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).

v = < x, y > 
(1)
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.

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).

| v |   ______
Öx2 + y2
(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).
 
 

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).

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.

v + w = < v1+ w1, v2+ w2
(3)

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).

k v
k < v1, v2
< k v1, k v2
(4)
In fact, this is the case for any real value of k. Consequently, vector subtraction can be thought of as the following.
v - w
v + (-1) w
< v1, v2 > + (-1) < w1, w2
< v1, v2 > + < -w1, -w2
< v1- w1, v2- w2
(5)

 
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.

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.

| v |2
| a w |2 + | a w - v |2
a2 w12 + a2 w22 + (a w1-v1)2 + (a w2-v2)2
and so we have
v12 + v22
a2 w12 + a2 w22 + a2 w12 -2 a v1 w1 + v12 + a2w22 - 2 a v2 w2 + v22
and so either a = 0 or a (w12 + w22)- v1 w1-v2 w2 = 0, the last of which gives equation (6).
a
v1 w1+v2 w2
w12 + w22
(6)
We'll define the dot product of two vectors a = < a1, a2 > and b = < b1, b2 > as the following scalar value.
a · b = a1b1 + a2 b2
(7)
Then, equation (6) becomes equation (8).
a
v · w
w
· w
(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.

u · (a v + b w
< u1, u2 > ·(a < v1, v2 > +b < w1, w2 > )
< u1, u2 > ·( < a v1, a v2 > + < b w1, b w2 > )
< u1, u2 > · < a v1 +b w1, a v2+b w2
< a u1 v1+b u1 w1, a u2 v2+b u2 w2
< a u1 v1, a u2 v2 > + < b u1 w1, b u2 w2
a < u1 v1, u2 v2 > + b < u1 w1, u2 w2
a( < u1, u2 > · < v1, v2 > ) + b( < u1, u2 > · < w1, w2 > )
a (u · v) + b (u · w)
(9)

Last of all, we note the relationship between the dot product and the magnitude of a vector.

v · v
< v1, v2 > · < v1, v2
v12 + v22
æ
è
  _______
Öv12 +v22
ö
ø
2
| v |2
(10)

 
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.

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

| v - w |2 = | v |2 + | w |2 - 2 | v || w | cos q
(11)
and by equations (9) and (10), we have the following.
| v - w |2
(v - w) · (v - w)
v · (v - w) - w · (v - w)
v · v - v · w  - w · v + w · w
| v |2 -2 v · w + | w |2
(12)
Equating equations (11) and (12) gives
| v |2 - 2 v · w + | w |2
| v |2 + | w |2 - 2 | v || w | cos q
which then leads to
- 2 v · w
- 2 | v || w | cos q
and solving for cos q gives equation (13).
cos q
v · w
| v || w |
(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.

v
< v1, v2, ¼, vn
w
< w1, w2, ¼, wn
Then, equations (2)-(5) and (7) are extended to the the following equivalent formulas..
| v |

Ö

v12 + v22 + ¼+ vn2
v + w
< v1+ w1, v2+ w2, ¼, vn+ wn
k v
< k v1, k v2, ¼, k vn
v - w
< v1 - w1, v2 - w2, ¼, vn- wn
v · w
v1 w1 + v2 w2 + ¼+ vn wn

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.
 
 

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.

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.

ê
ê
ê
ê
ê
i
j
k
v1
v2
v3
w1
w2
w3
ê
ê
ê
ê
ê
      (14)


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On 7 Aug 1999, 11:03.