# Find orthogonal complement of given set

Let $S=${$(1,0,1)$}$\cup${$(x,y,z)\in \mathbb{R}/y=z, x+y-z=0$} be a subset of $\mathbb{R^3}$. And let define an inner product as follows:$$\langle(x_1,x_2,x_3),(y_1,y_2,y_3)\rangle=x_1y_1+2x_2y_2 + 3x_3y_3$$

I need to find $S^{\perp}$.

I already know the answer beacuse i have been trying to solve this for my algebra final exam next week and i could not find the way to solve it.

The answer is : $S^{\perp}=${$(-6,-3,2)$

It is easy to see that $\langle(-6,-3,2),(1,0,1)\rangle=0$ with the inner product that i am given. But how can i use the information of the other set that i am given?

No, the answer is not $\{(-6,-3,2)\}$. Given any subset $S$ of $\mathbb{R}^3$, $S^\perp$ is a vector subspace of $\mathbb{R}^3$ which is not the case here.
It turns out that $S^\perp=\left\{\lambda(-6,-3,2)\,\middle|\,\lambda\in\mathbb{R}\right\}$. In order to see why, note that $S$ is the union of $\{(1,0,1)\}$ with that set$$\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,y=z\wedge x=0\right\}=\left\{\lambda(0,1,1)\,\middle|\,\lambda\in\mathbb{R}\right\}.$$But $\bigl\langle(-6,-3,-2),\lambda(0,1,1)\bigr\rangle=0$. Therefore, $(-6,-3,2)\in S^\perp$. On the other hand, if $(a,b,c)\in S^\perp$, then$$\bigl\langle(a,b,c),(1,0,1)\bigr\rangle=\bigl\langle(a,b,c),(0,1,1)\bigr\rangle=0.$$In other words,$$a+3c=2b+3c=0$$and a simple computation shows that then $(a,b,c)$ is a multiple of $(-6,-3,2)$.
• @TheNicouU That doesn't change the fact that, if $v\in S^\perp$, then obviously every multiple of $v$ will also belong to $S^\perp$. – José Carlos Santos Apr 26 '18 at 16:29
• I understand that, but how do i get to that result. What is the idea behind it? How do i know that $(-6,-3,2)$ is also perp to the other set i am given? – TheNicouU Apr 26 '18 at 16:31