# Linear Algebra - Orthogonal Vectors

1. May 1, 2007

### verd

I'm a bit confused, conceptually. This is the problem

Let v1=( 1, -1, 2) v2=( 2, 1, 3) v3=( 1, -4, 3)
Find a nonzero vector u that is orthogonal to all three vectors v1, v2, and v3.

I know how to find the projection matrix, P, which I can solve with v1, v2, and v3.
The equation for that is simply p=A(AT A)^-1AT
(Where AT is A transposed)

However, I'm not sure exactly what P is... I know it's the projection matrix, but if I solved this, would this give me a matrix that is orthogonal to A? (Assuming A is spanned by v1, v2, and v3). If so, would I just be able to take one of the column vectors from this matrix and assume that it is orthogonal to v1 v2 and v3??

If I'm going in the wrong direction, can someone tell me how to find a vector that is orthogonal to A?

Thanks!

2. May 1, 2007

### Dick

Two ways. Write the equations a.v1=0, a.v2=0 and a.v3=0 with 'a' unknown and solve the system for the components of a. Alternatively, row reduce the set of vectors to find the two that are linearly independent and then take the cross product. Your pick.

3. May 1, 2007

### verd

Thanks for responding... The first recommendation with choosing an 'a' sounds familiar. However, I don't exactly understand how that determines a vector that is perpendicular to the subspace formed by the three vectors.

I'm a bit lost on the concept... Having a hard time visualizing the problem.

v1*u=0
v2*u=0
v3*u=0

where u is a vector, right?

Could someone briefly explain why that works? I apologize if this is painfully obvious....

4. May 1, 2007

### Office_Shredder

Staff Emeritus
If the dot product of two vectors is zero, they are orthogonal (practically by definition).

So if the dot product of a vector with all three of the v's is zero, then it must be orthogonal to all three

5. May 1, 2007

### Dick

u.v1=0 means ux-uy+2*uz=0. u.v2=0 means 2*ux+uy+3*uz=0. You do the third one. So just three equations in three unknowns. You won't get a unique nonzero solution, but you can still find one, yes?

6. May 2, 2007

### verd

Thank you. I understand now what you did any why. However, now I'm having difficulty coming up with a non-zero vector through that 3x3 matrix. So I don't see how to find a unique nonzero solution-- I see why I can't just solve for one.

Any pointers?

Thanks again. This is my matrix:

1 -1 2
2 1 3
1 -4 3

7. May 2, 2007

### Dick

You can just solve for one. I don't really see the problem. Can you post more specifically what equations you are trying to solve? You will find you can't 'solve' for all of the variables. Just set x=1 and determine y and z.

8. May 2, 2007

### HallsofIvy

Have you noticed that the three vectors given are not independent? In fact, it is IMPOSSIBLE to have, in 3 dimensions, a non-zero vector that is orthogonal to each of the independent vectors. Do you remember anything from Calculus about finding a vector that is orthogonal to two given vectors.