# Thread: Finding a vector perpendicular to three vectors in R^4

1. ## Finding a vector perpendicular to three vectors in R^4

I'm trying to find all vectors in R^4 that are perpendicular to the following three vectors:

[ 1 1 1 1 ]
[ 1 2 3 4 ]
[ 1 9 9 7 ]

I'm not sure exactly how to set up a matrix that would help me to find these vectors. Any help setting this up would be appreciated. Thanks!

2. Here's what I would do: set up a system of three equations in four unknowns. You've got your three known vectors, call them $\mathbf{r}_{1}, \mathbf{r}_{2}, \mathbf{r}_{3}.$ The vector you're trying to produce that's perpendicular to all of them, we'll call $\mathbf{s}.$ Then you can get three equations by setting

$\mathbf{s}\cdot\mathbf{r}_{1}=0,$

$\mathbf{s}\cdot\mathbf{r}_{2}=0,$ and

$\mathbf{s}\cdot\mathbf{r}_{3}=0.$

Use normal Gaussian elimination to proceed. Does that make sense?

3. Another way of saying the same thing: call the vector <w, x, y, z>. Saying that is orthogonal to each of the given vectors means that
w+ x+ y+ z= 0
w+ 2x+ 3y+ 4z= 0
w+ 9x+ 9y+ 7z= 0

That is three equations in 4 unknow so you should be able to solve for three of them in terms of the fourth. That is, you should be able to find, say, w= az, x= bz, y= cz, for specific numbers a, b, and c, so that <w, x, y, z>= <az, bz, cz, z>= z<a, b, c, 1> and so the set of all such vectors is the one dimensional subspace, the set of all multiples of <a, b, c, z>.

4. Or you can just do a 4-dimensional cross-product!

(Note that while you can do the cross product of two vectors in three dimensional space, you need three vectors in four dimensional space.)

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# when three vectors are orthogonal

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