# Angle of Rotation to project a point

I am working on an Image Processing project and recently faced a situation which when put mathematically looks like:

Image

In the image, the planes AA' and BB' are parallel to each other and subtend an angle theta with the x-axis

A point (p) on the plane AA' is projected onto BB' as (q). At what angle should the plane AA' should be rotated such that when the point (p) is projected onto BB' falls at point (r)? The known parameters are the angle theta, x and y. Or Do I need any additional information to calculate the rotation angle?

• Projected parallel to the horizontal axis? AA’ and BB’ are planes seen on-end? You put r and x in your picture. Do those have something to do with your question? – rschwieb Mar 13 '18 at 18:05
• AA' and BB' are parallel to each other making an angle Theta with x-axis. Yes 'r' is the new Point and x is known value – Sakala Bhargava Ram Mar 13 '18 at 18:06
• But apparently what you're saying is that AA' will not be parallel to BB' after you have "rotated the plane". Does "rotating the plane" mean that you are turning the line in the picture about its intersection with the horizontal axis? – rschwieb Mar 13 '18 at 19:49
• Are you sure it wouldn't be better to refer to these "planes" as lines? I don't see what purpose it serves to call them planes. – rschwieb Mar 13 '18 at 19:50

Label the new angle between $AA'$ and the horizontal axis after rotation as $\phi$. Drop a parallel from the point of intersection $t$ of the rotated version with the horizontal through $r$, and label its point of intersection with the horizontal line as $u$.
Elementary trigonometry says that the side opposite to $\phi$ in $\triangle Atu$ should have length $y\sin{\theta}-x\sin(\theta)=(y-x)\sin(\theta)$. Relating that to $\phi$, we have $y\sin(\phi)=(y-x)\sin(\theta)$. Solving, you get $\phi=\arcsin(\frac{(y-x)\sin(\theta)}{y})$.
The amount you need to rotate, then, is the difference between $\theta$ and $\phi$.