# Homework Help: Linear Transformation

1. May 29, 2010

### Lunat1c

Hi,

I am trying to find an orthogonal transformation that maps the point (0,5) to the point (3,4).
Now, I found that the transformation matrix M for a reflection in the line y=mx is as follows:

$$M = \left( \begin{array}{cc} cos(2\theta) & sin(2\theta)\\ sin(2\theta) & -cos(2\theta) \end{array} \right)$$

$$\therefore \left( \begin{array}{cc} cos(2\theta) & sin(2\theta)\\ sin(2\theta) & -cos(2\theta) \end{array} \right) \left( \begin{array}{c} 0 \\ 5 \\ \end{array} \right)= \left( \begin{array}{c} 3 \\ 4\\ \end{array} \right)$$

However this means that
$$5sin(2\theta)=3$$
$$-5cos(2\theta)=4$$

$$\frac{5sin2\theta}{-5cos2\theta} = 3/4$$
$$\theta = arctan(-\frac{3}{4})$$
$$\therefore m=\frac{3}{4}$$

I noticed that if instead I find the angle by taking $$5sin(2\theta)=3$$
$$\theta = 18.43 and tan(\theta)=0.333$$
Or $$-5cos(2\theta)=4$$
$$\theta = 71.56$$ and $$tan(71.56)=3$$

Why don't they all yield the same result? isn't this like solving a system of linear equations?

Having said this, I tried to derive the matrix of the transformation myself. I drew the basis vectors i(1,0) and j(0,1) and checked what their new coordinates would be when reflected in a line that makes an angle $$\theta$$ with the x-axis.

When considering the j(0,1) vector, the angle between j and j' & that between i and i' is $$2\theta$$.

The new coordinates for i' would be:
$$x = cos(2\theta) y = sin(2\theta)$$

and those for j' would be:

$$x = sin(2\theta) y = cos(2\theta).$$

Why would you say that for j' $$y=-cos(2\theta)?$$
The only way j' will have negative y coordinates is if the gradient of the line is >45, and if this happens, cos(2x) will be negative (since 90 < 2x < 180, cos(2x) is negative).

Sorry for the very long post, I just wanted to show what I tried before asking any questions. With this being said, could someone please tell me what's wrong with the derivation I attempted? And most of all, why the first one doesn't work?

Thank you!

Last edited: May 29, 2010
2. May 29, 2010

### HallsofIvy

NO, this is 2$\theta$

$2\theta= 36.86$ and $tan(36.86)= 0.75$

NO, cos(71.56)= 0.3163, nowhere near -.8. I don't know how you got this. If $cos(\theta)= -.8$ then $\theta= 143$ or $\theta= -37$. tan(143)= -.75 but tan(-37)= .75.

3. May 29, 2010

### Lunat1c

The gradient of the line is [tex] tan(\theta) [/itex] not [tex] tan(2\theta) [/itex], or am I wrong?

[tex]2\theta [/itex] is the angle with which the vectors i and j will be rotated when they're reflected in the line [tex] y=(tan\theta)x [/itex].

4. May 29, 2010

### Lunat1c

Ok fair enough, I meant [tex] tan(2\theta) = -\frac{3}{4} [/itex]
[tex] \therefore \theta = \frac{arctan(-\frac{3}{4})}{2} [/itex].

[tex] -5cos(2\theta)=4 [/itex]
then [tex] cos(2\theta) = -4/5 = -0.8 [/itex]
therefore [tex] \theta = \frac{arccos(-0.8)}{2} = 71.56 [/itex]
and hence [tex] cos(2 * 71.56) = -0.8 [/itex]

Last edited: May 29, 2010
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