CHAPTER 1
1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find:
a) a unit vector in the direction of −M + 2N.
−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)
Thus
a = (26, 10, 4)
|(26, 10, 4)| = (0.92, 0.36, 0.14)
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. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7,−2, 1):
a) Specify the vector A extending from the origin to the point A.
A = (4, 3, 2) = 4ax + 3ay + 2az
b) Give a unit vector extending from the origin to the midpoint of line AB.
The vector from the origin to the midpoint is given by
M = (1/2)(A + B) = (1/2)(4 − 2, 3 + 0, 2 + 5) = (1, 1.5, 3.5)
The unit vector will be
m = (1, 1.5, 3.5)
|(1, 1.5, 3.5)| = (0.25, 0.38, 0.89)
c) Calculate the length of the perimeter of triangle ABC:
Begin with AB = (−6,−3, 3), BC = (9,−2,−4), CA = (3,−5,−1).
Then
|AB| + |BC| + |CA| = 7.35 + 10.05 + 5.91 = 23.32
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.
With A = (6,−2,−4) and B = 13B(2,−2, 1), we use the fact that |B − A| = 10, or
|(6 − 23B)ax − (2 − 23B)ay − (4 + 13B)az| = 10
Expanding, obtain
36 − 8B + 49B2 + 4 − 83B + 49B2 + 16 + 83B + 19B2 = 100
or B2 − 8B − 44 = 0. Thus B = 8±
√
64−176
2
= 11.75 (taking positive option) and so
B = 2
3
(11.75)ax − 2
3
(11.75)ay + 1
3
(11.75)az = 7.83ax − 7.83ay + 3.92az
1
1.4. given points A(8,−5, 4) and B(−2, 3, 2), find:
a) the distance from A to B.
|B − A| = |(−10, 8,−2)| = 12.96
b) a unit vector directed from A towards B. This is found through
aAB = B − A
|B − A| = (−0.77, 0.62,−0.15)
c) a unit vector directed from the origin to the midpoint of the line AB.
a0M = (A + B)/2
|(A + B)/2| =
(3,−1, 3)
√
19
= (0.69,−0.23, 0.69)
d) the coordinates of the poin