linear algebra - orthonormal basis

Question

Determine an orthonormal basis {e1, e2, e3} such that e1 is parallel to the vector u = (1, 2, -1).

Can you please help me find the ON-basis ?

Answer 1

If you don't know Gram-Schmidt, then you can go directly. Find the vectors that are orthogonal to $(1,2,-1)$: this means $$ (x,y,z)\cdot(1,2,-1)=0 $$ or $$ x+2y-z=0 $$ Set $z=0$, for instance, and $y=1$, so you find $(-2,1,0)$.

Now you have two orthogonal vectors and you need a third one. Add the equation $(x,y,z)\cdot(-2,1,0)=0$ getting $$ \begin{cases} x+2y-z=0\\ -2x+y=0 \end{cases} $$ Take $y=1$ that gives $x=2$ and $z=4$. Thus we have the three vectors $$ (1,2,-1),\quad (-2,1,0),\quad (2,1,4) $$ These form an orthogonal basis (why?). For an orthonormal one, just multiply each one of those vectors by the inverse of its norm: $$ \frac{1}{\sqrt{6}}(1,2,-1),\quad \frac{1}{\sqrt{5}}(-2,1,0),\quad \frac{1}{\sqrt{21}}(2,1,4) $$

Answer 2

You can start by picking any basis, and then applying the Gram-Schmidt process to get an orthonormal one.

For example, pick ${(1,2,-1),(0,1,0),(0,0,1)}$ as your Basis.

Project $(0,1,0)$ onto $(1,2,-1)$ then substract the result from $(0,1,0)$. Wikipedia gives this as follows: $\text{proj}_u = \frac{<u,v>}{<u,u>}v$.

$\text{proj}_{(1,2,-1)}(0,1,0) = \frac{<(1,2,-1), (0,1,0)>}{<(1,2,-1), (1,2,-1)>}(1,2,-1) = \frac{2}{6}(1,2,-1) = (\frac{1}{3}, \frac{2}{3}, \frac{-1}{3})$.

This gives a new second vector which is orthogonal to the first one, namely: $(0,1,0) - (\frac{1}{3}, \frac{2}{3}, \frac{-1}{3}) = (\frac{-1}{3}, \frac{1}{3}, \frac{-1}{3})$. Orthogonality is verified by $<(1,2,-1),(\frac{-1}{3}, \frac{1}{3}, \frac{-1}{3})> = \frac{-1+2-1}{3} = 0$.

But I'll have to side with Mnifldz here, the article on Wikipedia and this small example calculation should be enough to figure it out. (i.e. next you project $(0,0,1)$ onto each of the first vectors and substract the result.)