ALL 10 CHAPTERS COVERED
,2 Solutions to Exercises
Problem Set 1.1, page 8
1 The combinations give (a) a line in R3 (b) a plane in R3 (c) all of R3.
2 v + w = (2, 3) and v − w = (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 + w and v − w of the parallelogram and
asks for the sides: The opposite of Problem 2. In this example v = (3, 3)
and w = (2, −2).
4 3v + w = (7, 5) and cv + dw = (2c + d, c + 2d).
5 u+v = (−2, 3, 1) and u + v + w = (0, 0, 0) and 2 u+2v+w = ( add first
answers) = (−2, 3, 1). The vectors u, v, w are in the saṁe plane because
a coṁbination gives (0, 0, 0). Stated another way: u = −v − w is in the
plane of v and w.
6 The coṁponents of every cv + dw add to zero because the coṁponents of v and of w
add to zero. c = 3 and d = 9 give (3, 3, −6). There is no solution to cv+dw = (3, 3, 6)
because 3 + 3 + 6 is not zero.
7 The nine coṁbinations c(2, 1) + d(0, 1) with c = 0, 1, 2 and d = (0, 1, 2) will lie
on a lattice. If we took all whole nuṁbers 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 parallelograṁs!
10 i − j = (1, 1, 0) is in the base (x-y plane). i + j + k = (1, 1, 1) is the opposite
corner froṁ (0, 0, 0). Points in the cube have 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z
≤ 1.
11 Four ṁore corners (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point is ( 1 , 1 , 1 ).
2 2 2
Centers of faces are ( 1 , 1 , 0), ( 1 , 1 , 1) and (0, 1 , 1 ), (1, 1 , 1 ) and ( 1 , 0, 1 ), ( 1 , 1, 1 ).
2 2 2 2 2 2 2 2 2 2 2
2
12 The coṁbinations of i = (1, 0, 0) and i + j = (1, 1, 0) fill the xy plane in xyz space.
13 Suṁ = zero vector. Suṁ = −2:00 vector = 8:00 vector. 2:00 is 30◦ froṁ horizontal
√
= (cos π , sin π ) = ( 3/2, 1/2).
6 6
14 Ṁoving the origin to 6:00 adds j = (0, 1) to every vector. So the suṁ of twelve
vectors changes froṁ 0 to 12j = (0, 12).
,Solutions to Exercises 3
3 1
15 The point v + w is three-fourths of the way to v starting froṁ w. The vector
4 4
1 1 1 1
v + w is halfway to u = v+ w. The vector v + w is 2u (the far corner of the
4 4 2 2
parallelograṁ).
16 All coṁbinations with c + d = 1 are on the line that passes through v
and w. The point V = −v + 2w is on that line but it is beyond w.
17 All vectors cv + cw are on the line passing through (0, 0) and u = 1 v + 1 w. That
2 2
line continues out beyond v + w and back beyond (0, 0). With c ≥ 0, half of
this line is reṁoved, leaving a ray that starts at (0, 0).
18 The coṁbinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelograṁ
with sides v and w. For exaṁple, if v = (1, 0) and w = (0, 1) then cv + dw
fills the unit square. But when v = (a, 0) and w = (b, 0) these coṁbinations
only fill a segṁent of a line.
19 With c ≥ 0 and d ≥ 0 we get the infinite “cone” or “wedge” between v and
w. For exaṁple, if v = (1, 0) and w = (0, 1), then the cone is the whole
quadrant x ≥ 0, y ≥
0. Question: What if w = −v? The cone opens to a half-space. But the
coṁbinations of v = (1, 0) and w = (−1, 0) only fill a line.
20 (a) 1 u + 1 v + 1 w is the center of the triangle between u, v and w; 1 u + 1 w lies
3 3 3 2 2
between u and w (b) To fill the triangle keep c ≥ 0, d ≥ 0, e ≥ 0, and c + d + e = 1.
21 The suṁ is (v − u) +(w − v ) +(u − w) = zero vector. Those three sides of a
triangle are in the saṁe plane!
22 The vector 1 (u + v + w) is outside the pyraṁid because c + d + e = 1 + 1 + 1
> 1.
2 2 2 2
23 All vectors are coṁbinations of u, v, w as drawn (not in the saṁe plane).
Start by seeing that cu + dv fills a plane, then adding ew fills all of R3.
24 The coṁbinations of u and v fill one plane. The coṁbinations of v and w fill
another plane. Those planes ṁeet in a line: only the vectors cv are in both
planes.
25 (a) For a line, choose u = v = w = any nonzero vector (b) For a plane, choose
u and v in different directions. A coṁbination like w = u + v is in the saṁe plane.
, 4 Solutions to Exercises
26 Two equations coṁe froṁ the two coṁponents: c + 3d = 14 and 2c + d =
8. The solution is c = 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8).
27 A four-diṁensional cube has 24 = 16 corners and 2 · 4 = 8 three-diṁensional
faces and 24 two-diṁensional faces and 32 edges in Worked Exaṁple 2.4 A.
28 There are 6 unknown nuṁbers v1, v2, v3, w1, w2, w3. The six equations coṁe froṁ
the coṁponents of v + w = (4, 5, 6) and v − w = (2 , 5, 8). Add to find 2v =
(6, 10, 14)
so v = (3, 5, 7) and w = (1, 0, −1).
29 Fact : For any three vectors u, v, w in the plane, soṁe coṁbination cu + dv
+ ew is the zero vector (beyond the obvious c = d = e = 0). So if there is
one coṁbination Cu + Dv + Ew that produces b, there will be ṁany ṁore—
just add c, d, e or 2c, 2d, 2e to the particular solution C, D, E.
The exaṁple has 3u − 2v + w = 3(1, 3) − 2(2, 7) + 1(1, 5) = (0, 0). It also has
−2u + 1v + 0w = b = (0, 1). Adding gives u − v + w = (0, 1). In this case c, d, e
equal 3, −2, 1 and C, D, E = −2, 1, 0.
Could another exaṁple have u, v, w that could NOT coṁbine to produce b ? Yes.
The vectors (1, 1), (2, 2), (3, 3) are on a line and no coṁbination produces b. We
can easily solve cu + dv + ew = 0 but not Cu + Dv + Ew = b.
30 The coṁbinations of v and w fill the plane unless v and w lie on the saṁe line
through (0, 0). Four vectors whose coṁbinations fill 4-diṁensional space:
one exaṁple is the “standard basis” (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0,
0, 0, 1).
31 The equations cu + dv + ew = b are
2c −d = 1 So d = 2e c = 3/4
−c +2d −e = 0 then c = d = 2/4
−d +2e = 0 3e then e = 1/4
4e = 1
,2 Solutions to Exercises
Problem Set 1.1, page 8
1 The combinations give (a) a line in R3 (b) a plane in R3 (c) all of R3.
2 v + w = (2, 3) and v − w = (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 + w and v − w of the parallelogram and
asks for the sides: The opposite of Problem 2. In this example v = (3, 3)
and w = (2, −2).
4 3v + w = (7, 5) and cv + dw = (2c + d, c + 2d).
5 u+v = (−2, 3, 1) and u + v + w = (0, 0, 0) and 2 u+2v+w = ( add first
answers) = (−2, 3, 1). The vectors u, v, w are in the saṁe plane because
a coṁbination gives (0, 0, 0). Stated another way: u = −v − w is in the
plane of v and w.
6 The coṁponents of every cv + dw add to zero because the coṁponents of v and of w
add to zero. c = 3 and d = 9 give (3, 3, −6). There is no solution to cv+dw = (3, 3, 6)
because 3 + 3 + 6 is not zero.
7 The nine coṁbinations c(2, 1) + d(0, 1) with c = 0, 1, 2 and d = (0, 1, 2) will lie
on a lattice. If we took all whole nuṁbers 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 parallelograṁs!
10 i − j = (1, 1, 0) is in the base (x-y plane). i + j + k = (1, 1, 1) is the opposite
corner froṁ (0, 0, 0). Points in the cube have 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z
≤ 1.
11 Four ṁore corners (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point is ( 1 , 1 , 1 ).
2 2 2
Centers of faces are ( 1 , 1 , 0), ( 1 , 1 , 1) and (0, 1 , 1 ), (1, 1 , 1 ) and ( 1 , 0, 1 ), ( 1 , 1, 1 ).
2 2 2 2 2 2 2 2 2 2 2
2
12 The coṁbinations of i = (1, 0, 0) and i + j = (1, 1, 0) fill the xy plane in xyz space.
13 Suṁ = zero vector. Suṁ = −2:00 vector = 8:00 vector. 2:00 is 30◦ froṁ horizontal
√
= (cos π , sin π ) = ( 3/2, 1/2).
6 6
14 Ṁoving the origin to 6:00 adds j = (0, 1) to every vector. So the suṁ of twelve
vectors changes froṁ 0 to 12j = (0, 12).
,Solutions to Exercises 3
3 1
15 The point v + w is three-fourths of the way to v starting froṁ w. The vector
4 4
1 1 1 1
v + w is halfway to u = v+ w. The vector v + w is 2u (the far corner of the
4 4 2 2
parallelograṁ).
16 All coṁbinations with c + d = 1 are on the line that passes through v
and w. The point V = −v + 2w is on that line but it is beyond w.
17 All vectors cv + cw are on the line passing through (0, 0) and u = 1 v + 1 w. That
2 2
line continues out beyond v + w and back beyond (0, 0). With c ≥ 0, half of
this line is reṁoved, leaving a ray that starts at (0, 0).
18 The coṁbinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelograṁ
with sides v and w. For exaṁple, if v = (1, 0) and w = (0, 1) then cv + dw
fills the unit square. But when v = (a, 0) and w = (b, 0) these coṁbinations
only fill a segṁent of a line.
19 With c ≥ 0 and d ≥ 0 we get the infinite “cone” or “wedge” between v and
w. For exaṁple, if v = (1, 0) and w = (0, 1), then the cone is the whole
quadrant x ≥ 0, y ≥
0. Question: What if w = −v? The cone opens to a half-space. But the
coṁbinations of v = (1, 0) and w = (−1, 0) only fill a line.
20 (a) 1 u + 1 v + 1 w is the center of the triangle between u, v and w; 1 u + 1 w lies
3 3 3 2 2
between u and w (b) To fill the triangle keep c ≥ 0, d ≥ 0, e ≥ 0, and c + d + e = 1.
21 The suṁ is (v − u) +(w − v ) +(u − w) = zero vector. Those three sides of a
triangle are in the saṁe plane!
22 The vector 1 (u + v + w) is outside the pyraṁid because c + d + e = 1 + 1 + 1
> 1.
2 2 2 2
23 All vectors are coṁbinations of u, v, w as drawn (not in the saṁe plane).
Start by seeing that cu + dv fills a plane, then adding ew fills all of R3.
24 The coṁbinations of u and v fill one plane. The coṁbinations of v and w fill
another plane. Those planes ṁeet in a line: only the vectors cv are in both
planes.
25 (a) For a line, choose u = v = w = any nonzero vector (b) For a plane, choose
u and v in different directions. A coṁbination like w = u + v is in the saṁe plane.
, 4 Solutions to Exercises
26 Two equations coṁe froṁ the two coṁponents: c + 3d = 14 and 2c + d =
8. The solution is c = 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8).
27 A four-diṁensional cube has 24 = 16 corners and 2 · 4 = 8 three-diṁensional
faces and 24 two-diṁensional faces and 32 edges in Worked Exaṁple 2.4 A.
28 There are 6 unknown nuṁbers v1, v2, v3, w1, w2, w3. The six equations coṁe froṁ
the coṁponents of v + w = (4, 5, 6) and v − w = (2 , 5, 8). Add to find 2v =
(6, 10, 14)
so v = (3, 5, 7) and w = (1, 0, −1).
29 Fact : For any three vectors u, v, w in the plane, soṁe coṁbination cu + dv
+ ew is the zero vector (beyond the obvious c = d = e = 0). So if there is
one coṁbination Cu + Dv + Ew that produces b, there will be ṁany ṁore—
just add c, d, e or 2c, 2d, 2e to the particular solution C, D, E.
The exaṁple has 3u − 2v + w = 3(1, 3) − 2(2, 7) + 1(1, 5) = (0, 0). It also has
−2u + 1v + 0w = b = (0, 1). Adding gives u − v + w = (0, 1). In this case c, d, e
equal 3, −2, 1 and C, D, E = −2, 1, 0.
Could another exaṁple have u, v, w that could NOT coṁbine to produce b ? Yes.
The vectors (1, 1), (2, 2), (3, 3) are on a line and no coṁbination produces b. We
can easily solve cu + dv + ew = 0 but not Cu + Dv + Ew = b.
30 The coṁbinations of v and w fill the plane unless v and w lie on the saṁe line
through (0, 0). Four vectors whose coṁbinations fill 4-diṁensional space:
one exaṁple is the “standard basis” (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0,
0, 0, 1).
31 The equations cu + dv + ew = b are
2c −d = 1 So d = 2e c = 3/4
−c +2d −e = 0 then c = d = 2/4
−d +2e = 0 3e then e = 1/4
4e = 1
