Q. 245.0( 1 Vote )

# Find the coordinates of the foot of perpendicular, and the perpendicular distance from the point P(4, 3, 2) to the plane Also find the image of P in the plane.

Let Q(α, β, γ) be the image of the given point P(4, 3, 2) in the plane x + 2y + 3z = 2

PQ is the perpendicular to the plane and R is the midpoint of PQ and the foot of the perpendicular.

Here, direction ratios of PR are 1, 2, 3

x – 4 = λ, y – 3 = 2λ and z – 2 = 3λ

x = λ + 4, y = 2λ + 3 and z = 3λ + 2

Let the coordinate of R be (λ + 4, 2λ + 3, 3λ + 2)

If this point lies on plane, then

λ + 4 + 2(2λ + 3) + 3(3λ + 2) = 2

λ + 4 + 4λ + 6 + 9λ + 6 = 2

14 λ + 16 = 2

14 λ = 2 – 16

14 λ = -14

λ = -1

So, point R is (-1 + 4, 2(-1) + 3, 3(-1) + 2) = (3, 1, -1)

and above we suppose that Q is the image of P

By comparing both sides, we get

So, image point Q is (2, -1, -4)

Now, we have to find the perpendicular distance PR

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