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`