Review Exercise Set 19
Exercise 1:
Write the vector v, with its initial point at (-2, 3) and its terminal point at (6, 0), in terms
of i and j. Also, find the magnitude of the vector.
Exercise 2:
Find v + w, if v = 2i + 10j and w = -4i + 6j.
Exercise 3:
Find 2v - 3w, if v = 7i - 2j and w = 3i + 7j.
Exercise 4:
Find the unit vector that has the same direction as v = 6i - 11j.
Exercise 5:
A ship is traveling at 18 knots on a bearing of N15E. There is a strong water current
flowing at 7 knots from the northwest on a bearing of S80E. What is the actual course
and speed of the ship?
Review Exercise Set 19 Answer Key
Exercise 1:
Write the vector v, with its initial point at (-2, 3) and its terminal point at (6, 0), in terms
of i and j. Also, find the magnitude of the vector.
Write the vector
(x1, y1) = (-2, 3) and (x2, y2) = (6, 0)
v = (x2 - x1)i + (y2 - y1)j
v = (6 - (-2))i + (0 - 3)j
v = (6 + 2)i + (0 - 3)j
v = 8i - 3j
Find the magnitude
v =
( x2 − x1 ) + ( y2 − y1 )
=
( 6 − ( −2 ) ) + ( 0 − 3 )
2
2
=
(8) + ( −3)
=
64 + 9
2
2
2
2
= 73
Exercise 2:
Find v + w, if v = 2i + 10j and w = -4i + 6j.
v + w = (2i + 10j) + (-4i + 6j)
v + w = (2 + (-4))i + (10 + 6)j
v + w = -2i + 16j
Exercise 3:
Find 2v - 3w, if v = 7i - 2j and w = 3i + 7j.
Find 2v
2v = 2(7i - 2j)
2v = 14i - 4j
Find 3w
3w = 3(3i + 7j)
3w = 9i + 21j
Exercise 3 (Continued):
Find 2v - 3w
2v - 3w = (14i - 4j) - (9i + 21j)
2v - 3w = (14i - 4j) + (-1)(9i + 21j)
2v - 3w = (14i - 4j) + (-9i - 21j)
2v - 3w = (14 + (-9))i + (-4 + (-21))j
2v - 3w = (14 - 9)i + (-4 - 21)j
2v - 3w = 5i - 25j
Exercise 4:
Find the unit vector that has the same direction as v = 6i - 11j.
Find the magnitude of v
=
v
a 2 + b2
=
( 6 ) + ( −11)
=
36 + 121
2
2
= 157
Divide the vector by its magnitude
v
=
v
=
=
a
b
i+
j
v
v
6
i+
157
6
i−
157
−11
j
157
11
j
157
Exercise 5:
A ship is traveling at 18 knots on a bearing of N15°E. There is a strong water current
flowing at 7 knots from the northwest on a bearing of S80°E. What is the actual course
and speed of the ship? Round values to the nearest tenth.
Draw diagram of the problem using vectors
vector s will represent the speed and direction of the ship
vector c will represent the speed and direction of the current
Find the angle between the vectors and the x-axis (due East)
θ s= 90° − 15°= 75°
θ c= 90° − 80°= 10°
Find the vector components of vector s
a = s cos θ s
b = s sin θ s
= 18 cos 75° = 18 sin 75°
a ≈ 4.7
b ≈ 17.4
s ≈ 4.7i + 17.4j
Find the vector components of vector c
a = c cos θ c
=
7 cos10°
a ≈ 6.9
c ≈ 6.9i - 1.2j
(since c is in quadrant IV b must be negative)
b = − c sin θ c
=
− 7 sin10°
b ≈ −1.2
Exercise 5 (Continued):
Find the resultant vector of s + c
s + c ≈ (4.7i + 17.4j) + (6.9i - 1.2j)
s + c ≈ (4.7 + 6.9)i + (17.4 - 1.2)j
s + c ≈ 11.6i + 16.2j
Find the magnitude of the resultant vector
s+c =
a 2 + b2
≈
(11.6 ) + (16.2 )
2
2
≈ 135.46 + 262.44
≈ 397
≈ 19.9
Find the angle between the resultant vector and the x-axis
16.2
11.6
≈ 54.4°
θ r = tan −1
bearing ≈ 90° - 54.4° ≈ 35.6°
The ship is actually traveling 19.9 knots on a bearing of N35.6°E.