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Calculus III - Test # 1- Solutions
Professor Broughton December, 2003
Name: Box #
1. Vector operations
- Let u =5i + 6j− 4 k and v =2i − j + 3k.
Compute the following:
1.a u • v =5 • 2 − 6 • 1 − 4 • 3 = − 8
1.b ku − vk = kh 5 − 2 , 6 + 1, − 4 − 3 ik =
1.c u × v =
i j k
i−
j+
k =14i− 23 j−
17 k.
1.d Fill in the table
True False True Equation
u × v = v × u X u × v = −v × u
u • v = v • u X
u• (u × v) =0 X
u • v = kuk kvk sin θ X u • v = kuk kvk cos θ
u × v = kuk kvk sin θ X ku × vk = kuk kvk sin θ
2. Angles and projections
- Let u =4i + 4k and v =2i − j, θ is the angle between the two vectors.
2.a What is the angle θ between u and v?
cos θ =
u • v
kuk kvk
θ = arccos
μ 2 √ 10
2.b Compute the projection of v on u.
v 1 = projuv =
u • v
u • u
u =
(4i + 4k) = i + k
2.c Using part b., write v = v 1 + v 2 where v 1 is parallel to u and v 2 is
perpendicular to u.
v 2 = v − projuv =2i − j − (i + k) = i − j − k
v = 2i − j = (i + k) + (i − j − k) = v 1 + v 2
3. Lines and planes
- A tetrahedron has the following four vertices, P = (0, 0 , 0), Q = (0, 1 , 1), R =
(1, 0 , 1), S = (1, 1 , 0).
3.a Fill in the missing data on the faces in the table below.
Face Normal Equation of Plane
∆P QR h 1 , 1 , − 1 i x + y − z = 0
∆P RS h 1 , − 1 , − 1 i x − y − z = 0
∆P QS h 1 , − 1 , 1 i x − y + z = 0
∆QRS h 1 , − 1 , 1 i x + y + z = 2
Figure 4.1:
4. Torque
- A 0.5 foot tricycle crank is centered at the origin. Let r(θ) represent the end
of the pedal, where θ is the angle between crank arm and the horizontal. A
child stands on the pedal and exerting a force F of 60 lbs.
4.a Make a sketch showing r(θ) and F.
4.b What angles give the maximum forward and backward propulsion.
Your answer, written out in a sentence, should be justified by a torque
calculation, in addition to common sense.
r(θ) = 0 .5 (cos(θ)i + sin(θ)j)
F = − 60 j
T = 0 .5 (cos(θ)i + sin(θ)j) × (− 60 j)
= −30 cos(θ)i × j − 60 sin(θ)j × j
= −30 cos(θ)k
Thus kTk = 60 |cos θ.| The maximum torque value occurs when θ = 0
or π which is when the force will be perpendicular to the tricycle crank.
When θ = 0 we get propulsion assuming the tricyle is moving left to
right. θ = π gives maximum braking.
