = (x_3 - 2)/3` = λ
(λ + 4, 2λ + 3, 3λ + 2) lies in (1)
(λ + 4) + 2(2λ + 3) + 3(3λ + 2) – 2 = 0
λ + 4 + 4λ + 6 + 9λ + 6 – 2 = 0
14λ + 14 = 0
14λ = – 14
λ = – 1
Co-ordinates of the foot of the ⊥r is (3, 1, – 1).
Distance PQ = `sqrt((4 - 3)^2 + (3 - 1)^2 + (2 + 1)^2)`
= `sqrt(1^2 + 2^2 + 3^2)`
= `sqrt(1 + 4 + 9)`
= `sqrt(14)` units.
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