MCV 4U1 Grade 12 Calculus and Vectors
INTRODUCTION TO VECTORS
Review 2
1. If a𝑢
⃗ + b𝑣 = ⃗0 and 𝑢
⃗ and 𝑣 have different directions, what must a and b
be equal?
Answer
𝑢
⃗ and 𝑣 have different directions ⇒ 𝑢
⃗ and 𝑣 are linearly independent
∴ a linear combination a𝑢
⃗ + b𝑣 equals zero vector only if a = b = 0.
2. Show geometrically that ||𝑢
⃗ | – |𝑣|| ≤ |𝑢
⃗ + 𝑣|. Under what conditions
does equality hold?
Solution
Triangle inequalities:
|𝑣| ≤ |𝑢
⃗ + 𝑣| + |𝑢
⃗ | and |𝑢
⃗ | ≤ |𝑢
⃗ + 𝑣| + |𝑣| ⇒
|𝑣| – |𝑢
⃗ | ≤ |𝑢
⃗ + 𝑣| and |𝑢
⃗ | – |𝑣| ≤ |𝑢
⃗ + 𝑣| ⇒
||𝑢
⃗ |– |𝑣|| ≤ |𝑢
⃗ + 𝑣|
Equality holds if 𝑢
⃗ and 𝑣 are collinear vectors with the same direction.
3. A 3 kg mass is hanging from the end of a string. If a horizontal force of
12 N pulls the mass to the side, find the tension in the string and the angle
the string makes with the vertical.
Solution
Since the mass is in equilibrium, we get:
|Tx| = |𝐹 | = 12 N,
|Ty| = |𝐹 g| = (3 kg)(9.8 m/s2) = 29.4 N,
2
⃗ | = √|𝑇𝑥 |2 + |𝑇𝑦 |
T = |𝑇
= √122 + 29.42
= √1008.36 ≈ 31.75 N
|𝑇 |
12
tan θ = 𝑥 ≈
≈ 0.378
|𝑇𝑦 |
–1
31.75
θ ≈ tan (0.378) ≈ 20.7°
⃗ ≈ 31.75 N [20.7° with the vertical]
∴𝑇
4. Two forces 𝐹 1 and 𝐹 2 act on an object. Determine the magnitude of the
resultant if
a) |𝐹 1|= 54 N, |𝐹 2|= 34 N, and the angle between them is 55o.
Solution
|𝐹 1 + 𝐹 2|2 = |𝐹 1|2 +|𝐹 2|2 + 2|𝐹 1||𝐹 2|cos θ
= 542 + 342 + 2(54)(34) cos 55o
≈ 6178.1727
|𝐹 1 + 𝐹 2| ≈ 78.6 N
b) |𝐹 1|= 21 N, |𝐹 2|= 45 N, and the angle between them is 140o.
Solution
|𝐹 1 + 𝐹 2|2 = |𝐹 1|2 +|𝐹 2|2 + 2|𝐹 1||𝐹 2|cos θ
= 212 + 452 + 2(21)(45) cos 140o
≈ 1018.176
|𝐹 1 + 𝐹 2| ≈ 31.91 N
5. Two forces at an angle of 130o to each other act on an object. Determine
their magnitudes if the resultant has a magnitude of 480 N and makes an
angle 55o with one of the forces.
Solution
By the Sine Law, we get:
|𝐹1 |
sin 75°
|𝐹1 |=
|𝐹2 |
sin 55°
|𝐹2 |=
=
|𝐹1 + 𝐹2 |
sin 50°
⇒ |𝐹1 |=
|𝐹1 + 𝐹2 | sin 75°
sin 50°
=
|𝐹1 + 𝐹2 |
sin 50°
=
sin 50°
=
sin 50°
480 ∙ sin 75°
sin 50°
⇒ |𝐹2 |=
|𝐹1 + 𝐹2 | sin 55°
|𝐹1 + 𝐹2 | sin 75°
≈ 605.24 N
|𝐹1 + 𝐹2 | sin 55°
sin 50°
480 ∙ sin 55°
sin 50°
≈ 513.28 N
Answer: 605.24 N and 513.28 N
6. Forces of 5N, 2N, and 12N, all lying in the same plane, act on an object.
The 5N and 2N foeces lie on opposite sides of the 12N force at angles of
40o and 20o, respectively. Find the magnitude and direction of the
resultant.
Solution
𝑅⃗ = 𝐹 1 + 𝐹 2 + 𝐹 3
Rx = F1x + F2x + F3x
= 12 + 5 cos 40° + 2 cos 20° ≈ 17.71 N
𝑅⃗ = 𝐹 1 + 𝐹 2 + 𝐹 3
Ry = F1y + F2y + F3y = 0 + 5 sin 40° – 2 sin 20°
≈ 2.53 N
|𝑅⃗ | ≈ √17.712 + 2.532 ≈ 17.89 N
𝑅𝑦
2.53
𝑅𝑥
17.71
tan θ = | | ≈
≈ 0.1429 ⇒ θ = tan –1 (0.1429) ≈ 8.1°
∴ 𝑅⃗ ≈ 17.89 N (𝐹 1 8.1° 𝐹 2)
7. An airplane heads due south with an air speed of 480 km/h.
Measurements made from the ground indicate that the plane’s ground
speed is 528 km/h at 15o east of south. Calculate the wind speed.
Solution
p𝑣 w – plane velocity relative to the wind,
p𝑣 w = 480 km/h [S]
p𝑣 g – plane velocity relative to the ground,
o
p𝑣 g = 528 km/h [S15 E]
w𝑣 g – wind velocity relative to the ground
By the Cosine Law, determine |w𝑣g|:
|w𝑣g|2 = |p𝑣w|2 + |p𝑣g |2 – 2|p𝑣w |∙|p𝑣g | cos 15°
= 4802 + 5282 – 2 ∙ 480 ∙ 528 cos 15°
= 19575.5172 ⇒ |w𝑣g| ≈ 139.9 km/h
By the Sine Law, we get:
sin(180°− 𝜃)
𝑜𝑣
⃗ 𝑔|
|𝑝
=
sin 15°
𝑜𝑣
⃗ 𝑔|
|𝑤
⇒
sin 𝜃
528
=
sin 15°
139.9
⇒ sin θ =
528 sin 15°
139.9
≈ 0.9768
∴ θ ≈ sin–1 (0.9768)≈ 77.6° ∴ w𝑣g = 139.9 km/h [N77.6°E]
8. A coast guard cutter is steering west at 12 knots, when its radar detects a
tanker ahead at a distance of 9 nautical miles travelling with a relative
velocity of 19 knods, on heading of E 15o N. What is the actual velocity of
the tanker?
Solution
t𝑣 – velocity of the tanker
gc𝑣 = 12 knods [W] – velocity of the guard cutter
o
t𝑣 gc = 19 knods [E 15 N] – velocity of the tanker relative to the guard
cutter
t𝑣 = t𝑣 gc + gc𝑣
By the Cosine Law, determine |t𝑣|:
|t𝑣|2 = |gc𝑣|2 + |t𝑣gc|2 – 2|gc𝑣|∙|t𝑣gc| cos 15°
= 122 + 192 – 2(12) ∙ (19) cos 15°
≈ 64.54
∴ |t𝑣|≈ 8.03 knods
By the Sine Law, determine θ:
sin(180 − 𝜃) sin 15°
sin 𝜃
sin 15°
19 sin 15°
= 0⃗ ⇒
=
⇒ sin θ =
≈ 0.6124
𝑜⃗
| 𝑡𝑣𝑔𝑐 |
| 𝑡𝑣|
19
θ ≈ sin –1 (0.6124)≈ 37.8°
∴ t𝑣 ≈ 8.03 knods [E37.8°N]
8.03
8.03
9. A camp counsellor leaves a dock padding a canoe at 3 m/s. She heads
downstream at 30o to the current, which is flowing at 4 m/s.
a) How far downstream does she travel in 10 s?
b) What is the length of time required to cross the river if its width is 150
m?
Solution
𝑣 – canoe velocity relative to the ground
𝑢
⃗ – canoe velocity relative to the current,
𝑤
⃗⃗ – current velocity relative to the ground
𝑣=𝑢
⃗ +𝑤
⃗⃗
a) vx = ux + wx = |𝑢
⃗ |cos 30° + |𝑤
⃗⃗ |
= 3 cos 30° + 4 ≈ 6.6 m/s
∆dx = vx ∙ ∆t ≈ (6.6 m/s)(10 s) = 66 m
The canoe travels downstream about 66 m.
b) vy = uy + wy = |𝑢
⃗ |sin 30° + 0 = 3 sin 30° = 1.5 m/s
∆t =
∆𝑑𝑦
𝑣𝑦
=
150 𝑚
1.5 𝑚/𝑠
= 100 s = 1 min 40 s
The length of time required to cross the river is 1 min 40 s
10. The pilot of an airplane that flies at 800 km/h wishes to travel to a city
800 km due east. There is a 80 km/h wind from the northeast.
a) What should the plane’s heading be?
b) How long will the trip take?
Solution
p𝑣 w – plane velocity relative to the wind, |p𝑣 w| = 800 km/h
p𝑣 g [W] – plane velocity relative to the ground,
w𝑣 g = 80 km/h – wind velocity relative to the ground
a) By the Sine Law, we get:
sin 𝜃
𝑜𝑣
⃗ 𝑔|
|𝑤
=
sin 135°
⇒
𝑜𝑣
⃗ 𝑤|
|𝑝
sin θ =
𝑜𝑣
⃗ 𝑔 | sin 135°
|𝑤
𝑜𝑣
⃗ 𝑤|
|𝑝
=
(80) sin 135°
800
≈ 0.0707
∴ θ ≈ sin –1 (0.0707) ≈ 4°
The plane’s heading should be 4° north of east.
b) 180° – 135° – 4° = 41°
By the Sine Law, we get:
sin 41° sin 135°
= 𝑜⃗
⇒ |p𝑣g|
𝑜𝑣
⃗ 𝑔|
|𝑝
| 𝑝𝑣𝑤 |
800 𝑚
t=
742.25 m/s
=
𝑜𝑣
⃗ 𝑤 | sin 41°
|𝑝
sin 135°
≈ 1.078 h ≈ 1 h 5 min
The trip will take about 1 h 5 min
=
(800) sin 41°
sin 135°
≈ 742.25 km/h
11. Twice a week, a cruise ship carries vacationers from Miami, Florida to
Freeport in the Bahamas, and then on to Nassau before returning to Miami.
The distance from Miami to Freeport is 173 km on a heading of E 20o N.
The distance from Freeport to Nassau is 217 km on a heading of E 50o S.
Once a week the ship travels directly from Miami to Nassau. Determine
the displacement vector from Miami to Nassau.
Solution
By the cosine Law, determine MN:
MN2 = MB2 + BN2 – 2 ∙ MB ∙ BN cos ∠MBN
= 1732 + 2172 – 2 ∙ 173 ∙ 217 cos ∠110°
≈ 102697.5564
∴ MN ≈ 320.46 km
By the Sine Law, determine ∠BMN:
𝑀𝑁
sin 110°
=
𝐵𝑁
sin ∠𝐵𝑀𝑁
⇒
320.46
sin 110°
=
217
sin ∠𝐵𝑀𝑁
sin ∠BMN ≈ 0.6363 ⇒ ∠BMN ≈ sin
θ = ∠BMN – 20° ≈ 19.5°
⃗⃗⃗⃗⃗⃗⃗ = 320.46 km [E19.5°S]
∴ 𝑀𝑁
⇒ sin ∠BMN =
–1
217 sin 110°
(0.6363) ≈ 39.5°
320.46
12. A pilot whishes to reach an airport 350 km from his present position at
a heading of N 60o E. If the wind is from S 25o E with a speed of 73 km/h
and the plane has an airspeed of 450 km/h, find
a) what heading the pilot should steer
b) what the ground speed of the plane will be
c) how many minutes it will take for the plane to reach its destination
Solution
p𝑣 w – plane velocity relative to the wind, |p𝑣 w| = 450 km/h
o
p𝑣 g [N 60 E] – plane velocity relative to the ground,
w𝑣 g = 73 km/h [W 65° N] – wind velocity relative to the ground
a) By the Sine Law, determine α:
sin 𝛼
sin(180°− 65°− 60°)
=
𝑜⃗
𝑜⃗
| 𝑤𝑣𝑔 |
sin 𝛼
73
| 𝑝𝑣𝑤 |
=
sin 55°
450
73 sin 55°
sin α =
≈ 0.1329
450
α = sin –1 (0.1329) ≈ 7.6°
θ = 60° – α ≈ 52.4°
The pilot should steer [E52.4°N]
b) By the Sine Law, determine \ p𝑣g\:
sin(180°− 55° − 7.6°) sin 55°
sin 117.4° sin 55°
=
⇒
=
𝑜⃗
𝑜⃗
𝑜⃗
| 𝑝𝑣𝑔 |
| 𝑝𝑣𝑤 |
450 sin 117.4°
| 𝑝𝑣𝑔 |
450
|p𝑣g| =
≈ 487.7 km/h
sin 55°
The ground speed of the plane will be about 487.7 km/h
c) t =
∆𝑑
𝑜𝑣
⃗ 𝑔|
|𝑝
=
350 km
487.7 km/h
≈ 43 min
It will take for the plane about 43 min to reach its destination
13. A 10-kg mass is supported by two strings of length 5 m and 7 m
attached to two points in the ceiling 10 m apart. Find the tension in each
string.
Solution
1) By the Cosine Law, we get:
72 = 102 + 52 – 2(10)(5) cos α
49 = 100 + 25 – 100 cos α
100 cos α = 76
cos α = 0.76 ⇒
α = cos –1 (0.76) ≈ 40.54°
2) By the Cosine Law, we get:
52 = 102 + 72 – 2(10)(7) cos β ⇒ 25 = 100 + 49 – 140 cos β
31
31
140 cos β = 124 ⇒ cos β = ⇒ β = cos –1( )≈ 27.66°
35
35
⃗1+𝑇
⃗ 2. Since the mass is in equilibrium, we get:
Let 𝑅⃗ = 𝑇
|𝑅⃗ | = |𝐹 g| = mg = (10 kg)(9.8 m/s2) = 98 N.
3) By the Sine Law, we get:
⃗ 1|
|𝑇
=
⃗|
|𝑅
, 90° – β = 62.34°, α + β ≈ 68.20°,
sin(90°− 𝛽) sin(𝛼+ 𝛽)
(98 N) sin 62.34°
⃗ 1| =
∴ |𝑇
sin 68.20°
≈ 93.5 N
4) By the Sine Law, we get:
⃗ 2|
|𝑇
=
⃗|
|𝑅
, 90° – α = 49.46°, α + β ≈ 68.20°,
sin(90°− 𝛼) sin(𝛼+ 𝛽)
(98 N) sin 49.46°
⃗ 1| =
∴ |𝑇
≈ 80.2 N
sin 68.20°
Answer: 93.5 N and 80.2 N