2
Solutions to Exercises
Problem Set 1.1, page 8
1 The combinations give (a) a line in R
3
(b) a plane in R
3
(c) all of R
3
.
2 v Cw D .2; 3/ and v w D .6; 1/ will be the diagonals of the parallelogram with v
and w as two sides going out from .0; 0/.
3 This problem gives the diagonals v C w and v w of the parallelogram and asks for
the sides: The opposite of Problem 2. In this example v D .3; 3/ and w D .2; 2/.
4 3v C w D .7; 5/ and cv C d w D .2c C d; c C 2d /.
5 uCv D .2; 3; 1/ and uCvCw D .0; 0; 0/ and 2uC2vCw D . add first answers/ D
.2; 3; 1/. The vectors u; v; w are in the same plane because a combination gives
.0; 0; 0/. Stated another way: u D v w is in the plane of v and w.
6 The components of every cv C d w add to zero. c D 3 and d D 9 give .3; 3; 6/.
7 The nine combinations c.2; 1/ C d.0; 1/ with c D 0; 1; 2 and d D .0; 1; 2/ will lie on
a lattice. If we took all whole numbers c and d, the lattice would lie over the whole
plane.
8 The other diagonal is v w (or else w v ). Adding diagonals gives 2v (or 2w).
9 The fourth corner can be .4; 4/ or .4; 0/ or .2; 2/. Three possible parallelograms!
10 i j D .1; 1; 0/ is in the base (x-y plane). i Cj Ck D .1; 1; 1/ is the opposite corner
from .0; 0; 0/. Points in the cube have 0 x 1, 0 y 1, 0 z 1.
11 Four more corners .1; 1; 0/; .1; 0; 1/; .0; 1; 1/; .1; 1; 1/. The center point is .
1
2
;
1
2
;
1
2
/.
Centers of faces are .
1
2
;
1
2
; 0/; .
1
2
;
1
2
; 1/ and .0;
1
2
;
1
2
/; .1;
1
2
;
1
2
/ and .
1
2
; 0;
1
2
/; .
1
2
; 1;
1
2
/.
12 A four-dimensional cube has 2
4
D 16 corners and 2 4 D 8 three-dimensional faces
and 24 two-dimensional faces and 32 edges in Worked Example 2.4 A.
13 Sum D zero vector. Sum D 2:00 vector D 8:00 vector. 2:00 is 30
ı
from horizontal
D .cos
6
; sin
6
/ D .
p
3=2; 1=2/.
14 Moving the origin to 6:00 adds j D . 0; 1/ to every vector. So the sum of twelve vectors
changes from 0 to 12j D .0; 12/.
15 The point
3
4
v C
1
4
w is three-fourths of the way to v starting from w. The vector
1
4
v C
1
4
w is halfway to u D
1
2
v C
1
2
w. The vector v C w is 2u (the far corner of the
parallelogram).
16 All combinations with c C d D 1 are on the line that passes through v and w.
The point V D v C 2w is on that line but it is beyond w.
17 All vectors cv C cw are on the line passing through .0; 0/ and u D
1
2
v C
1
2
w. That
line continues out beyond v C w and back beyond .0; 0/. With c 0, half of this line
is removed, leaving a ray that starts at .0; 0/.
18 The combinations cv C dw with 0 c 1 and 0 d 1 fill the parallelogram with
sides v and w. For example, if v D .1; 0/ and w D .0; 1/ then cv C d w fills the unit
square.
19 With c 0 and d 0 we get the infinite “cone” or “wedge” between v and w. For
example, if v D .1; 0/ and w D .0; 1/, then the cone is the whole quadrant x 0,
y 0. Question: What if w D v? The cone opens to a half-space.
