# Find all orthogonal 3x3 matrices of the form

1. Mar 15, 2015

### pyroknife

1. The problem statement, all variables and given/known data
Find all orthogonal 3x3 matrices of the form

\begin{array}{cc}
a & b & 0 \\
c & d & 1\\
e & f & 0 \\\end{array}

2. Relevant equations
There are many properties of an orthogonal matrix. The one I chose to use is:
An n x n matrix is an orthogonal matrix IFF $$A^{T}A = I$$. That is, the transpose of A multiplied by A is equal to the n x n identity matrix.

3. The attempt at a solution
$$A^T*A$$ =
\begin{array}{cc}
a^2+c^2+e^2 & ab+cd+ef & c \\
ab+cd+ef & b^2+d^2+f^2 & d\\
c & d & 1 \\\end{array}
=
\begin{array}{cc}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1 \\\end{array}

From this we can see right away that c=d=0.

From the rest, we have left with 3 equations $$a^2+e^2=1; ab+ef=0; b^2+f^2=1$$

Is it possible to obtain a equation solution given the above 3 equations? I don't think so.
I am at a lost about how to solve this problem. Anyone have any suggestions?

2. Mar 15, 2015

### vela

Staff Emeritus
That's a good start. If you consider the columns of A to be vectors, $a^2+e^2=1$ says that the first column is normalized. Similarly, $b^2+f^2=1$ says the second column is normalized. Can you come up with an interpretation for $ab+ef=0$?

3. Mar 15, 2015

### LCKurtz

Actually, that third equation (equivalently normalizing the second column) should be $b^2+d^2+f^2 = 1$. Lots of extra variables here so there should be a solution.

4. Mar 15, 2015

### vela

Staff Emeritus
The OP omitted $d$ because the off-diagonal entries require $d=0$.

5. Mar 15, 2015

### Ray Vickson

Having $a^2+e^2=1$ suggests setting $a = \sin(\theta), e = \cos(\theta)$ --- or maybe $a = \cos(\theta), e = \sin(\theta)$ --- for some $\theta$, and having $b^2 + f^2 = 1$ suggests setting $b = \sin(\phi), f = \cos(\phi)$ for some $\phi$ (or swap the $\sin$ and $\cos$). Now, what would the remaining condition give you?

Last edited: Mar 15, 2015
6. Mar 15, 2015

### LCKurtz

Ahh.., missed that. Makes it easier too.

7. Mar 16, 2015

### pyroknife

8. Mar 16, 2015

### pyroknife

Hmmm I am still struggling with this. But if we use what Ray Vickson stated, then it would come out to be any of the following 4:
sin(θ)*sin(ϕ)+cos(θ)*cos(ϕ)=0
sin(θ)*cos(ϕ)+cos(θ)*sin(ϕ)=0
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

Actually, it reduces to the following 2:
cos(θ)*sin(ϕ)+sin(θ)*cos(ϕ)=0
cos(θ)*cos(ϕ)+sin(θ)*sin(ϕ)=0

9. Mar 17, 2015

### pyroknife

Also, a,b,c,d can be positive or negative sin or cos...

I am still unsure how to write the expressions for variables a,b,e,f

10. Mar 17, 2015

### Ray Vickson

Remember: addition formulas for sin and cos!
Can't you simplify $\cos(\theta) \sin(\phi) + \sin(\theta) \cos(\phi)$?

11. Mar 17, 2015

### pyroknife

Hmmm, I don't see a trig identity that would put this into a simpler form?

12. Mar 17, 2015

### pyroknife

Using a trig identity, I was able to obtain that
$$cos(\theta)sin(\phi)+sin(\theta)cos(\phi) = sin(\theta+\phi)-sin(\theta-\phi)$$

But going back to the question, since a,b,c,d, can be +/-cos or +/-sin, wouldn't there be many matrices that is a function of the 2 angles?

13. Mar 17, 2015

### LCKurtz

@pyroknife: You have $(a,e)$ and $(b,f)$ both on the unit circle. You have been asked what $ab +ef=0$ does for you, and as far as I can see, you haven't addressed that question. It is the missing link in finishing this question. Think about what that equation tells you about the position vectors to your two points.

14. Mar 17, 2015

### pyroknife

Oh hmmm. From post #12, we see that $$sin(\theta+\phi)=sin(\theta-\phi).$$
It seems this condition may only be satisfied if $$\phi=0$$

15. Mar 17, 2015

### Ray Vickson

No. See, eg., http://www.sosmath.com/trig/Trig5/trig5/trig5.html .

16. Mar 17, 2015

### LCKurtz

No. Here's what I asked from you:

17. Mar 17, 2015

### pyroknife

I'm having trouble with this. If a,e indicate the x coordinate on a unit circle, and b,f indicate the y-coordinate on a unit circle, then the equation is saying
the sum of the product of the 2 x-coordinates and the 2 y-coordinates will yield 0. But I don't think this is what you're looking for?

18. Mar 17, 2015

### pyroknife

I'm having trouble with this. If a,e indicate the x coordinate on a unit circle, and b,f indicate the y-coordinate on a unit circle, then the equation is saying
the sum of the product of the 2 x-coordinates and the 2 y-coordinates will yield 0. But I don't think this is what you're looking for?

19. Mar 17, 2015

### pyroknife

20. Mar 17, 2015

### LCKurtz

$(a,e)$ and $(b,f)$ are the coordinates of the two points. Anyway, do you recognize $ab + ef$ as a vector operation?