If there was one entity that has proved itself enormously useful in nearly ever area of mathematics and physics, it would be the vector. The term "vector" was coined by an Irish mathematician named William Rowan Hamilton.
In math and science we come across quantities that only have a magnitude, meaning they can be expressed with just a number. For example, the mass of an object is expressed with merely a number. Such quantities are called scalars. They are numbers and nothing more.
However, some quantities need something more than just a magnitude. Take for example the velocity of a car. Not only can a car drive at 30 miles per an hour, but it also moves in a direction. The car could travel north, south, east, west, or some combination of the four. Quantities that can only be described with a magnitude and a direction are called vectors. The manipulation of vectors is what most of this article will be about.
Notation of a Mathematical Quantities
The notation of the quantities introduced so far is quite important so that you can differentiate between scalars and vectors. A scalar will be written as a lower-case italic letter. A vector will be written as a lower-case bold letter. The case of the letter is important because later on when matrices and bases are introduced they will be written as upper-case bold letters.
Graphical Representation of a Vector
A vector is best draw as a directed line segment. In geometry you have line segments, which are just two points connected by a line. A directed line segment means that the regular line segment simply points in a direction.
The Anatomy of a Vector
In the picture of the vector that are a few terms to learn that relate to its anatomy.
Vectors are Independent from Position
It is important to note that the "position" of a vector has no meaning or relevance. Does the "position of a car's velocity" really mean anything to you? When talking about forces, which happens to be a vector, it is sometimes important to know the point on a body that the force is being applied, but that merely introduces another vector rather than the position of a vector. All of the following vectors are equivalent:
Most of the times it is tedious and an overall pain to deal with vectors in terms of their magnitude and direction. For this reason we deal with vectors as a coordinate where the tail of the vector is placed at the origin (0, 0) and the head at some coordinate (x, y). When we think of vectors like this we can write them down in the following fashion:
v = <x, y>
The x and y scalars measure how much along the x- and y-axis you have to travel down before you reach the head of the vector.
_________ ||v|| = \/ x2 + y2
The two bars on either side of v is used to denote that we are
take the magnitude of the vector. The magnitude of the vector is usually
called the norm of the vector, and is the name we will be using from this
point forward.
To find the direction of a vector we once again take note that the vector
is a right triangle with x and y as its legs. The angle
q can be found by taking the arctangent of
the ratio of y to x. This follows directly from the definition
of tangent and arctangent.
q = arctan (y / x)
Just as it is useful to find the norm and direction of a vector from knowing the components, it is also useful to find the components of a vector from knowing its norm and direction. We can do this from knowing the sine and cosine of the direction. Here is a picture of the situation:
sin q = y / ||v|| cos q = x / ||v||
The x and y values from the above equations can be found by multiplying boths by ||v||. Doing that gives the final equations for finding the components of a vector given its direction and norm.
x = ||v|| cos q y = ||v|| sin q
Vector Arithmetic
Just as we have done arithmetic with scalars since first grade, vectors
have a form a arithmetic themselves. We will first do addition. Vector
addition is usually done graphically first, but those methods are never
used in programming so I won't waste your time. For us, and for every
situation you will come accross, vector addition is simply the sum of
the two vectors' components.
a = <x1, y1> b = <x2, y2> a + b = <x1 + x2, y1 + y2>
You can visualize vector addition one of two ways. If you keep the tail of both vectors at the origin then the sum of the two vectors goes from the origin to the vertex of a parallelogram created by a and b.
Arithmetic of scalars also involves multiplication. With vectors there is no one defined way of doing vector multiplication; instead there are three different ways. The first way is the multiplication of a vector with a scalar. To do this you multiply each component of the vector by the scalar.
a = <x, y> a * s = <x * s, y * s>
Graphically this is only a scaling of the vector. If you multiply a vector by the scalar 2 then the new vector is twice the length of the old vector. Also, mutiplying a vector with a positive scalar does not change the direction of the vector. But, if you multiply by a negative number it will flip the vector, thus changing the direction.
a - b = a + (-b)
Therefore we define vector subtraction like this:
a = <x1, y1> b = <x2, y2> a - b = <x1 - x2, y1 - y2>
We can represent this graphically just as we did will vector addition:
There are two vectors that are given special names. These vectors have a length of one and point int the direction of the increasing x- and y-axis. Mathematically, they are defined like this:
i = <1, 0> j = <0, 1>
Many times it is convient to write a vector as a sum of its i
and j components. This is done by scaling the i and j
vectors until their sum is equal to the vector you are trying to obtain.
For example, the vector <3, 2> could be written as 3i + 2j.
In general, a vector <x, y> can also be written as xi + yj.
Individually i and j are called unit vectors, and grouped
together they are called the standard basis vectors. Bases will be something
covered in another tutorial and you will see why that name is given to
these vectors.
From the arithmetic of scalars you know that if you divide a number by
itself you get the number one. Well, a similar operation can be done to
vectors except the outcome is much more useful. If you multiply a vector
by the reciprocal of its norm then you get what is called a unit vector.
A unit vector has a norm of one.
v = <x, y>
u = (1 / ||v||) * v
________
= (1 / \/ x2 + y2) * v
Unit vectors are very useful in many different situations. The first peculiarity that we will look at with unit vectors are its components. By looking at the definition of a unit vector above we can see that the individual components come out like this:
________
ux = x / \/ x2 + y2
________
uy = y / \/ x2 + y2
Therefore the components of a unit vector are obtained by dividing each component by the norm of the vector. Doing this is actually the same as doing the "opposite over hypotenuse" or "adjacent over hypotenuse" familiar with sine and cosine. This means that the x-components of a unit vector is the cosine of the vector's direction, and the y-component of a unit vector is the sine of the vector's direction.
u = <cos q, sin q>Many times it is best to represent the "direction" of a vector with the vector's unit vector rather than its angle with the horizontal. For example, what if you wanted a vector in the same direction as v = <1, 2> but with a magnitude of four exactly. You may start of by finding the angle that v makes with the horizontal, then find the sine and cosine of that angle, and finally multiplying the sine and cosine by the new length to find the components of the new vector. However, with what we know about unit vectors we can simply take the unit vector of v, and scale it by four, thus giving us a new vector with a length of four.
_
||v|| = \/5
_
u = (1 / \/5) * <1, 2>
= <.447, .894>
w = 4 * u
= <1.788, 3.578>
The vector w is now in the same direction as v except with
a magnitude of four. You can double check that by find the norm of w.
In general, you can change the length of vector v to a by
doing this:
v
w = a * -----
||v||
Another form of multiplication for vectors is called the dot product. The dot product is defined as follows:
v = <vx, vy> w = <wx, wy> v · w = vxwx + vywy
Note that the dot product of two vectors is a scalar. A dot product does
not return a vector.
The best geometric interpretation of the dot product is the angle between
two vectors. The dot product relates the cosine of the angle between to
vectors with the following equation:
v · w = ||v|| * ||w|| * cos q
Although we are skipping all mathematical rigor for this tutorial it
can be said that the above is derived from the law of cosines.
By re-arranging the above equation we can explicitly solve for the cosine
of the angle between two vectors:
v · w
cos q = ---------------
||v|| * ||w||
Finally, we can also use the dot product to determine whether or not two vectors are perpendicular (orthogonal). Othogonal is used when two things meet at a right angle (90 degrees or p/2 radians). Plugging in the known angle into the dot product we see this:
v · w = ||v|| * ||w|| * cos 90
= ||v|| * ||w|| * 0
= 0
This implies that if the dot product of two vectors is zero then they
are orthogonal. This is a very important property and it will be used
later on.
A common situation that occurs when dealing with vectors is find the component
of one vector along another. This is known as a projection. There are
four types of projection that can be very useful: parallel scalar projection,
parallel vector projection, perpendicular scalar projection, and perpendicular
vector projection.
The parallel scalar projection is like dropping the shadow of one vector
onto another. The parallel vector projection is basically the same except
it is a vector in the direction of the vector being projected onto with
the norm of the scalar projection. The perpendicular scalar projection
is the shortest distance from the head of one vector to the vector being
projected onto. The perpendicular vector projection has a norm equal to
the perpendicular scalar projection and is orthogonal to the vector being
projected onto.
The picture below will help most with developing what these quantities
mean. Fortunately all are very simple to derive. To start off with we
will find the parallel and perpendicular projection equations. First here
is a picture:
// We can start by defining the cosine of the angle between the two vectors in two different ways
a · b
cos q = ---------------
||a|| * ||b||
proj b a
cos q = ----------
||a||
// Set the equations equal to each other and solve for proj b a
a · b proj b a
--------------- = ----------
||a|| * ||b|| ||a||
// Solve for proj b a and simplify
a · b
proj b a = -------
||b||
Therefore, the parallel scalar projection of a onto b is simply their dot product divided by b's magnitude. Knowing this you can then solve for the perpendicular scalar projection by using the Pythagorean Theorem since you know two sides of the right triangle.
_____________________ perp b a = \/ ||a||2 + (proj b a)2
Later on we are going to define another operator on vectors that will
help us with a more simple equation for calculating the perpendicular
projection.
The two scalar projections we have calculated so far are merely the norms
of the vector projections. The vector projections will add a direction
to the scalar projections we have given equations to so far. We will denote
the vector projections in the same way as the scalar projections except
"proj" and "perp" will now be in bold. The easiest
to do is the parallel vector projection since it is in the direction of
b. To calculate the vector projection scale the unit vector of
b by the scalar projection:
b
proj b a = proj b a * -----
||b||
a · b b
= ------- * -----
||b|| ||b||
And the parallel vector projection has been calculated.
Something similar can be done to find the perpendicular vector projection.
First, we must find a unit vector in the direction of the perpendicular
projection. This is equivalent to the parallel vector projection minus
a (remember the property of vector subtraction such that the difference
is a vector pointing from one vector head to another). Therefore the perpendicular
unit vector can be expressed in the rather cumbersome equation:
proj b a - a
u = ----------------
||proj b a - a||
Finally, scaling this unit vector by the perpendicular scalar projection you get the final equation for calculating the perpendicular vector projection (this equation is very long but still simple):
proj b a - a
perp b a = perp b a * ----------------
||proj b a - a||
_____________________ proj b a - a
= \/ ||a||2 + (proj b a)2 * ----------------
||proj b a - a||
The final operation that will be discussed for vectors with called the
cross product; it is like another way of multiplying vectors together.
Remember that the dot product was also like multiplication for two vectors
except it returned a scalar value. This time, a cross product returns
a vector. Usually the cross product is reserved for three dimensional
vectors. Not much is going to be said about 3D vectors because they are
the exact same as 2D. The addition, subtraction, scalar multiplication,
and dot product rules still hold. The only difference is that there are
now three components <x, y, z> instead of two.
The cross product is defined in a rather long equation:
v = <vx, vy, vz> w = <wx, wy, wz> v ´ w = < vywz - vzwy, vzwx - vxwz, vxwy - vywx >
One of the most important properties of the cross product is that the vector
it outputs is orthogonal to both v and w. You can verify
this by taking the dot products of the cross product with each of the
vectors v and w and seeing that they are zero.
Another geometric interpration is that the cross product relates the sine
of the angle between two vectors with the following equation:
||v ´ w|| = ||v|| * ||w|| * sin q
||v ´ w||
sin q = --------------
||v|| * ||w||
Now that we have a way of calculating the sine of the angle between two
vectors we can find the perpendicular scalar (and vector) projection much
easier. We can derive a new, and easier, equation for calculating the perpendicular
scalar projection by looking at the picture from above again:
// We can start by defining the sine of the angle between the two vectors in two different ways
||a ´ b||
sin q = ---------------
||a|| * ||b||
perp b a
sin q = ----------
||a||
// Set the equations equal to each other and solve for perp b a
||a ´ b|| perp b a
--------------- = ----------
||a|| * ||b|| ||a||
// Solve for perp b a and simplify
||a ´ b||
perp b a = ---------
||b||
You can now use this new calculation of the perpendicular projection to scale
the unit perpendicular vector and you wil get your perpendicular vector
projection.That is the end of this tutorial. You may be wondering where you would ever use this nonsense, but that is to come in the future. An upcoming tutorial on Ultrashock will be on making a camera viewing system for 3D rather than the usual coordinate rotations seen. To do this we will use nearly everything we have done so far, plus introduce a few new concepts from Linear Algebra (matrices and basis).