CSC 171/172/173 Laboratory Notes
Â3
Vector Operations
The types of numbers that we've traditionally used are often referred
to as scalars. Scalars are quite good at providing us with a sense
of magnitude, but give little indication as to direction. When studying
physical phenomena such as force, we need an entity that provides us with
both magnitude and direction. These will be referred to as vectors.
Our derivations will exploit the fact that vectors are readily understood
in the two dimensional plane (Â2
space) and their formulas are easily extended to higher dimensions (Ân
space). In all that follows, we will designate vectors using boldfaced
variables (such as x or y) and scalars using regular variables
(such as x or y).
Vector equivalence classes
Any vector of a given magnitude and direction is equal to any other
vector of the same magnitude and the same direction. For all of these vectors,
we will place their tail (the end without the arrow) at the origin of the
Euclidean plane and identify the vector with the point (x,y) that is touched
by their head (the end with the arrow). To avoid confusion, when we discuss
a point p in the Euclidean plane, we will use parentheses as in p = (x,y).
On the other hand, when we are discussing a vector that has its head at
the point (x,y) when its tail is at the origin, we will use the symbols
< x,y > as in equation (1).
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Figure 6.1 shows vectors u in black, v in orange
and w in purple. Vectors u and w have equal
directions but different magnitudes. Vectors u and v
have equal magnitudes but different directions. |
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Magnitude and direction
By creating a right reference triangle with its right angle at the
origin, the magnitude of the vector v is nothing more than the hypotenuse
of the right triangle and is denoted by equation (2).
The direction of the vector v can be determined by examining
the same right reference triangle and noticing that for the reference angle
q,
we have tan q = y/x and so, up to quadrant,
q
= Tan-1(y/x).
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Figure 6.2 shows vector v = <x, y> in purple and the
Cartesian coordinate axes in orange. The right reference triangle
for vector v is determined by the ordered pairs (0, 0), (x, 0) and
(x, y). |
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Zero and unit vectors
The zero vector 0 = < 0, 0 > clearly has magnitude 0 and we
say that it has direction q = 0 as well. A unit
vector u is any vector that has |u|
= 1.
Vector arithmetic
The sum of two vectors v and w is found by placing one
vector's tail at the other vector's head and forming a parallelogram with
two copies of each vector. The diagonal that begins at the corner with
two tails of the vectors and ends at the corner with two heads of the vectors
is the vector sum. It can be shown that for vectors v = < v1,
v2 > and w = < w1, w2 > , the
vector sum satisfies the following formula.
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| v + w = < v1+
w1, v2+ w2 > |
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(3) |
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Scalar multiples of vectors, kv can be thought of as stretching
(for |k| > 1) or
shrinking the vector (for |k|
< 1) and possibly changing it's direction (for k < 0). For instance,
when k is an integer, it is easy to see that kv creates k multiples
of the right reference triangle. In this case, it is easy to see by equation
(4).
In fact, this is the case for any real value of k. Consequently, vector
subtraction can be thought of as the following.
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| < v1, v2 > + (-1) < w1,
w2 > |
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| < v1, v2 > + < -w1,
-w2 > |
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(5) |
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Figure 6.3 shows the parallelogram method of determining the
sum of vector v in black and vector w in orange. The
vector sum v + w is the diagonal that travels from the corner with
the tails of v and w to the corner with the heads and is
shown in purple. |
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Vector Dot Product
One can motivate the creation of the vector dot product by asking the
following question. ``When a force that is not horizontal is exerted on
a horizontal plane, what is the resultant horizontal force?". For instance,
when we pull an object on a horizontal table with a force that is not horizontal,
how much of the force is parallel to the table? If the direction of the
table surface is given by the vector w and the direction of the
force is given by the vector v, then we are asking for the vector
that is a multiple of w that is generated by a perpendicular that
is "dropped" from the head of v to w. This is often referred
to as the vector projection of v onto
w which we will denote
as Projwv. Since this vector is in the same direction
as vector w, it must be equal to a w
for some real value of a. Then, assuming that
v
= < v1, v2 > and w = < w1,
w2 > , we have a right triangle that is bounded by vectors v,
a
w
and a w - v that due to Pythagoras,
satisfies the following relationship.
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| a2
w12 + a2 w22
+ (a w1-v1)2
+ (a w2-v2)2 |
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and so we have
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w12 + a2 w22
+ a2 w12 -2
a
v1
w1 + v12 + a2w22
- 2
a v2 w2 + v22 |
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and so either a = 0 or a
(w12
+ w22)- v1 w1-v2
w2 = 0, the last of which gives equation (6).
We'll define the dot product of two vectors a = < a1,
a2 > and b = < b1, b2 > as the
following scalar value.
Then, equation (6) becomes equation (8).
The dot product inherits commutativity and associativity from the multiplication
operation on the reals. The dot product satisfies a distributive property
such that, if u = < u1, u2 > ,
v
= < v1, v2 > and w = < w1,
w2 > with a, bÎÂ
we have the following.
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| < u1, u2 > ·(a
< v1, v2 > +b <
w1, w2 > ) |
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| < u1, u2 > ·( <
a
v1,
a
v2
> + <
b w1,
b
w2
> ) |
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| < u1, u2 > · <
a
v1
+b w1,
a
v2+b
w2
> |
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| < a u1 v1+b
u1
w1, a u2 v2+b
u2
w2 > |
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| < a u1 v1,
a
u2
v2 > + < b u1 w1,
b
u2
w2 > |
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| a < u1 v1,
u2 v2 > + b < u1
w1, u2 w2 > |
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| a( < u1,
u2 > · < v1, v2 > ) + b(
< u1, u2 > · < w1, w2
> ) |
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(9) |
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Last of all, we note the relationship between the dot product and the
magnitude of a vector.
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æ
è |
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_______
Öv12
+v22 |
ö
ø |
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(10) |
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Figure 6.4 shows the vector projection of vector v in
black onto vector w in orange. The vector projection Projwv
is shown in purple. |
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Angle q between two vectors
Let q be the counter-clockwise angle measured
from vector w to vector v. Then we have a triangle with sides
|v|,
|w|
and |v-w|.
The law of cosines states
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w
|2
= |
v
|2
+ |
w
|2
- 2 |
v
|| w
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cos
q |
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(11) |
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and by equations (9) and (10), we have the following.
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| v · (v - w) - w ·
(v - w) |
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| v · v - v · w
- w · v + w · w |
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(12) |
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Equating equations (11) and (12) gives
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| | v |2
- 2 v · w + | w |2 |
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+ | w |2
- 2 | v || w
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cos
q |
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which then leads to
and solving for cos q gives equation (13).
Equation (13) indicates that vectors v and w are perpendicular
(q = p/2), if and
only if
v · w = 0
Vectors in Ân
Although the geometry is difficult to visualize, most of the operations
that we've discussed extend quite easily to vectors in an n-space. So,
starting with equation (1), we'll let our vectors v and w
be defined as follows.
Then, equations (2)-(5) and (7) are extended to the the following equivalent
formulas..
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| < v1+ w1, v2+
w2,
¼, vn+ wn
> |
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| < v1 - w1, v2 -
w2,
¼, vn- wn
> |
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Equations (9), (10) and (13) extend to higher dimensional spaces with
no changes.
Vectors in Â3
Vectors in 3 space are worth making special mention of since they play
an important role in most of our physical examples. Since 3 distinct points
determine a unique plane, then any two non-parallel vectors will determine
a unique plane in 3 space. This can be visualized by choosing the 3 points
to be the two heads of the vectors and the point of intersection of their
two tails. Consequently, the parallelogram that we used to determine the
vector arithmetic is still available to us, however it is in the plane
determined by the vectors. The same holds for the angle between the two
vectors and the vector projection that led to the dot product.
For example, any two of the 3 canonical unit vectors, i = <
1,0,0 > , j = < 0,1,0 > or k = < 0,0,1 > determine
one of the coordinate planes in Â3.
i
and j are the familiar xy plane, i and k are the xz
plane and j and k are the yz plane.
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Figure 6.5 shows a vector v = <a, b, c> in purple.
The coordinate axes in 3 space are shown in orange. A surrounding
frame helps project the image of a 3 dimensional vector on the 2
dimensional screen. |
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Vector cross product
The cross product of any two Â3
vectors v = < v1, v2, v3 > and
w
= < w1, w2, w2 > is a vector that is
perpendicular to both v and w. We'll denote this as v
×
w.
By definition, v × w = (| v | | w | sin q)
n
where q is the angle between the vectors
v
and w and n is the vector that is normal (according to the
right-hand rule) to the plane that is determined by v and
w.
It can be shown to be equal to (v2 w3-v3
w2)i - (v1 w3 - v3 w1)
j
+ (v1 w2 - v2 w1)k which
may be easiest to remember as the value of the following determinant.
File translated from TEX by TTH,
version 2.32.
On 7 Aug 1999, 11:03.