Here's the '99 midterm solutions:

1a) P1 code is [1000] P2 is [0110]
it's neither a trivial accept, not reject, so must clip

1b) P1' [0000] P2 is still [0110] still have to clip
P1'_x = 2 (clipped against left egde where xwmin = 2)
P1'_y = P1_y + m (xwmin - P1_x) =
= 5 + (7/14)*(2-1) = 
= 5 + 1/2
= 5.5

1c) P1' still [0000]  P2' is clipped against the top, but code is still
[0100]  - have to clip

p2'_x = P1_x + (ywmax - P1_y)/m = 
= 1 + (10-5)/.5 =
= 1 + 10 
= 11

1d) P1' is [0000]  P2'' is [0000] - a trivial accept, done!

P2''_y = 9.5

1e) no further work needed


2a) Calculate the cross product of two vectors in the plane

N = (B-a) x (C - A) = [-1 1 0] x [-1 0 1] = [1 1 1]

the plane equation is x+y+z=1

2b) Calculate the cross product P of N and (C - A)

P = ( C - A ) x N
= [-1 0 1] x [1 1 1]
= [1 +2 -1]

The plane equation is -x+2y-z+d=0.
To figure out what d is, plug in a point that we know is on the
plane, eg. point A, we get -1+0+0+d = 0, hence d = -1.
The full equation is -x+2y-z-1=0.

3a)
The matrix is

 cosT   sinT   0  0
-sinT   cosT   0  0
  0      0     1  0
  0      0     0  1
  
  where T is a function of Z
  
3b) T = 90*z     (ask yourself: what do I multiply z with so that
I can get 45 when z is 1/2 ?)

3c) A twist is essectially a rotation, and a determinant of a rotation matrix
is always 1 (don't have to know determinants for this exam)

4) 

An alternative solution to what we did in the review session today:
Consider it a rotation of 45 degrees, followed by a shear in
the X direction of magnitude Alpha, followed by an inverse rotation
of 45 degrees, AKA
R(45), SHx(Alpha), R(-45)
the matrices are:

 cos45  sin45
-sin45  cos45
               1
			      1
----------------------
  1   0  0  0
Aplha 1  0  0
  0   0  1  0	  
  0   0  0  1	  
---------------------

 cos(-45)  sin(-45) 
-sin(-45)  cos(-45)
                     1
                        1
						
						

5)

Void CircleOf8 (float r, object o)
{
	float dtheta = 360/8;
	for (int i; i<8; i++) 
	{
		glPushMatrix();
		glRotate(dtheta*i, 0, 0, 1);
		glTranslate(r, 0, 0);
		drawObject(o);
		glPopMatrix();
	}
}


6a)
for orthogonal, just drop the z.
Pp:(15, 20, 0)


6b) cabinet is foreshortened by 1/2, so the length in Z
will be the half of what it used to be. X and Y dimensions
will change cos(30) and sin(30) respectively.
The matrix that accomplished this is:

1

            1
	 
	    
sqrt(3)     1 
-------    ---        1  
  2*2      2*2
  
  
                                1 
								
								
6c) Two ways to go about it:
- use the perpective projection matrix

1
   1
      1  1/10
          1
	  
to calculate it

- use similar triangles to see that for x,  15/(10+25) = x'/10
hence x'= 150/35.
Similarly, y'=200/35

---------------------------------------------------------------------------
--------------------------------------------------------------------------

Midterm '98

1)
When m>1, the angle of inclination of the line is > 45 degrees,
so we step in the y direction and calculate the corresponding x. 

In the given equation, we can substitute b with y_i-m*x_i (from 
y = m*x + b)
we also know m = delta_y/delta_x

so from 
P1  = -2*delta_y*x_i + 2*delta_x*(y_i-b) + 2*delta_x - delta_y
(algebra here)
= -delta_y +2*delta_x

P1 = d1-d2,
we know that of P1 < 0, that means d1<d2, ie, the pixel directly above
is closer to the line, so we move directly above, x, stays the same,
y is incremented by 1. 

otherwise, we move to the diagonal right, both x and y are incremented.


2) Cross product the vectors that define the planes

V1 = [2,3,4] 
V2 = [-3,2,-1]

V1xV2 = [5,14,13]
the plane is 5x+14y+13z=0


3a )
M1 swaps the values of x and y, which is the same as doing
a reflection about the x=y axis

M2 takes the x values and makes them negative.
you can say two things about it:
- it scales x by -1
- it reflects numbers about the y-axis

3b) draw a picture and you'll see,
it rotates objects 90 degrees.

3c) M_inv = M2_inv * M1_inv =  

0 -1  0
1  0  0
0  0  1

3d) det(M1) = det(M2) = 1

4)
To rotate in place, you need to bring object to origin,
rotate, translate back to where it came from.

M = T(x,y)*R(theta)*T(x,y)

T(x,y) = 
1 
    1
-x -y  1

R(theta) = 
 cos(theta) sin(theta) 
-sin(theta) cos(theta)
        
T(x,y) = 
1 
   1
x  y  1           
				

5)
void RotateInPlace (x, y, theta)
{
	glPushMatrix()
	glTranslate(x,y,0)
	glRotate(theta, 0, 0, 1);
	glTranslate(-x,-y, 0);
	draw_object();
	glPopMatrix();
}

6) has to be drawn.
you'll end up with two new vertices along the left edge and
two new on the right edge.