MATH 130 SOLUTIONS  PS1
1 pg. 35
(i) Without loss of generality, let x = (a1 ; a2 ), y = (b1 ; b2 ), and z = (c1 ; c2 ). Then, by the associativity
of the real numbers R and the denition of vector addition, we have
(x + y) + z = (a1 + b1 ; a2 + b2 ) + (c
MATH 130 SOLUTIONS  PS7
First, note that as bcc, we have that
b
c =
c
b . Thus, we have
a
b
c =
a
c
b
Then, as a and c are both perpendicular to b,
a
c is a rotation, and as b is perpendicular to them, we
conclude that
a
b
c is a glide reec
MATH 130 SOLUTIONS  PS2
18 pg. 36 Let T =
m
n be the translation. Then m and n are perpendicular to lH too, so that by
denition,
m
n = T is also a translation along lH
20 pg. 37 For all x P R2 , we have
v w (x) = v (x + w) = x + w + v = x + (w + v) =
MATH 130 SOLUTIONS  PS6
#1 First, we note that by Theorem 4, T (P ) = P . Further, by theorem 3, each of the direction vectors
of the three lines goes to a scalar multiple of itself. Finally, we note that the three direction vectors are all
distinct. Now
MATH 130 SOLUTIONS  PS3
# 1. Z=6Z is not a eld, as 2 f0; 1; 2; 3; 4; 5g = f0; 2; 4; 0; 2; 4g, so that 2 has no inverse.
Z=9Z is not a eld, as 3 Z=9Z = f0; 3; 6; 0; 3; 6; 0; 3; 6g, so that 3 has no inverse.
# 2. d = (3; 10) = 1. Now note that 3(3) + 10(1)
MATH 130 SOLUTIONS  PS5
p
p
p
#1 The sides of Triangle 1 have lengths 1:82 + 2:42 = :6 32 + 42 = 3, 3:22 + 2:42 = :8(5) = 4, and 5.
So do the sides of triangle 2. Now, we our isometry is given by the following picture
Where the rst map is a tranlation by
MATH 130 SOLUTIONS  PS8
22 pg. 122
#2 We know that for a projective plane of order n, the number of lines and points is n2 + n + 1. Thus,
the Fano plane is the projective plane of order 2. But from P3, we have at least 4 points. And if n = 1, then
the nu
MATH 130 SOLUTIONS  PS3
# 1. We have A = v , B = v , D = w, and E = w. Then the direction between B and D is v w, so
that the line passing through E parallel to BD is given by
w + [v w]
This intersects the line that A and B are on (which is given by [v ]
MATH 130 SOLUTIONS PS 11

Theorem 19: Suppose (x; y; z ) P H. Then (x; y; z ) = (x=z; y=z; 1). Further b( (x; y; z ); (x; y; z ) =
+ y 2 =z 2 1. Thus,
2
2
2
x + y z = 1 =
A x2 =z2 + y2 =z2 1 < 0
so that (x; y; z ) P D, as desired. To show surjectivity, t
MATH 130 SOLUTIONS  PS9
#2 pg. 179
Let = (1; 1; 1) and = (1; 0; 0). Then we see that ; are spacelike, = (0; 1; 1) is
lightlike, but that b(; )2 = 1 = 1 1 = b(; )b(; ).
3 pg. 179
Suppose that two lines l and m, with poles and are parallel(ultraparallel).