# Homework 6, due Thurs Oct 25

Hand in your solution on RPILMS. Each team should submit their solution under only 1 student's name. The other student's submission should just name the lead student. (This makes it easier for us to avoid grading it twice.) For programming exercises, hand in code and screen dumps. We won't run your code, but will use the screendumps to judge how it worked.

This homework has no programming. Use any tool that you wish for the multiplications, e.g., Maple or Mathematica. Include a copy of the relevant part of the session.

*(2 pts)*What is the quaternion for a rotation by 90 degrees about the axis (1,2,3)?- {$ q = \sqrt{1/2} + \sqrt{1/2} \frac(i+2j+3k){\sqrt{14}} = (0.707107 + 0.188982i + 0.377964j + 0.566947k) $}
- OK not to reduce from the expression with sqrts. OK to have only 2 significant digits. I used Mathematica.

*(2)*Apply it to the point (1,0,0). (0.0714286, 0.944641, -0.320237)*(2)*What is the quaternion for a rotation by 180 degrees about the axis (0,0,1)? k*(2)*What is the quaternion for the rotation by 90 degrees about the axis (1,2,3), followed by the rotation by 180 degrees about the axis (0,0,1)? -0.566947 -0.377964i + 0.188982j + 0.707107k*(2)*What is that rotation's axis and angle? 250 degrees, (-0.458831, 0.229416, 0.858395)*(2)*You wish to interpolate a smooth, constant speed, path on a sphere from (1,0,0) to {$ \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) $}. Give the midpoint of that path.- Use 3DInterpolation.
- {$ \cos(\alpha)=1/\sqrt{3} $}, so {$\alpha=55$} degrees, and {$\theta=27$}.
- The answer is {$ .563 (1,0,0) + .563 \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) $} = {$ (0.888074, 0.325058, 0.325058) $}.
- The two weights sum to more than 1 because we are interpolating on a circle, not a straight line.

*(2)*Give a homogeneous quadratic parametric equation for a circle of radius 2. {$ (x(t), y(t), w(t)) = \left( 2(t^2-1), 4t, (t^2+1)\right) $} corrected*(2 pt)*Consider the following projection:- The camera is at (0,0,0).
- The projection plane is 2x+y+3z=4.

*(4 pt)*Consider this projection:- The camera is at (0,0,0).
- The projection plane is z=-2.

- Write the view normalization equations for this projection.
- {$x' = {\large \frac{-2x}{z}}, y' = {\large\frac{-2y}{z}}, z' = {\large\frac{4}{z}} $}
- You can go through the derivation in the 10/15 notes, but with the plane at z=-2.
- Alternatively, you can scale everything by -1/2 to put the projection plane at z=1, apply the formula in the notes, then scale everything by -2 to restore the size.
- Note that points on the projection plane don't move.

- Consider the cube with opposite corners at (-4,-4,-2) and (4,4,-10). What does it transform to? Show the projection vertex by vertex. E.g., what does (4,4,-10) transform to?

- {$ \begin{array}{c} (-4,-4,-2)\\(-4,-4,-10)\\(-4,4,-2)\\(-4,4,-10)\\(4,-4,-2)\\(4,-4,-10)\\(4,4,-2)\\(4,4,-10) \end{array} \rightarrow \begin{array}{c} (-4,-4,-2)\\(-4/5,-4/5,-2)\\(-4,4,-2)\\(-4/5,4/5,-2)\\(4,-4,-2)\\(4/5,-4/5,-2)\\(4,4,-2)\\(4/5,4/5,-2) \end{array} $}

*Total: 20*