0.0.
1
Textbook Assignment 2
Your Name: LAST NAME, FIRST NAME (YOUR STUDENT ID: XXXX)
Your Instructors Name: Prof. FIRST NAME LAST NAME
YOUR SECTION: MATH 20300 XX
Due Date: NAME OF DAY, MONTH DAY, YEAR.
SECTION 13.2: do problems 8, 26, 29, 30, and 31.
Look up the terms: VECTOR SPACE, BASIS VECTORS, LINEARLY INDEPENDENT SET OF VECTORS, LINEARLY DEPENDENT VECTORS, LINEAR COMBINATION OF VECTORS, EUCLIDEAN
INNER PRODUCT, EUCLIDEAN INNER PRODUCT SPACE, EUCLIDEAN NORM, DETERMINANT
OF A 3 × 3 MATRIX
Formulas: If ~a = ha1 , a2 i and ~b = hb1 , b2 i, then
i)
ii)
~a + ~b = ha1 + b1 , a2 + b2 i.
p
k ~a k = a21 + a22 .
iii)
If θ is the angle³measured
in the counter-clockwise direction from the positive x-axis to the position vector ~a,
´
a2
then θ = arctan a1 .
iv)
If A(x1 , y1 ) and B(x2 , y2 ) are points in the xy-plane, then the directed segment from A to B is denoted
−−→
AB = hx2 − x1 , y2 − y1 i.
Theorems and ideas:
—-START SOLVING PROBLEMS ON A NEW PAGE (WRITE OUT THE QUESTION !!!!)—-
2
0.
−→
→
Exercise 8: Find a vector −
a with representation given by the directed line segment AB with A(−2, 2) and B(5, 3).
−→
Draw AB and the equivalent representation starting at the origin.
−−→
→
Solution. THE PLAN: It’s simple, Draw the directed segment AB in R2 . Then draw its representation −
a as
−→
a position vector in the same xy-plane. The directed line segment from A(−2, 2) to B(5, 3) is denoted AB and can
−−→
→
be represented by the position vector −
a = h5 − (−2), 3 − 2i = h7, 1i (see the picture below the red arrow is AB
and the green arrow ~a is a position vector that represents it.)
4
2
0
0
y
x
2
4
6
-2
t
u
Exercise 26: Find a vector that has the same direction h−2, 4, 2i but has length 6.
Solution.
1)
2)
THE PLAN:
→
First find a unit vector −
u in the direction of h−2, 4, 2i.
→
We will then scale the unit vector −
u found in step 1 by a factor of 6.
OKay, let’s do it: A unit vector in the direction of h−2, 4, 2i is given by
−
→
u =
1
h−2, 4, 2i =
k h−2, 4, 2i k
¿
2
1
1
−√ , √ , √
6
6
6
À
.
D
E
√ √ √
→
→
Now define the required vector as −
v = 6−
u = 6 − √16 , √26 , √16 = h − 6, 2 6, 6 i
t
u
0.0.
3
−
→
−
→
Exercise 29: Two forces F1 and F2 with magnitudes 10-lb and 12-lb act on an object at a point P as shown in
−
→
the figure. Find the resultant force F acting at P as well as its magnitude and its direction. (Indicate the direction
by finding the angle θ shown in the figure. The figure is not drawn here!)
Solution. THE PLAN: Find the projections of each vector onto the coordinate axes, i.e., find the following
−
→
−
→
−
→
−
→
proji F1 , projj F1 , proji F2 , and projj F2
This amounts to resolving the given vectors into components along the x and y coordinate axes as you do in your
physics courses. Then we will form the vector sum of the corresponding projections (This will gives us what we want,
−
→
−
→
−
→
the vector sum F of the two vectors F1 and F2 . Finally, we will calculate the angle θ as the inverse tan of the ratio
of the y and x components of the resultant
LET’S DO IT:
√
−→
−
→
−
→
F1x = proj~i F1 = k F1 k cos(135◦ ) i = −5 2 i
√
−→
−
→
−
→
F2x = proj~i F2 = k F2 k cos(30◦ ) i = 6 3 i
√
−→
−
→
−
→
F1y = proj~j F1 = k F1 k sin(135◦ ) i = 5 2 j
−→
−
→
−
→
F2y = proj~j F2 = k F2 k sin(30◦ ) j = 6 j
−
→
−
→
Now the vector sum or resultant of F1 and F2 is given by
−
→
−→
−→
−→
−→
F = F1x + F2x + F1y + F2y
√
√
√
= −5 2 i + 6 3 i + 5 2 j + 6 j
√
√
√
= (6 3 − 5 2) i + (6 + 5 2) j
√
√
√
= h 6 3 − 5 2, 6 + 5 2 i
Now we can calculate θ as
Ã
θ = arctan
√ !
6+5 2
√
√
≈ 75.7◦
6 3−5 2
t
u
Exercise 30: Velocities have both direction and magnitude and thus are vectors. The magnitude of a velocity
vector is called speed. Suppose that a wind is blowing from the direction N45◦ W at a speed of 50 km/h. (This means
that the direction from which the wind blows is 45◦ west of the northerly direction.) A pilot is steering a plane in
the direction N60◦ E at an airspeed (speed in still air) of 250 km/h. The true course, or track, of the plane is the
direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the
magnitude of the resultant. Find the true course and the ground speed of the plane.
Solution. THE PLAN: Let us denote by ~r the resultant of the velocity vectors of the plane and the wind. Let
−
−
→
v→
w denote the velocity of the wind and vp denotes the velocity of the plane. Then we are saying that
−
→
−
→
r = −
v→
w + vp .
→
→
Then the ground speed of the plane is k −
r k and the track is the direction of −
r , let’s called it θ (relative to the
◦
positive x-axis). Since the wind is blowing from a north westerly direction at 45 the velocity of the wind −
v→
w points
◦
◦
in a south easterly direction at 45 (i.e., the wind is blowing in a direction S45 E). This means that we can express
this velocity as −
v→
w as
Ã
√ !
√
√
√
2
2
−
→
−
→
−
→
◦
◦
i + −50
j = h25 2, −25 2i
vw = k vw k cos(315 ) i + k vw k sin(315 ) j = 50
2
2
and
Ã
−
−
→
−
→
◦
◦
v→
p = k vp k cos(30 ) i + k vp k sin(30 ) j =
√ !
µ
¶
√
3
1
250
i + 250
j = h125 3, 125i.
2
2
Now we have
√
√
√
√
√
√
−
→
−
→
r = −
v→
w + vp = h25 2, −25 2i + h125 3, 125i = h25 2 + 125 3, 125 − 25 2i
q √
√
√
→
and so the ground speed of the plane is k −
r k = (25 2 + 125 3)2 + (125 − 25 2)2 ≈ 267.34 km/h and the
´
³
√
2
√ − 25 √
≈ 19.6◦ (i.e., N 70.4◦ E) t
u
track of the plane is θ = arctan 25125
2 + 125 3
4
0.
Exercise 31: A woman walks due west on the deck of a ship at 3 mi/h. The ship is moving north at a speed of
22 mi/h. Find the speed and direction of the woman relative to the surface of the water.
Solution. THE PLAN: Do vector addition in the usual xy-plane and then calculate magnitude and direction of
our result.
−−→
LET’S DO IT: Let w =surface of the water and s = ship and o=woman and V=velocity and Vxy is to be read as
the velocity of x relative to y. Furthermore, we let θ denote the angle measured in the counter-clockwise direction
−−→
from the positive x-axis to the position vector Vow . Then we denote the velocity of the ship relative to the surface
−−→
−−→
of the water by Vsw = h0, 22i and the velocity of the woman relative to the surface of the water by Vow = ? and
−−→
the velocity of the woman relative to the ship by Vos = h−3, 0i. Thus
p
√
−−→
−−→ −−→
−−→
Vow = Vos + Vsw = h−3, 0i + h0, 22i = h−3, 22i thus k Vow k = (−3)2 + 222 = 493 mi/h ≈ 22.2 mi/h
and so the direction
µ
θ = arctan
22
−3
¶
≈ −82.2◦ = 180◦ − 82.2◦ = 97.8◦ or nearly N 8◦ W.
20
10
y
0
-5
x
0
t
u
0.0.
5
Definitions of the terms
VECTOR SPACE: A vector space is a non-empty set V along with an addition (+) on V and a scalar multiplication
on V such that the following properties hold:
commutativity
u + v = v + u for all u, v ∈ V ;
associativity
(u + v) + w = u + (v + w) and (αβ)v = α(βv) for all u, v, w ∈ V and all α, β ∈ R;
additive identity
there exists an element 0 ∈ V such that v + 0 = v for all v ∈ V ;
additive inverse
for every v ∈ V , there exists w ∈ V such that v + w = 0;
multiplicative identity
1v = v
for all v ∈ V ;
distributive properties
α(u + v) = αu + αv and (α + β)u = αu + βu
for all α, β ∈ R
and all u, v ∈ V
LINEAR COMBINATION: A linear combination of a set S = {v1 , . . . , vm } of vectors in a vector space V is a
vector of the form
a1 v1 + a2 v2 + · · · + am vm , where the scalarsa1 , a2 , . . . , am ∈ R.
LINEARLY INDEPENDENT SET OF VECTORS: A set S = {v1 , v2 , . . . , vm } of vectors in a vector space
V is called linearly independent if the only choice of scalars a1 , a2 , . . . , am ∈ R that satisfies the equation
a1 v1 + a2 v2 + · · · + am vm = 0
is a1 = a2 = · · · = am = 0.
LINEARLY DEPENDENT SET OF VECTORS: A set S = {v1 , v2 , . . . , vm } of vectors in a vector space
V is called linearly dependent if it is not linearly independent. In other words, a set S = {v1 , v2 , . . . , vm } of
vectors in a vector space V is called linearly dependent if there exist scalars a1 , a2 , . . . am ∈ R, not all 0, such that
a1 v1 + a2 v2 + · · · + am vm = 0.
BASIS OF A VECTOR SPACE: A basis of a vector space V is a set B = {b1 , b2 , . . . , bm } of vectors in V
that is linearly independent and that every vector v ∈ V is a linear combination of the vectors in B. For example,
our vector space is R3 and our basis is B = {i, j, k}. That is, for any a ∈ R3 we always have the linear combination
a = a1 i + a2 j + a3 k where the scalars a1 , a2 , a3 ∈ R. Scalar products allows us to define geometric notions such as
the angle between vectors or other notions such as length.
INNER-PRODUCT: An inner-product also called (scalar product) is a function, h i : V × V → R which
assigns a scalar (a real number) to each pair of vectors u, v in a vector space V . hu, vi = u · v is a scalar product
called the dot product or the Euclidean inner product.
INNER-PRODUCT SPACE: An inner-product space is a vector space V with an inner product defined. A
vector space with a Euclidean inner product defined is called a Euclidean inner-product space.
NORM: A norm is a function which assigns a positive length or size to all vectors in a vector space, other than
the zero vector. If V is a vector space and a is a vector in V , then the norm of a√is denoted k a k. In aPEuclidean
n
inner product space the Euclidean norm of a vector x ∈ Rn is defined as k x k = x · x where x · x = j=1 x2j