# Finding perpendicular vector in a skewed coordinate system

1. Mar 7, 2016

### AllenFaust

1. The problem statement, all variables and given/known data
I have an a-b coordinate system which is skewed with an angle = 60 deg. I also have a particle position defined by vector V1 (a1, b1, 0) which follows the coordinate system.

The problem I have is that I need to get V2 (a2, b1, 0) which is perpendicular to V1.

2. Relevant equations

3. The attempt at a solution

I tried a normal cross product to generate V2

V2 = V1 (a1, b1, 0) x Z (0, 0, 1) = |b1| (a^hat) - |a1| (b^hat) = V2 (b1, -a1, 0)

which works sometimes but not at all points in space. I was thinking this way is only applicable for orthogonal coordinate system.

2. Mar 7, 2016

### geoffrey159

switch to an orthonormal coordinate system with the appropriate change of basis matrix.
In this new coordinate system, if $V_1 = P v_1$, then you know that $V_2 = ( B_1, -A_1, 0)$ is orthogonal to $V_1$, and switch again : $v_2 = P^{-1} V_2$

3. Mar 7, 2016

### AllenFaust

That is actually a good suggestion, however, I am actually thinking of a solution that operates in the skewed coordinate system only. Meaning to say, I can only use the skewed system without the transformation to an orthonormal coordinate system. Thank you.

4. Mar 7, 2016

### Brian T

Well, one way to do it is take the inner product of V2 and V1 and set that to 0. Note that since you're not in orthogonal coordinates, the product will contain some off diagonal terms. That will give you one equation, you have two unknowns so you'll need one more

5. Mar 9, 2016

### geoffrey159

In an orthonormal coordinate system : $<x,y> = {}^T X Y$
In a skewed coordinate system $<x,y > = {}^T X' ({}^T PP ) Y'$,
where $X,Y$ (resp. $X',Y'$) are the coordinates of $x,y$ in the orthonormal coordiate system (resp. skewed coordinate system), and $P$ the change of basis matrix from orthonormal to skewed.

No matter which coordinate system you choose, $x$ and $y$ are orthogonal iff $<x,y> = 0$.