Download MATH 203 Lab 1 solutions Spring 2005

MATH 203
Lab 1 solutions
Spring 2005
(1) Use vectors to prove that the line joining the midpoints of 2 sides of a triangle is parallel to the third
side and measures half its length.
A
F
E
C
B
~ + 1 AC = 1 (BA
~ + AC)
~ = 1 BC
~
~ = EA
~ + AF
~ = 1 BA
EF
2
2
2
2
~ = 1 BC,
~ therefore, EF is parallel to BC and measures half its length.
Since EF
2
~ + OD
~ + OF
~ is
(2) Let OABCDEFG be a cube, labelled as shown in the diagram below. Show that OB
~
parallel to OG.
F
G
E
D
A
O
~ + OD
~ + OF
~
OB
B
C
~ + (OC
~ + CD)
~ + OF
~
= OB
~ + CD)
~ + (OC
~ + OF
~ )
= (OB
~ + BG)
~ + (OC
~ + F~G)
= (OB
~ + OG
~ = 2OG.
~
= OG
~ + OD
~ + OF
~ is parallel to OG.
~
Hence OB
(3) A constant force with vector representation F =< 5, 4, −3 > moves an object along a straight line from
the point (2, 3, 0) to the point (4, 6, 8). Find the work done, if the distance is measured in meters and
the magnitude of the force is measured in newtons.
Distance vector D =< 4 − 2, 6 − 3, 8 − 0 >=< 2, 3, 8 >. Hence
W D = F · D =< 5, 4, −3 > · < 2, 3, 8 >= 5(2) + 4(3) − 3(8) = −2
(4) Velocities have both magnitude and direction and thus are vectors. The magnitude of a velocity vector
is called its speed. Suppose that a wind is blowing from the direction N45◦ W at a speed of 40 km/h
(this means the direction from which the wind blows is 45◦ west of the northerly direction). A pilot
is steering a plane in the direction of S60◦ E at an airspeed (speed in the air) of 350 km/h. The true
course, or track of the plane is the direction of the resultant velocity vectors of the plane and wind.
MATH 203
Lab 1 solutions
Spring 2005
The ground speed of the plane is the magnitude of the resultant. Find the true course and ground
speed of the plane.
Let p be the velocity vector of the plane and w be the velocity vector of the plane. Then
√
p = 175 3i − 175j
√
√
w = 20 2i − 20 2j
√
√
√
The resultant is the vector v = p + w = (175 3 + 20 2)i − (175 + 20 2)j.
The ground speed of the plane is |v| = |p + w| = 388.76km/h. The direction is SθE where θ ≈ 1.021 ≈
58.47◦ .
(5) Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square.
The decoration has a mass of 5kg. The ropes, fastened at different heights, make angles of 52◦ and 40◦
with the horizontal. Find the tension in each wire and the magniture of each tension.
40
52
5m
3m
5kg
Let T3 and T5 be the tensions on the 3m and 5m string. Then,
T3
T5
=
=
−|T3 | cos 52◦ i + |T3 | sin 52◦ j
−|T5 | cos 40◦ i + |T5 | sin 40◦ j
The total tension is equal to the force exerted by the decoration, which has a force 5gj = 5(9.8)j = 49j
(in Newtons).
Therefore, T3 + T5 = 49j. Resolving in the i and j direction, we have
−|T3 | cos 52◦ − |T5 | cos 40◦
|T3 | sin 52◦ + +|T5 | sin 40◦
= 0
= 49
Solving for |T3 | and |T5 |, we have |T3 | ≈ 38N and |T5 | ≈ 30N , and
T3
T5
≈
≈
−23i + 30j
23i + 19j.
(6) Show that the vector
v = b−
a·b
a
|a|2
is orthogonal to a. Given vectors a, b, can you draw on a diagram what the vector v is?
Note that proja a =
to a.
a·b
|a|2 a,
hence v is the vector as shown in the diagram (hence it should be orthogonal
MATH 203
Lab 1 solutions
b
Spring 2005
v
a
To show the result algebraically, we show that v · a = 0 as follows
a·b
a
·a
v·a =
b−
|a|2
a·b
= b·a−
aa
|a|2
a·b
|a|2
= a·b−
|a|2
= a · b − a · b = 0.
(7) The Cauchy-Bunyakovskii-Schwarz (CBS) inequality states that
|a · b| ≤ |a| |b|.
(a) Prove the CBS inequality from the definition of the dot product.
Since a · b = |a| |b| cos θ. Hence
|a · b| = |a| |b| | cos θ| ≤ |a| |b|
(b) What conditions must satisfy if the CBS inequality becomes a equality?
The inequality becomes a equality when | cos θ| = 1, that is, θ = 0 or π. That is, when a and b
are parallel.
(8) The Triangle Inequality states that
|a + b| ≤ |a| + |b|.
(a) Give a geometric interpretation why the above inequality is called the triangle inequality.
b
a
a+b
(b) Prove |a + b|2 = (a + b) · (a + b)
Since |a|2 = a · a, we must have
|a + b|2 = (a + b) · (a + b)
(c) Use the dot product, the CBS inequality, and the above to show the triangle inequality.
MATH 203
Lab 1 solutions
|a + b|2
= (a + b) · (a + b)
= a·a+b·a+a·b+b·b
= |a|2 + 2a · b + |b|2
≤ |a|2 + 2|a · b| + |b|2
≤ |a|2 + 2|a| |b| + |b|2
≤ (|a| + |b|)2
The result follows.
Spring 2005