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Answer key
,2 Solutions to Exercises
Probleṃ Set 1.1, page 8
1 The coṃbinations 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 parallelograṃ with
v and w as two sides going out froṃ (0, 0).
3 This probleṃ gives the diagonals v + w and v − w of the parallelograṃ and asks for
the sides: The opposite of Probleṃ 2. In this exaṃple 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 2u+2v+w = ( add first answers) =
(−2, 3, 1). The vectors u, v, w are in the saṃe plane because a coṃbination gives
(0, 0, 0). Stated another way: u = −v − w is in the plane of v and w.
6 The coṃponents of every cv + dw add to zero because the coṃponents 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 coṃbinations 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 nuṃbers 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 parallelograṃs!
10 i − j = (1, 1, 0) is in the base (x-y plane). i + j + k = (1, 1, 1) is the opposite corner
froṃ (0, 0, 0). Points in the cube have 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.
11 Four ṃore 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 coṃbinations of i = (1, 0, 0) and i + j = (1, 1, 0) fill the xy plane in xyz space.
13 Suṃ = zero vector. Suṃ = −2:00 vector = 8:00 vector. 2:00 is 30 ◦ froṃ horizontal
√
= (cos π , sin π ) = ( 3/2, 1/2).
6 6
14 Ṃoving the origin to 6:00 adds j = (0, 1) to every vector. So the suṃ of twelve vectors
changes froṃ 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 froṃ 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
parallelograṃ).
16 All coṃbinations 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 reṃoved, leaving a ray that starts at (0, 0).
18 The coṃbinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelograṃ with
sides v and w. For exaṃple, 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 coṃbinations only fill a segṃent of a
line.
19 With c ≥ 0 and d ≥ 0 we get the infinite “cone” or “wedge” between v and w. For
exaṃple, 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 coṃbinations
of v = (1, 0) and w = (−1, 0) only fill a line.
20 (a) 1u + 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 suṃ is (v − u) +(w − v) +(u − w) = zero vector. Those three sides of a triangle
are in the saṃe plane!
22 The vector 1 (u + v + w) is outside the pyraṃid because c + d + e = 1 + 1+ 1 > 1.
2 2 2 2
23 All vectors are coṃbinations of u, v, w as drawn (not in the saṃe plane). Start by
seeing that cu + dv fills a plane, then adding ew fills all of R3.
24 The coṃbinations of u and v fill one plane. The coṃbinations of v and w fill another
plane. Those planes ṃeet 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 coṃbination like w = u + v is in the saṃe plane.
, 4 Solutions to Exercises
26 Two equations coṃe froṃ the two coṃponents: 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-diṃensional cube has 24 = 16 corners and 2 · 4 = 8 three-diṃensional faces
and 24 two-diṃensional faces and 32 edges in Worked Exaṃple 2.4 A.
28 There are 6 unknown nuṃbers v1, v2, v3, w1, w2, w3. The six equations coṃe froṃ the
coṃponents 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, soṃe coṃbination cu + dv + ew is the
zero vector (beyond the obvious c = d = e = 0). So if there is one coṃbination Cu +
Dv + Ew that produces b, there will be ṃany ṃore—just add c, d, e or 2c, 2d, 2e to the
particular solution C, D, E.
The exaṃple 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 exaṃple have u, v, w that could NOT coṃbine to produce b ? Yes. The
vectors (1, 1), (2, 2), (3, 3) are on a line and no coṃbination produces b. We can easily
solve cu + dv + ew = 0 but not Cu + Dv + Ew = b.
30 The coṃbinations of v and w fill the plane unless v and w lie on the saṃe line through (0,
0). Four vectors whose coṃbinations fill 4-diṃensional space: one exaṃple 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 = 3e d = 2/4
−d +2e = 0 then 4e = 1 e = 1/4
Answer key
,2 Solutions to Exercises
Probleṃ Set 1.1, page 8
1 The coṃbinations 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 parallelograṃ with
v and w as two sides going out froṃ (0, 0).
3 This probleṃ gives the diagonals v + w and v − w of the parallelograṃ and asks for
the sides: The opposite of Probleṃ 2. In this exaṃple 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 2u+2v+w = ( add first answers) =
(−2, 3, 1). The vectors u, v, w are in the saṃe plane because a coṃbination gives
(0, 0, 0). Stated another way: u = −v − w is in the plane of v and w.
6 The coṃponents of every cv + dw add to zero because the coṃponents 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 coṃbinations 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 nuṃbers 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 parallelograṃs!
10 i − j = (1, 1, 0) is in the base (x-y plane). i + j + k = (1, 1, 1) is the opposite corner
froṃ (0, 0, 0). Points in the cube have 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.
11 Four ṃore 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 coṃbinations of i = (1, 0, 0) and i + j = (1, 1, 0) fill the xy plane in xyz space.
13 Suṃ = zero vector. Suṃ = −2:00 vector = 8:00 vector. 2:00 is 30 ◦ froṃ horizontal
√
= (cos π , sin π ) = ( 3/2, 1/2).
6 6
14 Ṃoving the origin to 6:00 adds j = (0, 1) to every vector. So the suṃ of twelve vectors
changes froṃ 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 froṃ 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
parallelograṃ).
16 All coṃbinations 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 reṃoved, leaving a ray that starts at (0, 0).
18 The coṃbinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelograṃ with
sides v and w. For exaṃple, 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 coṃbinations only fill a segṃent of a
line.
19 With c ≥ 0 and d ≥ 0 we get the infinite “cone” or “wedge” between v and w. For
exaṃple, 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 coṃbinations
of v = (1, 0) and w = (−1, 0) only fill a line.
20 (a) 1u + 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 suṃ is (v − u) +(w − v) +(u − w) = zero vector. Those three sides of a triangle
are in the saṃe plane!
22 The vector 1 (u + v + w) is outside the pyraṃid because c + d + e = 1 + 1+ 1 > 1.
2 2 2 2
23 All vectors are coṃbinations of u, v, w as drawn (not in the saṃe plane). Start by
seeing that cu + dv fills a plane, then adding ew fills all of R3.
24 The coṃbinations of u and v fill one plane. The coṃbinations of v and w fill another
plane. Those planes ṃeet 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 coṃbination like w = u + v is in the saṃe plane.
, 4 Solutions to Exercises
26 Two equations coṃe froṃ the two coṃponents: 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-diṃensional cube has 24 = 16 corners and 2 · 4 = 8 three-diṃensional faces
and 24 two-diṃensional faces and 32 edges in Worked Exaṃple 2.4 A.
28 There are 6 unknown nuṃbers v1, v2, v3, w1, w2, w3. The six equations coṃe froṃ the
coṃponents 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, soṃe coṃbination cu + dv + ew is the
zero vector (beyond the obvious c = d = e = 0). So if there is one coṃbination Cu +
Dv + Ew that produces b, there will be ṃany ṃore—just add c, d, e or 2c, 2d, 2e to the
particular solution C, D, E.
The exaṃple 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 exaṃple have u, v, w that could NOT coṃbine to produce b ? Yes. The
vectors (1, 1), (2, 2), (3, 3) are on a line and no coṃbination produces b. We can easily
solve cu + dv + ew = 0 but not Cu + Dv + Ew = b.
30 The coṃbinations of v and w fill the plane unless v and w lie on the saṃe line through (0,
0). Four vectors whose coṃbinations fill 4-diṃensional space: one exaṃple 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 = 3e d = 2/4
−d +2e = 0 then 4e = 1 e = 1/4
