CSE 152 – Homework 1 solution
Question 1
a.
In Euclidean coordinates, denote the line as y = ax + b, we have
1 = a + b
pass through (1,1)
2 = 3a + b
pass through (3,2)
=> a = 1/2, b = 1/2 => y = x/2 + 1/2 => -x + 2y -1 = 0
b.
In homogeneous coordinates, the line connecting (1,1) and (3,2) is
(1,1,1)
×
(3,2,1) = (-1,2,-1)
=> line equation: -x + 2y -1 = 0
c.
Four corners: A=(1,3), B=(2,4), C=(5,2), D=(2,1) => 4 edges
AB: (1,3,1)
×
(2,4,1) =
(-1,1,-2)
=> -x +y -2 = 0
BC: (2,4,1)
×
(5,2,1) =
(2,3,-16)
=> 2x + 3y -16 = 0
CD: (5,2,1)
×
(2,1,1) =
(1,-3,1)
=> x – 3y +1 = 0
DA: (2,1,1)
×
(1,3,1) =
(-2,-1,5)
=> -2x -y + 5 = 0
Two vanishing points
AB
×
CD = (-1,1,-2)
×
(1,-3,1) = (-5,-1,2)
=> (-5/2, -1/2) in Euclidean
BC
×
DA = (2,3,-16)
×
(-2,-1,5) = (-1,22,4)
=> (-1/4, 22/4) in Euclidean
Question 3
Homogeneous coordinates of the rectangle (4 corners):
P
1
= (-6, -3, 6, 1)
T
, P
2
= (3, -3, 6, 1)
T
, P
3
= (3, 3, 9, 1)
T
, P
4
= (-6, 3, 9, 1)
T
a.
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
=
0
3
1
0
0
0
0
1
0
0
0
0
1
e
perspectiv
M
. The projection of the rectangle:
M
perspective
* P
1
= (-6, -3, 2)
T
=> Euclidean (-3, -1.5)
T
M
perspective
* P
2
= (3, -3, 2)
T
=> Euclidean (1.5, -1.5)
T
M
perspective
* P
3
= (3, 3, 3)
T
=> Euclidean (1, 1)
T
M
perspective
* P
4
= (-6, 3, 3)
T
=> Euclidean (-2, 1)
T
b.