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Question
Find the coordinates of the foot of perpendicular and perpendicular distance from the point P(4,3,2) to the plane x + 2y + 3z = 2. Also find the image of P in the plane.
Solution
Let `pi : "x" + 2"y" + 3"z" = 2.`
The d.r.'s of its normal are 1, 2, 3.
Draw PR ⊥ plane `pi`.
So, equation of PR : `("x"-4)/(1) = ("y"-3)/(2) = ("z"-2)/(3) = λ "say"`
Coordinates of any random point on the line PR is,
R(λ+ 4,2λ +3λ +2).
As R lies on the plane so,
λ + 4 +2(2λ + 3) + 3(3λ + 3) = 2
⇒ -1.
So, foot of perpendicular is R(3. 1, -1).
And, perpendicular distance, PR = `sqrt((3-4)^2 + (1-3)^2 + (-1-2)^2)`
∴ PR = `sqrt14 "units"`.
Let Q(h, p,s) be the image of point P in the plane.
So, R must be the mid-point of PQ.
That is, `"R" (("h"+4)/(2),("p"+3)/(2),("s"+2)/(2)) = "R"(3,1,-1)`
On comparing the coordinates, `("h"+4)/(2)=3,("p"+3)/(2)=1,("s"+2)/(2)=-1`
⇒ `"h" = 2, "p" =-1, "s"=-4`
Hence the image is Q(2, -1, -4).
Perpendicular distance PR = sqrt14`
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