Point on a plane?

Question

Cant figure out this question, can anyone help me out please?

Find the point P on the plane 0 x + 2 y + 1 z = -5 closest to the point Q(-4,2,1).

P = P( ?,?,?).

Answer 1

The closest point P on the plane to the point Q(-4,2,1) will be on the line whose directional vector is also the plane's normal vector n.

n = <0, 2, 1>

The equation of the line L is:

L = OQ + tn = <-4, 2, 1> + t<0, 2, 1>
L = <-4, 2 + 2t, 1 + t>
where t is a constant ranging over the real numbers

The equation of the line and plane will be equal at the point of intersection. Rewrite the equation of the plane in terms of t.

0x + 2y + z = -5
2y + z = -5
2(2 + 2t) + (1 + t) = -5
4 + 4t + 1 + t = -5
5t + 5 = -5
5t = -10
t = -2

Find the point of intersection P.

P:
x = -4 + 0t = -4
y = 2 + 2t = 2 - 4 = -2
z = 1 + t = 1 - 2 = -1

P(-4, -2, -1)

Answer 2

The closest point to Q on the plane will lie on a line perpendicular to the plane (a "normal"). So first we find the equation for this line, then find the intersection.

First, note that a normal to the plane that passes through Q is given by (since the line

s(0,2,1)+(-4,2,1)

Plug each coordinate into equation for plane and solve for s (y coordinate is 2s+2, z coordinate is s+1):

2(2s+2)+s+1=-5

or

5s=-10

so

s= -2

Thus, the closest point P on the plane is -2*(0,2,1)+(-4,2,1)

or

(-4,-2,-1).

Hope this helps!