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:

$V^⊥ = {(x,y,z)|(x,y,z)(3,12,-1)=0}$


$V^⊥ = $ $sp{(-4,1,0)(\frac13,0,1)}$

Now I turn the basis of $V^⊥$ to an orthogonal basis:



$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.

  • $\begingroup$ ok, I already understood that I need to normalize the vectors in order to get an orthonormal basis, but I still see that I have some kind of mistake in my calculations as the multiplication of my $V^⊥$ span does not return 0. $\endgroup$ – durian Sep 13 '14 at 18:20

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).

  • $\begingroup$ are you just using another vector as an example or saying I can actually do it like this? $\endgroup$ – durian Sep 13 '14 at 23:27
  • $\begingroup$ I'm saying you can do it like this; avoiding fractions makes the calculations easier, and won't change the answer. $\endgroup$ – user84413 Sep 13 '14 at 23:38
  • $\begingroup$ First of all thanks a lot! second I have also found the mistake in my calculations. If I may ask another question about the normalization, its supposed to be the vector divided by its normal and the new vectors normal should be = 1? because for some reason after normalization (of mine and yours answer) i don't get the 1 $\endgroup$ – durian Sep 13 '14 at 23:51
  • $\begingroup$ for examples id i try to normalize (−18,5,6) its supposed to be: $\frac{(−18,5,6)}{||(−18,5,6)||}$ right? so that is: $\frac{(−18,5,6)}{\sqrt{385}}$ and that is $(−18,5,6){(\sqrt{385})}$ and the absolute value of this does not equal 1, or am I missing something here? $\endgroup$ – durian Sep 14 '14 at 0:02
  • 1
    $\begingroup$ Right, the vector is $(-\frac{18}{\sqrt{385}},\frac{5}{\sqrt{385}},\frac{6}{\sqrt{385}})$, and its length is equal to 1. $\endgroup$ – user84413 Sep 14 '14 at 19:54

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