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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?

Templeton Peck
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  • can we use calculus? – janmarqz Mar 15 '22 at 00:19
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    I don't know that it would necessarily be something that I could use for this assignment, but I would definitely like to learn about how to use it. – Templeton Peck Mar 15 '22 at 00:22
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    The orthogonal compliment of a vector is the subspace spanned by all vectors orthogonal to it. This is identical to the orthogonal compliment of the subspace spanned by that vector. The orthogonal compliment of a one dimensional subspace like this is called a hyperplane so they're important enough to get their own name. – CyclotomicField Mar 15 '22 at 00:22
  • Thank you, this makes sense. – Templeton Peck Mar 15 '22 at 00:24
  • @CyclotomicField - You don't need the span. The orthogonal complement of a vector is the set of all vectors orthogonal to it. This is in fact a subspace. – mr_e_man Mar 15 '22 at 03:52
  • @mr_e_man I know. That's the first sentence I wrote in my comment. – CyclotomicField Mar 15 '22 at 11:49
  • @CyclotomicField - No, you referred to the span of all orthogonal vectors. You could have referred to the set of all orthogonal vectors. Of course they're the same; my point is that you're doing extra work, taking the (linear) closure of something that's already closed. – mr_e_man Mar 16 '22 at 00:07
  • @mr_e_man You seem to already know that we're talking about exactly the same set of vectors so it's unclear to me where your confusion lies. Can you be more explicit? – CyclotomicField Mar 16 '22 at 00:19
  • @CyclotomicField - I'm not the OP. – mr_e_man Mar 16 '22 at 00:20
  • @mr_e_man I'm talking to you because you made these comments. I really don't understand what you're talking about and I'm trying to figure it out. – CyclotomicField Mar 16 '22 at 00:27
  • @mr_e_man ok I think I figured it out. I used the word span twice to reinforce the fact that the orthogonal compliment of the vector was the orthogonal compliment of its spanning set since that's where they were making the mistake. It's just boring old repetition to encourage retention. – CyclotomicField Mar 16 '22 at 00:30
  • @CyclotomicField - I'm not confused. When you mentioned my confusion, it seemed that you thought I had asked a question, which only the OP did. – mr_e_man Mar 16 '22 at 00:39
  • There are two spans which don't need to be taken: $${v}^\perp=(\text{span}{v})^\perp$$ $$={w\mid v\cdot w=0}=\text{span}{w\mid v\cdot w=0}$$ I was mainly pointing out the last equation. – mr_e_man Mar 16 '22 at 00:46

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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 $$x_0=\frac{26}{7}\,\ y_0=\dfrac{41}{14} \ ,\ z_0=\dfrac{19}{14}.$$

janmarqz
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