Math 138 Problem Set 10 Solutions
Problem 1
a. Given distinct lines l and l , and a point C not on either line, we dene
perspectivity [C ; l l ] has a mapping that sens each point P l to the
intersection of line P C and l .
i. Proof of well-denition: Take
Math E-138 Problem Set 2 Solutions
February 25, 2010
1
Ane Multiplication
We have (Emma) (Neil) = (Neil) (Emma) = Dave (so Emma and Neil are multiplicative inverses).
We show the multiplication in Figure 1.1. Note that Committees s = ACE and u = BDD inter
Math E-138 Problem Set 9 Solutions
May 1, 2010
1
Spherical Glide Reections
Suppose that and intersect at points P and P . Let l be the line through P perpendicular
to and let F be the point of intersection of l and that lies on the segment of that forms
a
Problem 1:
a.
b. Prove that , B , B = , B 2 . Proof: By construction, the line joining poles
and is a lightlike line, whose pole is the lightlike intersection of r and s. Thus the pole of
the line p through and , must lie on p. That is, we have that B 1
Problem Set 122 Solutions
1. a) As the triangles are equilateral, all of their angles are equal. The angles around the
common point must sum to 2 . Since all triangles are congruent, each angle is
=
2
7
(0.1)
Now by the dual cosine law, the side-length a
Math E-138 Problem Set 1 Solutions
February 24, 2010
1
Practice with Axioms
(a) First we show that if p + q = p + r = s, then q = r. Given vectors p and s, Axiom A2
guarantees the existence of a vector q such that p + q = s. Suppose there exists another v
Math E-138 Problem Set 8 Solutions
April 21, 2010
1
Spherical Geography
(a)
p = cos1 P, Q = cos1 (cos 1.38 cos0.91) = 1.4541 radians.
(b)
Q = cos1
cos q cos p cos l
sin p sin l
= cos1
cos 0.91 cos 1.4541 cos 1.38
sin 1.4541 sin 1.38
= 0.9189 radians
L = c
Math 138 Problem Set 7 Solutions
April 5, 2010
The rst three problems are from Problem Set 6.
1
Galilean Theorem of Menelaus
(a) The diagram below illustrates the theorem:
y
C
F
E
A
T1 t1
T2
B
t2
D
T
t
(b) The theorem is the Galilean version of the theore
Math E-138 Problem Set 3 Solutions
March 28, 2010
1
Reection Matrices
(a) We have
sin 2 = 2 sin cos =
sin
2 cos
cos
2 sin cos
=
sin
cos2 + sin2 cos
1 + cos
1
cos2 sin2 cos
=
cos 2 = cos sin =
2 + sin2
cos
cos
1+
2
2
2
=
sin 2
cos
sin 2
cos
2 ta
MATH 130 PROBLEM SET 4 SOLUTIONS
Problem 1
a) A reection in the x-axis followed by a reection in a line of slope 1 is a rotation twice the
3
counter-clockwise angle formed between the x-axis and the line, about the point where the line
intersects the x-ax
Math E-138 Problem Set 5 Solutions
March 23, 2010
1
Determining an Ane Transformation
Let P = (1, 1), P = (4, 2), Q = (1, 1), Q = (2, 0), R = (1, 1), and R = (0, 6). Because
cfw_Q P, R P = cfw_(2, 0), (0, 2) and cfw_Q P , R P = cfw_(6, 2), (4, 8) are ba