# Help with making an orthonormal basis

1. Jun 22, 2004

### graphic7

I'm wanting to form an orthonormal basis from two non-parallel vectors.

$$a = \left(\begin{array}{cc}3 & 4\end{array}\right)$$

$$b = \left(\begin{array}{cc}2 & -6\end{array}\right)$$

Could someone please walk me through the calculations needed? Much appreciated.

2. Jun 22, 2004

### graphic7

I think I've finally figured out. I had it explained to me visually, which didn't make sense. By visually, I am referring to the whole vector "arrow" idea which always troubles me.

3. Jun 22, 2004

### Muzza

Look up the Gram-Schmidt process.

Or consider the image I've attached. The vectors are probably pointing in the wrong directions, but it doesn't really matter. The red vector (call it $$a_o$$) is orthogonal to b, and the green vector (call it $$a_p$$) is parallel to b. Now, a_p is the projection of a onto b, and there's a nice formula for finding it:

$$a_p = \frac{a.b}{b.b} * b$$

(It's easy to prove, see any linear algebra text). After some calculations, we find that $$a_p = \left(\begin{array}{cc}-9/10, & 27/10\end{array}\right)$$. But since $$a = a_p + a_o$$, we have that $$a_o = a - a_p = \left(\begin{array}{cc}3, & 4\end{array}\right) - \left(\begin{array}{cc}-9/10, & 27/10\end{array}\right) = \left(\begin{array}{cc}39/10, & 13/10\end{array}\right)$$. And so, we have "created" a vector that's orthogonal to b, so you can chose a_o and b as the basis vectors. (Of course, you've got to normalize them first, but that's trivial).

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Last edited: Jun 22, 2004
4. Jun 23, 2004

### graphic7

Thanks very much for the explanation. Orthogonalization makes quite a bit more sense now. I've done some visual graphs with Mathematica, and that's helped quite a bit, also.

5. Aug 16, 2004

### mathwonk

the reason the projection of a onto b is ghiven by that formula is because of the familiar formula a.b = |a| |b| cos(C) where C is the angle between a and b. I.e. from triangle trig, the projection of a onto b has length |a| cos(C). so the vector of that length in that direction is obtained by multiplying a unit vector by that length. Now of course b/|b| is a unit vector in the direction of b,

so the vector parallel to b, with length |a| cos(C), is the product

|a|cos(C) b/|b| = |a||b| cos(C)/|b|^2 b = (a.b/b.b) b.