Math 255- Math Fundamentals for Game

What Math Do Games Need?

Every moving object in a game is a position, a direction, and a rule for changing them. Vectors describe positions and directions, trigonometry converts between angles and directions, physics updates motion each frame, rotations orient objects in 3D, and collision tests decide when objects touch.

A useful approach: Draw the situation → name the vectors → choose the formula → substitute numbers → check units and sign → test against the game (does it look right?).

1. Vectors

A vector stores a magnitude and a direction as components. In a game it can be a position, a velocity, or a direction.

v = (v_{x}, v_{y}, v_{z})

Core Vector Operations

Operation	Formula	Game Use
Addition	$a + b = (a_{x} + b_{x}, a_{y} + b_{y})$	Combine movement and wind
Scaling	$k v = (k v_{x}, k v_{y})$	Change speed
Magnitude	$\| v \| = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}$	Speed or distance
Normalize	$\hat{v} = \frac{v}{\| v \|}$	Direction only, length 1
Move	$p_{n e w} = p + v Δ t$	Update position each frame

Example: Normalize

v = (3, 4)

Step 1: Find the magnitude.

| v | = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5

Step 2: Divide each component by 5.

\hat{v} = (\frac{3}{5}, \frac{4}{5}) = (0.6, 0.8)

Step 3: Check:

\sqrt{0.36 + 0.64} = 1

Conclusion: The unit direction is $(0.6, 0.8)$ .

Example: A player is at

p = (2, 1)

with velocity

v = (4, - 2)

units/s. Find the position after

Δ t = 0.5

Step 1: Scale the velocity.

v Δ t = 0.5 (4, - 2) = (2, - 1)

Step 2: Add to the position.

p_{n e w} = (2, 1) + (2, - 1) = (4, 0)

Conclusion: The new position is $(4, 0)$ .

Always normalize a direction before multiplying by speed. Otherwise diagonal movement is faster than straight movement:

(1, 1)

has length

\sqrt{2} \approx 1.41

Practice: Find the magnitude of

v = (2, 3, 6)

| v | = \sqrt{4 + 9 + 36} = \sqrt{49} = 7

Practice: Normalize

v = (0, 5)

| v | = 5

, so

\hat{v} = (0, 1)

2. Dot Product and Cross Product

These two products measure how vectors relate: the dot product returns a number about alignment, and the cross product returns a vector perpendicular to both inputs.

a \cdot b = a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z} = | a | | b | \cos θ

a \times b = (a_{y} b_{z} - a_{z} b_{y}, a_{z} b_{x} - a_{x} b_{z}, a_{x} b_{y} - a_{y} b_{x})

Result	Dot Product Meaning	Game Use
$a \cdot b > 0$	Vectors point roughly the same way	Target is in front
$a \cdot b = 0$	Vectors are perpendicular	Right-angle check
$a \cdot b < 0$	Vectors point roughly opposite	Target is behind

Example: Find the angle between

a = (2, 0)

and

b = (3, 3)

Step 1: Dot product:

2 \cdot 3 + 0 \cdot 3 = 6

Step 2: Magnitudes:

| a | = 2

and

| b | = \sqrt{18} \approx 4.243

Step 3: Solve for the angle.

\cos θ = \frac{6}{2 (4.243)} \approx 0.7071 \to θ = 45^{°}

Conclusion: The angle is $45^{°}$ .

Example: Find the surface normal of a triangle with edge vectors

a = (2, 3, 4)

and

b = (5, 6, 7)

Step 1: First component:

3 \cdot 7 - 4 \cdot 6 = - 3

Step 2: Second component:

4 \cdot 5 - 2 \cdot 7 = 6

Step 3: Third component:

2 \cdot 6 - 3 \cdot 5 = - 3

Conclusion: $a \times b = (- 3, 6, - 3)$ . Normalize it to use it as a lighting or collision normal.

Example: Project

a = (3, 4)

onto the x-axis direction

b = (1, 0)

Step 1: Use

{proj}_{b} a = \frac{a \cdot b}{b \cdot b} b

Step 2: Substitute:

\frac{3}{1} (1, 0) = (3, 0)

Conclusion: The projection is $(3, 0)$ . Projection is how you slide along walls and ramps.

Practice: Compute

a \cdot b

for

a = (1, 2, 3)

and

b = (4, - 5, 6)

4 - 10 + 18 = 12

Practice: Compute

(1, 0, 0) \times (0, 1, 0)

(0 \cdot 0 - 0 \cdot 1, 0 \cdot 0 - 1 \cdot 0, 1 \cdot 1 - 0 \cdot 0) = (0, 0, 1)

3. Trigonometry

Trigonometry connects an angle to a direction. For a right triangle with hypotenuse $r$ :

\sin θ = \frac{y}{r} \cos θ = \frac{x}{r} \tan θ = \frac{y}{x}

Rearranged, a direction from an angle is $(x, y) = (r \cos θ, r \sin θ)$ , and radians relate to degrees by $rad = \deg \cdot \frac{π}{180}$ .

Angle	Radians	Sine	Cosine
$0^{°}$	$0$	$0$	$1$
$30^{°}$	$\frac{π}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$
$45^{°}$	$\frac{π}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$
$60^{°}$	$\frac{π}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$
$90^{°}$	$\frac{π}{2}$	$1$	$0$

Example: A projectile launches at speed

10

m/s and angle

30^{°}

. Find its velocity components.

Step 1: Horizontal:

v_{x} = 10 \cos 30^{°} = 10 \cdot \frac{\sqrt{3}}{2} \approx 8.66

Step 2: Vertical:

v_{y} = 10 \sin 30^{°} = 10 \cdot \frac{1}{2} = 5

Conclusion: $v = (8.66, 5)$ m/s.

Example: An enemy is at offset

(x, y) = (- 1, 1)

from the player. Find the facing angle.

Step 1: Use the two-argument arctangent, which respects the quadrant:

θ = atan2 (y, x)

Step 2: The point is in quadrant II, so

θ = atan2 (1, - 1) = 135^{°} = \frac{3 π}{4}

Conclusion: The enemy is at $135^{°}$ . Plain $\arctan (y / x) = - 45^{°}$ would point the wrong way.

Most game engines use radians in their trig functions. Convert degrees before calling

\sin

\cos

Practice: Convert

60^{°}

to radians.

60 \cdot \frac{π}{180} = \frac{π}{3} \approx 1.047

Practice: Find the direction vector for a heading of

90^{°}

with speed

4

(4 \cos 90^{°}, 4 \sin 90^{°}) = (0, 4)

4. Physics for Movement

Physics in games is a loop: compute forces, update velocity, update position. Constant acceleration $a$ (such as gravity $g = 9.8$ m/s²) follows these kinematic equations:

Quantity	Formula	Meaning
Velocity	$v = v_{0} + a t$	Speed after time $t$
Position	$y = y_{0} + v_{0} t + \frac{1}{2} a t^{2}$	Position after time $t$
Speed–distance	$v^{2} = v_{0}^{2} + 2 a Δ y$	No time needed
Range	$R = \frac{v^{2} \sin 2 θ}{g}$	Flat-ground projectile
Force	$F = m a$	Acceleration from force

Example: A character jumps upward at

v_{0} = 6

m/s with

g = 9.8

m/s². Find the time to the peak and the jump height.

Step 1: At the peak

v = 0

, so

0 = 6 - 9.8 t

and

t = \frac{6}{9.8} \approx 0.61

Step 2: Use

v^{2} = v_{0}^{2} - 2 g h

with

v = 0

h = \frac{v_{0}^{2}}{2 g} = \frac{36}{19.6} \approx 1.84 m

Conclusion: The peak occurs at $0.61$ s at a height of about $1.84$ m.

Example: Use semi-implicit Euler for a ball at

y = 10

v = 0

a = - 9.8

Δ t = 0.1

Step 1: Update velocity first:

v = 0 + (- 9.8) (0.1) = - 0.98

Step 2: Then update position with the new velocity:

y = 10 + (- 0.98) (0.1) = 9.902

Conclusion: After one frame, $v = - 0.98$ and $y = 9.902$ . Repeat every frame.

Example: Find the range of a projectile with

v = 20

m/s at

45^{°}

Step 1:

\sin 2 θ = \sin 90^{°} = 1

Step 2:

R = \frac{20^{2} (1)}{9.8} = \frac{400}{9.8} \approx 40.8

Conclusion: The range is about $40.8$ m.

Practice: How fast is a dropped object moving after

2

s of free fall (

g = 9.8

v = 0 + 9.8 (2) = 19.6

m/s downward.

Practice: A

2

kg object receives a force of

10

N. Find its acceleration.

a = \frac{F}{m} = \frac{10}{2} = 5

m/s².

5. Rotations in 2D and 3D

Rotating a point is multiplying it by a rotation matrix. In 2D, a counter-clockwise rotation by $θ$ is:

(\begin{matrix} x ‘ \\ y ‘ \end{matrix}) = (\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}) (\begin{matrix} x \\ y \end{matrix})

In 3D, there is one matrix for each axis. The rotation about the x-axis is:

R_{x} (θ) = (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos θ & - \sin θ \\ 0 & \sin θ & \cos θ \end{matrix})

Example: Rotate

(1, 1)

90^{°}

in 2D.

Step 1:

\cos 90^{°} = 0

and

\sin 90^{°} = 1

Step 2:

x ‘ = 0 \cdot 1 - 1 \cdot 1 = - 1

Step 3:

y ‘ = 1 \cdot 1 + 0 \cdot 1 = 1

Conclusion: The result is $(- 1, 1)$ .

Example: Rotate

(0, 1, 0)

90^{°}

about the x-axis.

Step 1:

x ‘ = 0

Step 2:

y ‘ = \cos 90^{°} (1) - \sin 90^{°} (0) = 0

Step 3:

z ‘ = \sin 90^{°} (1) + \cos 90^{°} (0) = 1

Conclusion: The result is $(0, 0, 1)$ : up turns into forward.

Quaternions

Chaining Euler-angle rotations can cause gimbal lock, where two axes align and a degree of freedom is lost. Engines use quaternions to avoid this. A rotation of $θ$ about a unit axis $n$ is:

q = (\cos \frac{θ}{2}, n \sin \frac{θ}{2})

Example: Build the quaternion for

90^{°}

about the z-axis.

Step 1: Half angle:

\frac{θ}{2} = 45^{°}

Step 2:

w = \cos 45^{°} \approx 0.707

and

n \sin 45^{°} = (0, 0, 0.707)

Conclusion: $q \approx (0.707, 0, 0, 0.707)$ .

Practice: Rotate

(2, 0)

90^{°}

x ‘ = 2 \cdot 0 - 0 = 0

y ‘ = 2 \cdot 1 + 0 = 2

, so the result is

(0, 2)

Practice: Why is

\frac{θ}{2}

used in a quaternion?

A quaternion rotates by sandwiching a vector,

q v q^{- 1}

, which applies the rotation twice in effect, so each side carries half the angle.

6. Collision Detection

Collision tests are chosen to be cheap. Compare squared distances to avoid square roots.

Test	Condition	Notes
Circle / Sphere	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} \leq (r_{1} + r_{2})^{2}$	Very fast
AABB (box)	$a_{m a x} \geq b_{m i n}$ and $b_{m a x} \geq a_{m i n}$ on every axis	Check x, y (and z)
Ray / Plane	$t = \frac{(p_{0} - o) \cdot n}{d \cdot n}$	Hit if $t \geq 0$
Reflection	$v ‘ = v - 2 (v \cdot n) n$	Bounce off a surface

Example: Do circles at

(0, 0)

with

r = 2

and

(3, 4)

with

r = 3

collide?

Step 1: Squared distance:

3^{2} + 4^{2} = 25

Step 2: Squared radius sum:

(2 + 3)^{2} = 25

Step 3: Compare:

25 \leq 25

Conclusion: The circles are exactly touching, which counts as a collision.

Example: A ball with velocity

v = (3, - 4)

hits a floor with normal

n = (0, 1)

. Find the bounce velocity.

Step 1:

v \cdot n = - 4

Step 2:

v ‘ = (3, - 4) - 2 (- 4) (0, 1) = (3, - 4) + (0, 8)

Conclusion: $v ‘ = (3, 4)$ : the vertical part flips and the horizontal part is kept.

Example: A ray starts at

o = (0, 5, 0)

heading

d = (0, - 1, 0)

. The floor plane passes through

p_{0} = (0, 0, 0)

with

n = (0, 1, 0)

. Where does it hit?

Step 1:

(p_{0} - o) \cdot n = (0, - 5, 0) \cdot (0, 1, 0) = - 5

Step 2:

d \cdot n = - 1

, so

t = \frac{- 5}{- 1} = 5

Step 3: Hit point:

o + t d = (0, 5, 0) + 5 (0, - 1, 0) = (0, 0, 0)

Conclusion: The ray hits the floor at $(0, 0, 0)$ after $t = 5$ .

Practice: Do boxes A:

x \in [0, 4]

y \in [0, 2]

and B:

x \in [3, 6]

y \in [3, 5]

overlap?

No. The x-ranges overlap, but the y-ranges

[0, 2]

and

[3, 5]

do not, so there is no collision.

Practice: Do spheres at

(0, 0, 0)

r = 1

and

(2, 2, 1)

r = 1.5

collide?

Squared distance is

4 + 4 + 1 = 9

. Squared radius sum is

{2.5}^{2} = 6.25

. Since

9 > 6.25

, they do not collide.

Key Takeaways

Vectors store positions, velocities, and directions; normalize directions before scaling by speed.
The dot product measures alignment and angles; the cross product gives perpendicular normals.
Trigonometry turns angles into directions; use atan2 to turn directions into angles.
Update velocity, then position, each frame using $Δ t$ .
Rotation matrices rotate points; quaternions avoid gimbal lock in 3D.
Collision tests compare squared distances or ranges; reflection uses the surface normal.