CS465 Fall 2006
Homework 5 Solution
-------------------

1.1

The main idea is to first rotate V into the plane spanned by two canonical axes (here, we 
use the Y-Z plane). Then, rotate to align the resulting vector with a canonical axis (here,
the Z axis). Now apply the 45 degree canonical rotation around that axis (here, Z), and then 
undo the first 2 rotations (which can be thought of as rotations by the negation of the 
original rotations).

R_axis_angle = Rz(-atan(2)) * Rx(-atan(2*sqrt(5)) * Rz(pi/4) * Rx(atan(2*sqrt(5)) * Rz(atan(2))

where Rz(theta) and Rx(theta) are the canonical rotations around the z and x axes. Numerically,

Rz(atan(2)) = 

    0.4472   -0.8944         0
    0.8944    0.4472         0
         0         0    1.0000

Rx(atan(2*sqrt(5)) = 

    1.0000         0         0
         0    0.2182    0.9759
         0   -0.9759    0.2182

Rz(pi/4) = 

    0.7071   -0.7071         0
    0.7071    0.7071         0
         0         0    1.0000

The resulting matrix is

R_axis_angle = 

    0.9303   -0.0427    0.3644
    0.2659    0.7629   -0.5893
   -0.2528    0.6451    0.7211


1.2

First, choose a vector that is not parallel with V, like [0 0 1]'.
Taking the cross product of V and this vector yields a vector perpendicular
to V:

V2 = normalize(V x [0 0 1]') = [0.4472  -0.8944  0]'.

The cross product of V and V2 will produce a direction perpendicular to both:

V3 = V x V2 = [0.1952    0.0976   -0.9759]'.

So Rbasis = [V2 V3 V] =  

    0.4472    0.1952    0.8729
   -0.8944    0.0976    0.4364
         0   -0.9759    0.2182

And we can express the axis-angle rotation as

R_axis_angle = Rbasis * Rz(pi/4) * Rbasis'

and the answer is the same as in Problem 1.1.

Note that cross(u,v) = -cross(v,u). If you aren't careful to perform cross products
in the right order, your basis can be left-handed, which leads to an incorrect answer.


2.1 (B-Spline)

a) matrix is [-1 3 -3 1; 3 -6 3 0; -3 0 3 0; 1 4 1 0] / 6. This gives 

x(t) = 1/2 + t/2 - 3t^2/2 + 2t^3/3
y(t) = 1/6 + t/2 + t^2/2 -2t^3/3

b) t=0 -> (x,y) = (1/2,1/6).  t=1 -> (x,y) = (1/6,1/2)

c) dx/dt = 1/2 - 3*t + 2*t^2.

   Solutions t = (3 +- sqrt(9-4*.5*2))/4 = (3 +- sqrt(5))/4 = 0.191, and something > 1.
   So answer = (.191,.545)
   
   dy/dt = 1/2 + t - 2*t^2.
   
   Solutions t = (-1 +- sqrt(1+4*2*.5))/-4 = (1 -+ sqrt(5))/4 = 0.809, and something < 0.
   So answer = (.809,.545)

d) see b_spline_poly.png

e) see b_spline.png


2.2 (Bezier)

a) matrix is [-1 3 -3 1; 3 -6 3 0; -3 3 0 0; 1 0 0 0]. This gives

x(t) = -1 + 6*t - 9*t^2 + 4*t^3
y(t) =  0 +   0 + 3*t^2 - 4*t^3

b) t=0 -> (x,y) = (-1,0). t=1 -> (x,y) = (0,-1)

c) dx/dt = 6 - 18*t + 12*t^2.

   Solutions t = (18 +- sqrt(324-4*6*12))/24 = (18+-6)/24 = 1/2 and 1
   So answer = (1/2, 1/4) and (1,0)
   
   dy/dt = 6*t - 12*t^2.
   Solutions t = (-6 +- sqrt(36+4*12*0))/-24 = (6-+6)/24 = 0 and 1/2
   So answer = (0,0) and (1/2,1/4)

d) see bezier_poly.png

e) see bezier.png