Textbook pdf's

Section 6.7, pp. 332–333
1. a. 1i ! 2j ! 4k !
b. about
2. OB ! 4.58
13, 4, 42,
OB ! 6.40
3. 3
4. a. 11,
b. OA ! 6, 112
13, OB ! 3,
OP ! 12.57
c. AB ! ,
AB ! 15, 2, 132
,
AB ! 14.07
represents the vector from the
tip of OA !
to the tip of OB ! . It is the
difference between the two vectors.
5. a.
b.
c.
11, 3, 32
17, 16, 82
a 13 2 , 2, 3 2 b
d. 12, 30, 132
6. a. i ! 2j ! 2k !
b. 3i ! 0j ! 0k !
c. 9i ! 3j ! 3k !
d. 9i ! 3j ! 3k !
7. a. about 5.10
b. about 1.41
c. about 5.39
8.
d. about
x ! ,
y ! i ! 11.18
2i ! 4j !
2j ! k !
6k !
9. a. The vectors OA ! , OB !
, and OC !
represent the xy-plane, xz-plane,
and yz-plane, respectively. They
are also the vector from the origin
to points 1a, b, 02, 1a, 0, c2, and
10, , respectively.
b. OA ! b, c2
,
OB ! ai ! ,
OC ! ai ! bj !
0i ! 0j ! 0k !
bj ! ck !
ck !
c. OA ! ,
OB ! a 2 b 2
a 2 c 2 ,
OB ! b 2
d. , AB ! c 2
10, b, c2 is a direction vector
from A to B.
10. a. 7
b. 13
c. (5, 2, 9)
d. 10.49
e. 15, 2, 92
f. 10.49
11. In order to show that ABCD is a
parallelogram, we must show that
AB ! DC !
or BC ! AD ! . This will
show they have the same direction,
thus the opposite sides are parallel.
AB! 13, 4, 122
!
DC 13, 4, 122
We have shown and
BC ! AD ! AB ! DC !
, so ABCD is a parallelogram.
12. a 2 , b 7,
c 0
3
13. a.
z
x
OB
O
OA
OC
b. V 1 10, 0, 02,
V 2 12, 2, 52,
V 3 10, 4, 12,
V 4 10, 5, 12,
V 5 12, 6, 62,
V 6 12, 7, 42,
V 7 10, 9, 02,
V 8 12, 11, 52
14. (1, 0, 0)
15. 4.36
Section 6.8, pp. 340–341
1. They are collinear, thus a linear
combination is not applicable.
2. It is not possible to use 0 !
in a spanning
set. Therefore, the remaining vectors
only span R 2 .
3. The set of vectors spanned by (0, 1)
is m10, 12. If we let m 1, then
4. i !m10, 12 10, 12. spans the set m11, 0, 02. This is any
vector along the x-axis. Examples:
(2, 0, 0), 121, 0, 02.
5. As in question 2, it isn’t possible to use
0 ! in a spanning set.
6. 511, 22, 11, 126, 512, 42, 11, 126,
511, 12, 13, 626 are all the possible
spanning sets for with 2 vectors.
7. a. 14i !
b. 7i ! 43j ! R 2
23j ! 40k !
14k !
8. 511, 0, 02, 10, 1, 026:
11, 2, 02 111, 0, 02 210, 1, 02
13, 4, 02 311, 0, 02 410, 1, 02
511, 1, 02, 10, 1, 026
11, 2, 02 111, 1, 02 310, 1, 02
13, 4, 02 311, 1, 02 10, 1, 02
A
B
C
y
9. a. It is the set of vectors in the
xy-plane.
b. 211, 0, 02 410, 1, 02
c. By part a., the vector is not in the
xy-plane. There is no combination
that would produce a number other
than 0 for the z-component.
d. It would still only span the
xy-plane. There would be no
need for that vector.
10. a 2, b 24,
c 3
11. 110, 342 2 11, 32 811, 52
12. a. a x y,
b x 2y
b. 12, 32 112, 12 411, 12
1124, 52 11912, 12
11411, 12
14, 112 712, 12
1811, 12
13. a. The statement a11, 2, 32
b14, 1, 22 114, 1, 162 does
not have a consistent solution.
b. 311, 3, 42 510, 1, 12
13, 14, 72
14. 7
15. m 2, n 3; Non-parallel vectors
cannot be equal, unless their
magnitudes equal 0.
16. Answers may vary. For example:
p 6 and q 1,
p 25 and q 0,
p 13 and q 2 3 3
17. As in question 15, non-parallel vectors.
Their magnitudes must be 0 again to
make the equality true.
m 2 2m 3 1m 121m 32
m 1, 3
m 2 m 6 1m 221m 32
m 2, 3
So, when m 3, their sum will be 0.
Review Exercise, pp. 344–347
1. a. false; Let , then:
0a ! b ! b ! a !
0 0a !
1a ! 0
20
00 0
b. true; @a ! 0
b ! 6 0a ! 0
@ and @a ! c ! @ both
represent the lengths of the
diagonal of a parallelogram, the
first with sides a ! and and the
second with sides a ! b !
and c ! ; since
both parallelograms have a ! as a
side and diagonals of equal length
@b ! @ @c ! @.
c. true; Subtracting from both sides
shows that b ! c !a! .
668 Answers
NEL