Shortest distance between plane and point

[physics_197]

Homework Statement

Given the equation of the plane to be: P = 0 + a[1,1,1]^T + b[x1,x2,x3]^T

and the point Y = [y1, y2, y3]^T

Show: (F)^T(Y-F)=0, where F is the point on the plane closest to Y

Homework Equations

The Attempt at a Solution

Y = F + FY
Y - F = FY

Now if somehow by multiplying both sides by F^T
F^T(Y - F) = F^T(FY)
and if F^T(FY) = 0, then I would be set, but I don't think it does(or atleast I don't have any properties that say it does)

 

[HallsofIvy]

The shortest line from the point to the plane is perpendicular to the plane. That is, if F is the point on the plane closest to Y then the vector Y- F is perpendicular the plane and so is perpendicular to any vector in the plane.

You don't seem to be using the fact that plane is given by P = 0 + a[1,1,1]^T + b[x1,x2,x3]^T which is surely important! One thing that tells us is that the origin is in the plane so the vector F, the vector from the origin to the point F, is itself in the plane.

 

[physics_197]

Thanks, got it