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Scalars are one dimensional, they are simply numbers, though they might have a certain metric associated with them, like meters per second or degrees. They are used for the speed of a car, the angle of a wheel or a uniform zoom factor for example.


A two dimensional point, defined as \((x, y)\) describes a position or coordinate in a 2D world. It can be the position of a tree, the points of a polygon describing a region, the waypoints for a character path or the points of an animation curve. However 2D points are not limited to games with a 2D world. Even in a 3D world, screen space coordinates will be two dimensional.

World and screen space

Screen space is where the pixels of your screen live, while world space is where the objects of your game live. In anything but the simplest game, world and screen space will not perfectly match. When they don’t match, world space points have to be transformed from world space to screen space.

For example, most mobile games need to scale from world space to screen space. Your game’s world space coordinates might be set in a fixed resolution, which are scaled to the full screen size of the phone or tablet it runs on. If the aspect ratio of the phone or tablet doesn’t match, the world might have to be translated as well to center it on the screen. Sometimes the world is larger than the screen, think side-scroller or a world map in an RTS or RPG game. In this case we need to transform our world so that a piece of it is both translated and scaled to fit on the screen.

Note that we are not scaling pixels, but rather the corners of sprites or the points of a polygon. Once these positions are transformed, we can render the actual world to pixels. Also note that we can not really talk about scaling an individual point, because what scales is their position relative to each other.

Scaling from the origin

While we can’t translate yet, we can do a uniform scale in order to do some zoom operations. To transform a point with a uniform scale factor s, a scalar, we simply multiply both \(x\) and \(y\) with s.

\[(x',y') = s(x, y)=(s*x, s*y)\]

function mul(s, x, y)
    return x * s, y *s

To determine the scale in case of matching the world with a device screen, we simply use min(screen_width/world_width, screen_height/world_height). This gives us a scale which makes both width and height fit inside the screen, preserving the aspect ratio.

We can also do a non-uniform scale by using different factors for x and y. While this stretches the image, it is very often useful in situations where you need to make a certain element longer or higher depending on the size of something else, for example a button in a GUI. In this case the corners will have a uniform scale while the sides and center have a non-uniform scale.

Until now, all our scaling will work from the center. The point \((0,0)\) is the only point which doesn’t change. As whatever scaling factor we use, when multiplied with zero, it gives us zero


If the scale is larger than 1, all other points drift away from from the origin. If the scale is smaller than 1, all other points will drift towards \((0, 0)\). All this is not very useful without being able to translate though, so let’s move on.


When we subtract two points, we get a vector. Vectors are also defined as a pair \(\langle x, y\rangle\), however the x and y are no longer a position, but rather define a direction and distance from the origin. A vector \(\langle x, y\rangle\) can be thought of as an arrow starting from \((0, 0)\) towards \((x, y)\). The further a vector is from the origin, the greater its magnitude. If we have two points a and b, we can build the vector \(\vec{ab}\) by subtracting a from b.

function sub(x1, y1, x2, y2)
    return x1 - x2, y1 - y2

The point at the arrow of the vector should always come first.

If you have trouble remembering this, it might be a good idea to make a function to build a vector.

function vector(x1, y1, x2, y2)
    return x2 - x1, y2 - y1

Notice that calculating the vector from for example \((0, 0)\) to \((1, 4)\) or from \((4, 5)\) to \((5, 9)\) both give the vector \(\langle 1, 4\rangle\). Because to move from any of the two origins towards their destination you need to move in the same direction, crossing the same distance.

Adding the vector to its origin gives us the destination.

\[a + \vec{ab}=b\]

This is quite logical

\[a + \vec{ab} = a + (b - a)=a + b - a=b\]

A vector can move or translate a point by a given distance in a given direction simply by adding it to the point.

function add(x1, y1, x2, y2)
    return x1 + x2, y1 + y2

Since a point and a vector are so similar, should we really distinguish them? Especially since any point can also be seen as a vector starting from the origin

\[(x, y) - (0, 0) = \langle x, y\rangle\]

Yes and no. In code, we almost never really distinguish the two. In our functional lua code, we use a plain pair of floats x and y for both, and in an object oriented framework we would just use a point or vector class, not both.

However there is an important difference between them. A point defines a coordinate, a vector defines a direction and distance. When we transform points and vectors to a different coordinate system, points are translated, rotated and scaled. Vectors however are only ever rotated and scaled, never translated.

While a point gets translated by adding a vector, a vector doesn’t get translated by adding another vector. Vector addition changes the direction and magnitude of the vector.

\[\langle x_3, y_3\rangle=\langle x_1,y_1\rangle+\langle x_2,y_2\rangle\]

For example if we have the vector \(\langle 1,0\rangle\) pointing along the positive x axis, and the vector \(\langle 0,1\rangle\) pointing along the positive y axis, we get when adding them

\[\langle 1, 1\rangle=\langle 1,0\rangle+\langle 0,1\rangle\]

This new vector \(\langle 1,1\rangle\) is pointing diagonally towards the positive x, y plane, and its magnitude is no longer 1. Since we only translated the world, directions and magnitudes shouldn’t change. We will later see how we can enforce not having vectors translated when using matrices.

Scaling from a given center

Now that we can translate, we can devise a method to scale from any given point, not just from the origin. If we want to scale from a point \((c_x, c_y)\), we need to place that point at the origin. We can do this by translating the point by adding the vector \(\langle -c_x, -c_y\rangle\) or subtracting the vector \(\langle c_x, c_y\rangle\)

\[(x',y')=(x, y) - \langle c_x, c_y\rangle\]

If we scale now, the point \((c_x, c_y)\) will be the center of our scale. \((c_x, c_y)\) will not be affected by the scale, and all other points will drift away from or towards it

\[(x',y')=s((x, y) - \langle c_x, c_y\rangle)\]

However, everything will also be translated, which is not what we wanted, so we need to translate everything back again after scaling

\[(x',y')=s((x, y) - \langle c_x, c_y\rangle) + \langle c_x, c_y\rangle\]
function scale(x, y, s, cx, cy)
    cx = cx or 0
    cy = cy or 0
    return s*(x-cx)+cx, s*(y-cy)+cy

Besides adding or subtracting vectors, or doing scalar multiplication or division, we can also take the product of two vectors. There are two different kinds of products.

The dot product

The dot product is defined as the scalar obtained by

\[\langle x_1,y_1\rangle.\langle x_2,y_2\rangle=x_1*x_2+y_1*y_2\]
function dot(x1, y1, x2, y2)
    return x1*x2+y1*y2

The dot product can be used to calculate the length (magnitude or norm) of a vector. But it can also give you the cosine of the angle between two vectors or project one vector onto another one.

The cross product

The cross product if defined as

\[\langle x_1,y_1\rangle\times\langle x_2,y_2\rangle=x_1*y_2-y_1*x_2\]
function cross(x1, y1, x2, y2)
    return x1*y2-y1*x2

In 3 dimensions this product would produce a vector, namely the vector perpendicular to the plane defined by the two vectors. In 2D however, we get a scalar, as there is no third dimension. The cross product can give you the sine of the angle between two vectors. Combined, the dot and cross products can be used to rotate points. As these two products have many applications. I would suggest that if you want to memorize anything, memorize these two products, as they can be used in many ways.

Note that unlike for the dot product, the order of the vectors in the cross product matters. If the order of the vectors is switched, the sign is flipped. This is because the order determines which angle between the vectors is chosen, the smaller or the greater angle.

The length of a vector

If we take the dot product of a vector with itself, we obtain its length squared.

\[{\left|\langle x, y\rangle\right|}^2 = \langle x,x\rangle.\langle x,y\rangle=x*x+y*y\]

When we need the actual length, we need to take the square root

\[\left|\langle x, y\rangle\right| = \sqrt{\langle x,x\rangle.\langle x,y\rangle}=\sqrt{x*x+y*y}\]
function length(x, y)
    return math.sqrt(x*x+y*y)

Being able to obtain the length of a vector allows us to calculate the distance between two points. Given the points a and b, we can create the vector \(\vec{ab}\) and take it’s length. It gives us the distance to travel from a to b.

function distance(x1, y1, x2, y2)
    local dx, dy = x1-x2, y1-y2
    return math.sqrt(dx*dx+dy*dy)

Note that when comparing magnitudes, you can often do with the length squared, which saves you a square root calculation.


\[\sqrt{a} \leq\sqrt{b}\]


\[a\leq b\]

If only one side has a square root, we can square the expression, so if

\[\sqrt{a} \leq b\]


\[a\leq b*b\]

A useful application of this is testing whether a point is inside a circle. To know the answer, we need to compare the distance between the point and the center of the circle with the radius of the circle. However instead of using a square root, we can compare the distance squared with the radius squared

function inCircle(x, y, cx, cy, r)
    local dx, dy = x-cx, y-cy
    return dx*dx+dy*dy <= r*r

Being able to determine the length of a vector, we can see now that for example \(\langle 2, 0\rangle\) and \(\langle 0, 2\rangle\) have the same length. We are now able to compare the magnitudes of vectors while disregarding their direction.

\[\sqrt{2*2+0*0} = \sqrt{0*0+2*2}=\sqrt{4}=2\]

The direction of a vector

While previously we disregarded direction in order to compare lengths, we will now concentrate on the direction dimension of a vector instead. If we erase the magnitude from our vectors, we can determine whether they have the same direction. We can do this by dividing a vector by its length. This process is also called normalization.

\[\frac{\langle x, y\rangle}{\left|\langle x, y\rangle\right|}\]
function normalize(x, y)
    local l = math.sqrt(x*x+y*y)
    return x/l, y/l

Normalized vectors have always 1 as length, since their original length has been erased. We can now see that \(\langle 1, 3\rangle\) an \(\langle 2, 6\rangle\) point in the same direction, as normalization gives for both \(\langle 0.31622776601, 0.94868329805\rangle\), while \(\langle -1, -3\rangle\) points in the opposite direction \(\langle -0.31622776601, -0.94868329805\rangle\).

Vector negation

Negating or multiplying a vector by -1 gives the vector pointing in the opposite direction.

function negate(x, y)
    return -x, -y

Clockwise and counterclockwise perpendicular vector

Sometimes we need the vector perpendicular to a certain vector. There are two possibilities, one in the clockwise direction, and one in the counterclockwise direction.

function ccw(x, y)
    return -y, x

function cw(x, y)
    return y, -x

A note on notation

A point \((x, y)\) is written with round brackets, while I use square brackets to indicate that we’re talking about a vector \(\langle x, y\rangle\). When working with named vectors, I use an arrow \(\overrightarrow{a}\) or in case of a unit vector a hat \(\hat{a}\).

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