C H A P T E R 1
Vectors
1.1 VECTORS IN TWO AND THREE DIMENSIONS
1. Here we just connect the point (0, 0) to the points indicated :
y
3
b
2.5
2
c
1.5
1 a
0.5
x
-1 1 2 3
2. Although more difficult for students to represent this on paper, the figures should look something like the following.
Note that the origin is not at a corner of the frame box but is at the tails of the three vectors.
3
2 a
z
1 b
c
0
-2 -2 0 2
0
2
x y
In problems 3 and 4, we supply more detail than is necessary to stress to students what properties are being used :
3. (a) 3, 1 C −1, 7 D 3 C [−1], 1 C 7 D 2, 8.
(b) −28, 12 D −2 · 8, −2 · 12 D −16, −24.
(c) 8, 9 C 3−1, 2 D 8 C 3−1, 9 C 32 D 5, 15.
(d) 1, 1 C 52, 6 − 310, 2 D 1 C 5 · 2 − 3 · 10, 1 C 5 · 6 − 3 · 2 D −19, 25.
(e) 8, 10 C 38, −2 − 24, 5 D 8 C 38 − 2 · 4, 10 C 3−2 − 2 · 5 D 8, −26.
4. (a) 2, 1, 2 C −3, 9, 7 D 2 − 3, 1 C 9, 2 C 7 D −1, 10, 9.
1
(b) 8, 4, 1 C 2 5, −7, 1 D 4, 2, 1 C 10, −14, 1 D 14, −12, 1.
2 4 2 2
(c) −2 2, 0, 1 − 6 1 , −4, 1 D −22, 0, 1 − 3, −24, 6 D −2−1, 24, −5 D 2, −48, 10.
2
5. We start with the two vectors a and b. We can complete the parallelogram as in the figure on the left. The vector
from the origin to this new vertex is the vector a C b. In the figure on the right we have translated vector b so
that its tail is the head of vector a. The sum a C b is the directed third side of this triangle.
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. 1
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 2 Chapter 1 Vectors
y y
7 7
6 6
a+b a+b
5 5 b translated
b 4 b 4
3 3
2 a 2 a
1 1
x x
-2 -1.5 -1 -0.5 0.5 1 -2 -1.5 -1 -0.5 0.5 1
6. a D 3, 2 b D −1, 1
1
a − b D 3 − −1, 2 − 1 D 4, 1 a D 3 , 1 a C 2b D 1, 4
2 2
y
a+2b
4
3
2 a
b 1 a-b
(1/2)a
x
-2 -1 1 2 3 4 5
-1
−
' −
' −'
7. (a) AB D −3 − 1, 3 − 0, 1 − 2 D −4, 3, −1 BA D −AB D 4, −3, 1
−
'
(b) AC D 2 − 1, 1 − 0, 5 − 2 D 1, 1, 3
−
'
BC D 2 − −3, 1 − 3, 5 − 1 D 5, −2, 4
−
' −'
AC C CB D 1, 1, 3 − 5, −2, 4 D −4, 3, −1
(c) This result is true in general:
B
A
C
Head-to-tail addition demonstrates this.
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Section 1.1 Vectors in Two and Three Dimensions 3
8. The vectors a D 1, 2, 1, b D 0, −2, 3 and a C b D 1, 2, 1 C 0, −2, 3 D 1, 0, 4 are graphed below.
Again note that the origin is at the tails of the vectors in the figure.
Also, −11, 2, 1 D −1, −2, −1. This would be pictured by drawing the vector (1, 2, 1) in the opposite
direction. Finally, 41, 2, 1 D 4, 8, 4 which is four times vector a and so is vector a stretched four times as
long in the same direction.
4
b
a+b
z
2
a
0
-2 0
1 0 2
x y
9. Since the sum on the left must equal the vector on the right componentwise:
−12 C x D 2, 9 C 7 D y, and z C −3 D 5. Therefore, x D 14, y D 16, and z D 8.
10. If we drop a perpendicular
√ from√ (3, 1) to the x-axis we see that by the Pythagorean Theorem the length of the
vector 3, 1 D 32 C 12 D 10.
y
1
0.8
0.6
0.4
0.2
x
0.5 1 1.5 2 2.5 3
11. Notice that b (represented by the dotted line) D 5a (represented by the solid line).
y
10
8 b
6
4
2
a
x
1 2 3 4 5
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 4 Chapter 1 Vectors
12. Here the picture has been projected into two dimensions so that you can more clearly see that a (represented by
the solid line) D −2b (represented by the dotted line).
a 8
6
4
2
-8 -6 -4 -2 2 4
-2
b
-4
13. The natural extension to higher dimensions is that we still add componentwise and that multiplying a scalar by a
vector means that we multiply each component of the vector by the scalar. In symbols this means that:
a C b D a1 , a2 , ... , an C b1 , b2 , ... , bn D a1 C b1 , a2 C b2 , ... , an C bn
and ka D ka1 , ka2 , ... , kan .
In our particular examples, 1, 2, 3, 4 C 5, −1, 2, 0 D 6, 1, 5, 4,
and 27, 6, −3, 1 D 14, 12, −6, 2.
14. The diagrams for parts (a), (b) and (c) are similar to Figure 1.12 from the text. The displacement vectors are:
(a) (1, 1, 5)
(b) −1, −2, 3
(c) 1, 2, −3
(d) −1, −2
Note: The displacement vectors for (b) and (c) are the same but in opposite directions (i.e., one is the negative
of the other). The displacement vector in the diagram for (d) is represented by the solid line in the figure below:
y
1
0.75 P1
0.5
0.25
x
0.5 1 1.5 2 2.5 3
-0.25
-0.5
P2
-0.75
-1
15. In general, we would define the displacement vector from a1 , a2 , ... , an to b1 , b2 , ... , bn to be b1 − a1 , b2 −
a2 , ... , bn − an .
In this specific problem the displacement vector from P1 to P2 is 1, −4, −1, 1.
−'
16. Let B have coordinates (x, y, z). Then AB D x − 2, y − 5, z C 6 D 12, −3, 7 so x D 14, y D 2, z D 1
so B has coordinates (14, 2, 1).
17. If a is your displacement vector from the Empire State Building and b your friend’s, then the displacement vector
from you to your friend is b − a.
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Vectors
1.1 VECTORS IN TWO AND THREE DIMENSIONS
1. Here we just connect the point (0, 0) to the points indicated :
y
3
b
2.5
2
c
1.5
1 a
0.5
x
-1 1 2 3
2. Although more difficult for students to represent this on paper, the figures should look something like the following.
Note that the origin is not at a corner of the frame box but is at the tails of the three vectors.
3
2 a
z
1 b
c
0
-2 -2 0 2
0
2
x y
In problems 3 and 4, we supply more detail than is necessary to stress to students what properties are being used :
3. (a) 3, 1 C −1, 7 D 3 C [−1], 1 C 7 D 2, 8.
(b) −28, 12 D −2 · 8, −2 · 12 D −16, −24.
(c) 8, 9 C 3−1, 2 D 8 C 3−1, 9 C 32 D 5, 15.
(d) 1, 1 C 52, 6 − 310, 2 D 1 C 5 · 2 − 3 · 10, 1 C 5 · 6 − 3 · 2 D −19, 25.
(e) 8, 10 C 38, −2 − 24, 5 D 8 C 38 − 2 · 4, 10 C 3−2 − 2 · 5 D 8, −26.
4. (a) 2, 1, 2 C −3, 9, 7 D 2 − 3, 1 C 9, 2 C 7 D −1, 10, 9.
1
(b) 8, 4, 1 C 2 5, −7, 1 D 4, 2, 1 C 10, −14, 1 D 14, −12, 1.
2 4 2 2
(c) −2 2, 0, 1 − 6 1 , −4, 1 D −22, 0, 1 − 3, −24, 6 D −2−1, 24, −5 D 2, −48, 10.
2
5. We start with the two vectors a and b. We can complete the parallelogram as in the figure on the left. The vector
from the origin to this new vertex is the vector a C b. In the figure on the right we have translated vector b so
that its tail is the head of vector a. The sum a C b is the directed third side of this triangle.
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. 1
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 2 Chapter 1 Vectors
y y
7 7
6 6
a+b a+b
5 5 b translated
b 4 b 4
3 3
2 a 2 a
1 1
x x
-2 -1.5 -1 -0.5 0.5 1 -2 -1.5 -1 -0.5 0.5 1
6. a D 3, 2 b D −1, 1
1
a − b D 3 − −1, 2 − 1 D 4, 1 a D 3 , 1 a C 2b D 1, 4
2 2
y
a+2b
4
3
2 a
b 1 a-b
(1/2)a
x
-2 -1 1 2 3 4 5
-1
−
' −
' −'
7. (a) AB D −3 − 1, 3 − 0, 1 − 2 D −4, 3, −1 BA D −AB D 4, −3, 1
−
'
(b) AC D 2 − 1, 1 − 0, 5 − 2 D 1, 1, 3
−
'
BC D 2 − −3, 1 − 3, 5 − 1 D 5, −2, 4
−
' −'
AC C CB D 1, 1, 3 − 5, −2, 4 D −4, 3, −1
(c) This result is true in general:
B
A
C
Head-to-tail addition demonstrates this.
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Section 1.1 Vectors in Two and Three Dimensions 3
8. The vectors a D 1, 2, 1, b D 0, −2, 3 and a C b D 1, 2, 1 C 0, −2, 3 D 1, 0, 4 are graphed below.
Again note that the origin is at the tails of the vectors in the figure.
Also, −11, 2, 1 D −1, −2, −1. This would be pictured by drawing the vector (1, 2, 1) in the opposite
direction. Finally, 41, 2, 1 D 4, 8, 4 which is four times vector a and so is vector a stretched four times as
long in the same direction.
4
b
a+b
z
2
a
0
-2 0
1 0 2
x y
9. Since the sum on the left must equal the vector on the right componentwise:
−12 C x D 2, 9 C 7 D y, and z C −3 D 5. Therefore, x D 14, y D 16, and z D 8.
10. If we drop a perpendicular
√ from√ (3, 1) to the x-axis we see that by the Pythagorean Theorem the length of the
vector 3, 1 D 32 C 12 D 10.
y
1
0.8
0.6
0.4
0.2
x
0.5 1 1.5 2 2.5 3
11. Notice that b (represented by the dotted line) D 5a (represented by the solid line).
y
10
8 b
6
4
2
a
x
1 2 3 4 5
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 4 Chapter 1 Vectors
12. Here the picture has been projected into two dimensions so that you can more clearly see that a (represented by
the solid line) D −2b (represented by the dotted line).
a 8
6
4
2
-8 -6 -4 -2 2 4
-2
b
-4
13. The natural extension to higher dimensions is that we still add componentwise and that multiplying a scalar by a
vector means that we multiply each component of the vector by the scalar. In symbols this means that:
a C b D a1 , a2 , ... , an C b1 , b2 , ... , bn D a1 C b1 , a2 C b2 , ... , an C bn
and ka D ka1 , ka2 , ... , kan .
In our particular examples, 1, 2, 3, 4 C 5, −1, 2, 0 D 6, 1, 5, 4,
and 27, 6, −3, 1 D 14, 12, −6, 2.
14. The diagrams for parts (a), (b) and (c) are similar to Figure 1.12 from the text. The displacement vectors are:
(a) (1, 1, 5)
(b) −1, −2, 3
(c) 1, 2, −3
(d) −1, −2
Note: The displacement vectors for (b) and (c) are the same but in opposite directions (i.e., one is the negative
of the other). The displacement vector in the diagram for (d) is represented by the solid line in the figure below:
y
1
0.75 P1
0.5
0.25
x
0.5 1 1.5 2 2.5 3
-0.25
-0.5
P2
-0.75
-1
15. In general, we would define the displacement vector from a1 , a2 , ... , an to b1 , b2 , ... , bn to be b1 − a1 , b2 −
a2 , ... , bn − an .
In this specific problem the displacement vector from P1 to P2 is 1, −4, −1, 1.
−'
16. Let B have coordinates (x, y, z). Then AB D x − 2, y − 5, z C 6 D 12, −3, 7 so x D 14, y D 2, z D 1
so B has coordinates (14, 2, 1).
17. If a is your displacement vector from the Empire State Building and b your friend’s, then the displacement vector
from you to your friend is b − a.
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
