Finding an Orthogonal Transformation for Mapping Points

[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(<br /> \begin{array}{cc}<br /> cos(2\theta) &amp; sin(2\theta)\\<br /> sin(2\theta) &amp; -cos(2\theta)<br /> \end{array}<br /> \right)<br />

\therefore \left(<br /> \begin{array}{cc}<br /> cos(2\theta) &amp; sin(2\theta)\\<br /> sin(2\theta) &amp; -cos(2\theta)<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c} 0 \\<br /> 5 \\<br /> \end{array}<br /> \right)=<br /> \left(<br /> \begin{array}{c} 3 \\<br /> 4\\<br /> \end{array}<br /> \right)<br /> <br />

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}<br />

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)<br /> y = sin(2\theta)

and those for j' would be:

x = sin(2\theta)<br /> y = cos(2\theta).<br />

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!

 

[HallsofIvy]

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(<br /> \begin{array}{cc}<br /> cos(2\theta) &amp; sin(2\theta)\\<br /> sin(2\theta) &amp; -cos(2\theta)<br /> \end{array}<br /> \right)<br />

\therefore \left(<br /> \begin{array}{cc}<br /> cos(2\theta) &amp; sin(2\theta)\\<br /> sin(2\theta) &amp; -cos(2\theta)<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c} 0 \\<br /> 5 \\<br /> \end{array}<br /> \right)=<br /> \left(<br /> \begin{array}{c} 3 \\<br /> 4\\<br /> \end{array}<br /> \right)<br /> <br />

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

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

NO, this is 2\theta

\therefore m=\frac{3}{4}<br />

I noticed that if instead I find the angle by taking 5sin(2\theta)=3
\theta = 18.43 and tan(\theta)=0.333

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

Or -5cos(2\theta)=4
\theta = 71.56

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.

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)<br /> y = sin(2\theta)

and those for j' would be:

x = sin(2\theta)<br /> y = cos(2\theta).<br />

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!

 

[Lunat1c]

The gradient of the line is tan(\theta) [/itex] not tan(2\theta) [/itex], or am I wrong?&lt;br /&gt; &lt;br /&gt; 2\theta [/itex] is the angle with which the vectors i and j will be rotated when they&amp;amp;#039;re reflected in the line y=(tan\theta)x [/itex].

 

[Lunat1c]

NO, this is 2\theta2\theta= 36.86 and tan(36.86)= 0.75

Ok fair enough, I meant tan(2\theta) = -\frac{3}{4} [/itex]<br /> \therefore \theta = \frac{arctan(-\frac{3}{4})}{2} [/itex].&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; NO, cos(71.56)= 0.3163, nowhere near -.8. I don&amp;#039;t know how you got this. If cos(\theta)= -.8 then \theta= 143 or \theta= -37. tan(143)= -.75 but tan(-37)= .75. &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;&lt;br /&gt; -5cos(2\theta)=4 [/itex]&amp;lt;br /&amp;gt; then cos(2\theta) = -4/5 = -0.8 [/itex]&amp;amp;lt;br /&amp;amp;gt; therefore \theta = \frac{arccos(-0.8)}{2} = 71.56 [/itex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; and hence cos(2 * 71.56) = -0.8 [/itex]