# Finding all vectors normal to a given set of vectors

1. Sep 24, 2008

### mds9668

How do you find all vectors perpendicular to a set of three vectors in R^4? I know that the dot product of a normal vector and each given vector will be equal to 0. How could I set up the system of equations in matrix form?

2. Sep 25, 2008

### Defennder

Hmm, you just have to write out the equations for each dot product. For example, suppose the three vectors are $v_1, v_2, v_3$. Then let the perpendicular vector be $u$. So we know that $u \cdot v_i = 0$. Let $u = (a_1, a_2, a_3, a_4)^T$ for concreteness.

So if $$v_1 = (x_1, x_2, x_3, x_4)^T$$, then we have, after applying the dot product of u and v1:
$$a_1 x_1 + a_2 x_2 +a_3 x_3 + a_4 x_4 = 0$$. Do the same for the other two vectors and you have a system of equations which you can then express in matrix form.

3. Sep 25, 2008

### ||spoon||

could you not cross product??

4. Sep 25, 2008

### danago

I believe the cross product is defined only in R^3, well as far as i know :P

5. Sep 25, 2008

### Defennder

6. Sep 25, 2008

### sal

As the Wiki article says, there's more than one way to extend the cross product to higher dimensions. One very straightforward way extends it to all dimensions of Rn, but it's a little peculiar: In dimension n it becomes a function of n-1 vectors rather than a simple product of two vectors. This is mentioned in Spivak's classic monograph, Calculus on Manifolds, chapter 4, p83 in the old Benjamin edition. I'll quote the relevant snippet here:

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7. Sep 25, 2008

### tiny-tim

Welcome to PF!

Hi mds9668! Welcome to PF!

You use the wedge product, ⋀, of all three vectors (with x⋀x = 0).

That will give you a 3-form, which is a linear combination a.x⋀y⋀z + b.y⋀z⋀t + c.z⋀x⋀t + d.x⋀y⋀t.

The vectors perpendicular to all three will be the multiples of a.t + b.x + c.y + d.z

(because (a.t + b.x + c.y + d.z)(a.x⋀y⋀z + b.y⋀z⋀t + c.z⋀x⋀t + d.x⋀y⋀t) = a² + b² + c² + d²)