Find the distance from the point (-4, 0, 2) to the plane!?

Question

-2x+5y-2z=-5

Answer 1

The unit normal to this plane is (-2,5,-2)/(33)^(1/2) = u. A vector in this direction with length d, is then du (u is a vector). We then want the point p = (-4,0,2) + du to be a point on the plane, q = (x,y,-x +5/2y + 5/2) (where I've subsituted in for z)

So p + du = q or (-4 -2d/33^.5, 0 + 5d/33^.5, 2 -2d/33^.5) = (x, y, -x +5/2y + 5/2)

At this point, you have 3 unknowns d,x,y, and 3 equations:

-4 -2d/33^.5 = x
5d/33^.5 = y
2-2d/33^.5 = -x +5/2y +5/2

Solve for d. The sign of d tells you whether the point is above or below the plane.

2 -2d/33^.5 = 4 +2d/33^.5 + (5/2)5d/33^.5 +5/2

d = -9/33^.5

The distance itself is then 9/33^.5 (the magnitude of du)

To check, this means p + du = (-4,0,2) + -9/33(-2, 5, -2) =
(-114/33, -45/33, 84/33) should be a point on the plane.

Check: -2(-114/33) + 5(-45/33) -2(84/33) = -5 Q.E.D.

Vote this best answer because the algebra took me forever to get right.

Answer 2

Find the distance from the point P(-4, 0, 2) to the plane
-2x + 5y - 2z = -5

Rewrite the equation of the plane as

-2x + 5y - 2z + 5 = 0

Use the distance formula.

distance = |(-2)(-4) + 5(0) - 2(2) + 5| / √[(-2)² + 5² + (-2)²]
distance = | 8 + 0 - 4 + 5| / √(4 + 25 + 4)
distance = 9/√33