This is the problem that I have been asked:
Find the orthogonal complement of the transpose of the vector = [3,4,1]. Also find the point on the plane 2x-3y+z=0 which is closest to (3,4,1).
I know how to do the second part of the this problem, and I don't have any issues with it. The first sentence doesn't quite make sense to me though. I don't know how to find the orthogonal complement of a single vector that is not a subspace. I have tried looking this up online, but every place that I check, it always prefaces the explanation by saying that this is how you find the orthogonal complement to some subspace. Am I wrong in thinking that it does not make sense to ask about the orthogonal complement of a non-subspace?
Best Answer
A solution based on calculus uses partial differentiation.
First we construct a function called "distance from position $(3,4,1)$"
This is $d=\sqrt{x-3)^2+(y-4)^2+(z-1)^2}$ which gives the distance from generic point $(x,y,z)$ just to $(3,4,1)$.
Now we take the auxiliar function $$F=2x-3y+z+\lambda d^2,$$ The we seek the minimum of $F$ which will be attained a critical point, that is, at a point $(x_0,y_0,z_0)$ which solves $$\frac{\partial F}{\partial x}(x_0,y_0,z_0)=0,$$ $$\frac{\partial F}{\partial y}(x_0,y_0,z_0)=0,$$ $$\frac{\partial F}{\partial z}(x_0,y_0,z_0)=0.$$ So we get $$2+\lambda 2(x-3)=0,$$ $$-3+\lambda 2(y-4)=0,$$ $$1+\lambda 2(z-1)=0,$$ which can be arranged as $$\lambda=\dfrac{1}{3-x}=\dfrac{3}{2(y-4)}=\dfrac{1}{2(1-z)},$$ these, together with the restriction $$2x-3y+z+0,$$ would imply $$x_0=\frac{26}{7}\,\ y_0=\dfrac{41}{14} \ ,\ z_0=\dfrac{19}{14}.$$