Find the length and the foot of the perpendicular from the point P (7, 14, 5) to the plane2x + 4y - z = 2. Also find the image of point P in the plane.

Given: a point P (7, 14, 5), plane 2x + 4y - z = 2

To find: the length and the foot of the perpendicular from the point P to the given plane. Also find the image of point P in the plane.

Let M(α, β, γ) be the foot of the perpendicular from the point P (7, 14, 5) to the plane 2x + 4y - z = 2.

Then PM is the normal to the plane.

So its direction ratios are 2, 4,-1

Now as PM passes through the point P(7, 14, 5). So its equation will be

Then from this

Similarly,

Then these are the coordinates of point M(2k+7, 4k+14, -k+5)

Now as M lies on plane 2x+4y-z=2, so

2α+4β-γ=2

Substituting the values, of α, β, γ, we get

2(2k+7)+4(4k+14)-(-k+5)=2

⇒ 4k+14+16k+56+k-5=2

⇒ 21k+65=2

⇒ 21k=2-65

⇒ 21k=-63

⇒ k=-3

Substituting the value of k in coordinates of M, we get

M(2k+7, 4k+14, -k+5)

⇒ M(2(-3)+7, 4(-3)+14, -(-3)+5)

⇒ M(-6+7, -12+14, 3+5)

⇒ M(1, 2, 8)

Hence the coordinate of the foot of the perpendicular from the point P(7, 14, 5) to the given plane is (1,2, 8)

Now we will find the length of this perpendicular

So

Or The length of the perpendicular is √189 or 3√21 units.

Now Let P’(x₁, y₁, z₁) be the image of point P (7, 14, 5) in the plane 2x+4y-z=2

Now from figure it is clear that M(1, 2, 8) is the midpoint of line PP’

Hence now applying the formula

Hence the image of the given point P(7, 14, 5) is P’(-5, -10, 11).