Question

Find an orthonormal basis for the subspace
S = {(x1, x2, x3) : x1 + 2x2 − x3 = 0} ⊂ R^3

Answer 1

[-1,1,1]
[1,0,1]

Answer 2

hmm, did I miss something?

Answer 3

that is just orthogonal..you need to normalize it

Answer 4

\[x_1+2x_2-x_3=0\] is just a plane in \[R^3\]

find two vectors in the plane that are orthonormal.

Answer 5

thanks

Answer 6

Definitely got that backwards. What we have is
\[[1 -2 -1]\left[\begin{matrix}x1 \\ x2 \\x3\end{matrix}\right]=0\]
Now find the basis for the null space of [1 -2 1], and normalize, which is what Zarkon did.

Answer 7

Yeah I understand it better when you rewrote it phi, how do you normalize it then...

Answer 8

\[\|<a,b,c>\|=\sqrt{a^2+b^2+c^2}\]

so our unit vector in the sam direction of \[<a,b,c>\]

is \[\frac{<a,b,c>}{\|<a,b,c>\|}\]

Answer 9

'sam' should be same