# Find an orthogonal basis for a set of vectors S

Find a basis for the set S of vectors $\left(x,y,z\right)$ in $\mathbb{R}^{3}$ with $z=2x-5y$. Find an orthogonal basis for S. Is the vector $\left(-4,10,2\right)$ in $S^{\bot}$? Find all vectors in $S^{\bot}$

So far I have found that $S$ is spanned by the vectors $\left\{\left(1,0,2\right),\left(0,1,-5\right)\right\}$ and these vectors are also linearly independent therefore they form a basis for S.

I have then gone onto turn this basis into an orthogonal basis using the Gram-Schmidt Process:

$$f_1=\left(1,0,2\right)$$ $$f_2=\left(0,1,-5\right)-\dfrac{\left(0,1,-5\right)\cdot\left(1,0,2\right)}{\left(1,0,2\right)\cdot\left(1,0,2\right)}\left(1,0,2\right)=\left(2,1,-1\right)$$ From this I get the basis $\left\{\left(1,0,2\right),\left(2,1,-1\right)\right\}$ which is indeed orthogonal.

Edit: You forgot a minus in front of the $-5$ in the projection. Fixed now.

• You forgot a minus in the projection, in front of the $-5$, I have edited. May 6, 2016 at 15:55
• Thanks, @B.Pasternak. That now makes them orthogonal. May 6, 2016 at 19:26

here is one approach (there are many more). Pick one vector in $z$, for example $(1,1,-3)$ Another vector on $z$ is of the form $(a,b,2a-5b)$. Now for them to be perpendicular, we find through the dot product $1a+1b-3(2a-5b)=0$ which results in $5a=16b$. Now choose $b=5$ which gives $a=16$ and so a second perpendicular vector is ultimately found: $(16,5,7)$. You can verify that both vectors are perpendicular, they are both lin independent and they are in plane $z$, they form a basis. You did not show how you did your Gram Schmidt process so hence my alternative approach...