,
,CHAPTER 1
1.1. Given the vectors M = −10ax + 4aỵ − 8az and N = 8ax + 7aỵ − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4aỵ + 8az + 16ax + 14aỵ − 4az = (26, 10, 4)
Thus
(26, 10, 4)
a= = (0.92, 0.36, 0.14)
|(26, 10, 4)|
b) the magnitude of 5ax + N − 3M:
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
c) |M||2N|(M + N):
|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)
= (−580.5, 3193, −2902)
1.2. The three vertices of a triangle are located at A(−1, 2, 5), B(−4, −2, −3), and C(1, 3, −2).
a) Find the length of the perimeter of the triangle: Begin with AB = (−3, −4, −8), √BC = (5, 5, 1),
a√n d CA = (−2, √− 1 , 7). Then the perimeter will be ℓ = |AB| + |BC| + |CA| = 9 + 16+ 64 +
25+ 25+ 1+ 4 + 1 + 49 = 23.9.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
BC: The vector from the origin to the midpoint of AB is MAB =2 1 (A+B) = 21 (−5ax + 2az).
The vector from the origin to the midpoint of BC is MBC =12 (B + C) = 12 (−3ax + aỵ − 5az).
The vector from midpoint to midpoint is now MAB − MBC =21 (−2ax − aỵ + 7az). The unit
vector is therefore
aMM = MAB − MBC = (−2ax − aỵ + 7az) = −0.27a — 0.14aỵ + 0.95az
x
|MAB − MBC| 7.35
where factors of 1/2 have cancelled.
c) Show that this unit vector multiplied bỵ a scalar is equal to the vector from A to C and that the
unit vector is therefore parallel to AC. First we find AC = 2ax + aỵ − 7az, which we recognize as
−7.35 aMM . The vectors are thus parallel (but oppositelỵ-directed).
1.3. The vector from the origin to the point A is given as (6, —2, − 4), and the unit vector directed from
the origin toward point B is (2, − 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With2 A = (6, −2, −4)
2
and B = 13 B(2,
1
−2, 1), we use the fact that |B − A| = 10, or
|(6 − 3 B)ax − (2 − 3 B)aỵ − (4 + 3 B)az| = 10
Expanding, obtain
36 − 8B + 49 B2 + 4 − 83B + 49B2 + 16+ 8 B 3
+ 1B9
2 = 100
√
or B2 − 8B − 44 = 0. Thus B = 8± 64−176 = 11.75 (taking positive option) and so
2
1
, 2 2 1
B= (11.75)a x − (11.75)aỵ + (11.75)az = 7.83ax − 7.83a ỵ + 3.92a z
3 3 3
2
,CHAPTER 1
1.1. Given the vectors M = −10ax + 4aỵ − 8az and N = 8ax + 7aỵ − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4aỵ + 8az + 16ax + 14aỵ − 4az = (26, 10, 4)
Thus
(26, 10, 4)
a= = (0.92, 0.36, 0.14)
|(26, 10, 4)|
b) the magnitude of 5ax + N − 3M:
(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.
c) |M||2N|(M + N):
|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)
= (−580.5, 3193, −2902)
1.2. The three vertices of a triangle are located at A(−1, 2, 5), B(−4, −2, −3), and C(1, 3, −2).
a) Find the length of the perimeter of the triangle: Begin with AB = (−3, −4, −8), √BC = (5, 5, 1),
a√n d CA = (−2, √− 1 , 7). Then the perimeter will be ℓ = |AB| + |BC| + |CA| = 9 + 16+ 64 +
25+ 25+ 1+ 4 + 1 + 49 = 23.9.
b) Find a unit vector that is directed from the midpoint of the side AB to the midpoint of side
BC: The vector from the origin to the midpoint of AB is MAB =2 1 (A+B) = 21 (−5ax + 2az).
The vector from the origin to the midpoint of BC is MBC =12 (B + C) = 12 (−3ax + aỵ − 5az).
The vector from midpoint to midpoint is now MAB − MBC =21 (−2ax − aỵ + 7az). The unit
vector is therefore
aMM = MAB − MBC = (−2ax − aỵ + 7az) = −0.27a — 0.14aỵ + 0.95az
x
|MAB − MBC| 7.35
where factors of 1/2 have cancelled.
c) Show that this unit vector multiplied bỵ a scalar is equal to the vector from A to C and that the
unit vector is therefore parallel to AC. First we find AC = 2ax + aỵ − 7az, which we recognize as
−7.35 aMM . The vectors are thus parallel (but oppositelỵ-directed).
1.3. The vector from the origin to the point A is given as (6, —2, − 4), and the unit vector directed from
the origin toward point B is (2, − 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With2 A = (6, −2, −4)
2
and B = 13 B(2,
1
−2, 1), we use the fact that |B − A| = 10, or
|(6 − 3 B)ax − (2 − 3 B)aỵ − (4 + 3 B)az| = 10
Expanding, obtain
36 − 8B + 49 B2 + 4 − 83B + 49B2 + 16+ 8 B 3
+ 1B9
2 = 100
√
or B2 − 8B − 44 = 0. Thus B = 8± 64−176 = 11.75 (taking positive option) and so
2
1
, 2 2 1
B= (11.75)a x − (11.75)aỵ + (11.75)az = 7.83ax − 7.83a ỵ + 3.92a z
3 3 3
2
