I need to find the orthonormal basis for the orthogonal complement of $U = sp{(3,12,-1)}$
(U is a subset of $R^3$).
This is what I have done so far:
$V^⊥ = {(x,y,z)|(x,y,z)(3,12,-1)=0}$
$3x+12y-z=0$
$V^⊥ = $ $sp{(-4,1,0)(\frac13,0,1)}$
Now I turn the basis of $V^⊥$ to an orthogonal basis:
$(-4,1,0)-\frac{(\frac13,0,1)(-4,1,0)}{||(-4,1,0)||^2}(-4,1,0)$
$=(\frac{-272}{3},\frac{68}{3},1)$
$V^⊥ = $ $sp{(-4,1,0)(\frac{-272}{3},\frac{68}{3},1)}$
1.) I'm not sure I solved it the right way so far?
2.) And also , how do I find the Orthonormal basis, I know that each of the vectors absolute value needs to = 1.
Best Answer
To simplify the calculations, let $v_1=(1,0,3)$ and $v_2=(-4,1,0)$.
Then to get an orthogonal basis, we can use (as you did)
$w_2=v_2-\frac{v_2\cdot v_1}{v_1\cdot v_1}v_1=(-4,1,0)-\frac{-4}{10}(1,0,3)=(-4,1,0)+(\frac{2}{5},0,\frac{6}{5})=(-\frac{18}{5},1,\frac{6}{5}).$
Now we can replace $w_2$ by $5w_2=(-18,5,6)$ for convenience, and
then normalize the vectors to get an orthonormal basis (as you remarked).