Find the coordinates of the pt P in the plane that is the foot of the perpendicular from the origin to plane?

Question

A rectangular equation of a plane is 3x-2y-4z-12=0. Find the coordinates of the pt P in the plane that is the foot of the perpendicular from the origin to the plane.

Answer 1

1) The perpendicular minimizes the distance:

Min: x^2 + y^2 + z^2 subject to 3x - 2y - 4z - 12 = 0

Using Lagrange multipliers, set all four partial derivatives to zero in f(x, y, z, lambda1) = x^2 + y^2 + z^2 -lambda1(3x - 2y - 4z - 12)

dx: 2x - 3lambda1 = 0
dy: 2y + 2lambda1 = 0
dz: 2z + 4lambda1 = 0
dlambda1: 3x - 2y - 4z - 12 = 0

4.5lambda1 + 2lambda1 + 8lambda1 = 12
14.5lambda1 = 12
29lambda1 = 24
lambda1 = 24/29
x = 36/29
y = -24/29
z = -48/29

2) From the equation we know that [ 3, -2, -4 ]^T is perpendicular to the plane. We want the pependicular through the origin, which is the line [ x, y, z ]^T = [ 0, 0, 0 ]^T + t [ 3, -2, -4 ]^T. That is on the plane when 3x - 2y - 4z = 12, or

3(0 + 3t) - 2(0 - 2t) - 4(0 - 4t) = 12
9t + 4t + 16t = 12
29t = 12
t = 12/29
x = 36/29
y = -24/29
z = -48/29

Dan

Answer 2

A rectangular equation of a plane is 3x - 2y - 4z - 12 = 0. Find the coordinates of the point P in the plane that is the foot of the perpendicular from the origin to the plane.

The point P will be on the line thru the origin whose directional vector is the normal vector to the plane. The equation of this line is:

L = t[3, -2, -4]^T
where t is a scalar ranging over the real numbers

At the point P the values of the line and plane will be equal. Rewrite the equation of the plane in terms of t.

3x - 2y - 4z - 12 = 0
3(3t) - 2(-2t) - 4(-4t) -12 = 0
9t + 4t + 16t - 12 = 0
29t = 12
t = 12/29

x = 3t = 3(12/29) = 36/29
y = -2t = -2(12/29) = -24/29
z = -4t = -4(12/29) = -48/29

P(36/29, -24/29, -48/29)

Answer 3

They are all in Mathematics! Merry Christmas ;)!