Vectors are a foundational mathematical component to any 3D game engine.


Base Notation

Vectors take the following notation:

\[ \vec{V} = \langle V_1, V_2, V_3, \ldots, V_n \rangle \]

or, as a 4D Point. 3D game engines will be primarily concerned with its 3D and 4D representations, not just any arbitrary n.

\[ \vec{P} = \langle P_x, P_y, P_z, P_w \rangle \]

Each \( V_i \) are alled the components of \( \vec{V} \).

Vectors notated as a Matrix

Column vector:

\[ \vec{V} = \begin{bmatrix}V_1 \\ V_2 \\ V_3 \\ \vdots \\ V_n\end{bmatrix} \]

Row vector:

\[ \vec{V^T} = \begin{bmatrix}V_1 & V_2 & V_3 & \cdots & V_n\end{bmatrix} \]

The T superscript in the row vector indicates that column and row vectors are transpositions of each other.

Vector Space

All vectors belong to a vector space denoted \( ℝ^n \), where n refers to the n-dimension of the vectors the vector space owns.

For example, the vector space consisting of all 3D vectors is \( ℝ^3 \).

A vector space can be defined by its basis vectors. There will be n such basis vectors for any given \( ℝ^n \). Basis vectors must be linearly independent, which means that there is no scalar a that can be multiplied against one basis vector that will generate another basis vector. Basis vectors must also be orthogonal (and preferably orthonormal).

There are infinite choices for a set of n vectors to serve as the basis for any \( ℝ^n \). As soon as you find n vectors, each orthogonal to each other, they can form yet another basis for the same vector space.

The basis vectors can, with scalar multiplication, generate any other vector in their entire vector space.

Gram-Schmidt Orthogonalization can be used to transform a set of n linearly independent vectors into a set of orthogonal vectors.

  • All orthonormal sets of vectors are also orthogonal.
  • All orthogonal sets of vectors are also linearly independent.
  • All linearly independent sets of vectors are not necessarily orthogonal.
  • All orthogonal sets of vectors are not necessarily orthonormal.

Vector Operations

Scalar Multiplication

Scalars apply to each vector component.

\[ \mathit{a}\vec{V} = \langle \mathit{a}V_1,\;\mathit{a}V_2,\;\ldots,\;\mathit{a}V_n\rangle \]

Addition and Subtraction

Vectors add and subtract component-wise. Subtraction is a special case of addition that involves multiplying one vector by a scalar -1 before adding.

\[ \vec{P} + \vec{Q} = \langle P_1+Q_2, \; P_2+Q_2, \; \ldots, \; P_n+Q_n \rangle \]


The magnitude of a vector is also its length and can be found by summing the squares of its components and taking the square root.

The magnitude is also known as the norm or the length of a vector.

\[ \lVert V \rVert = \sqrt{\sum_{i=1}^n{V_i^2}} \]

Dot Product

The dot product of two vectors is also known as the scalar product or the inner product.

This measures the difference between the direction in which two vectors point.

\[ \vec{P}\cdot\vec{Q} = \sum_{i=1}^n{P_i Q_i} \]

\[ \vec{P^T}\vec{Q} = \begin{bmatrix}P_1 & P_2 & \cdots & P_n \end{bmatrix} \begin{bmatrix}Q_1 \\ Q_2 \\ \vdots \\ Q_n \end{bmatrix} \]

\[ \vec{P}\cdot\vec{Q} = \lVert P \rVert \lVert Q \rVert cos\alpha \]

Interpreting the Dot Product

\( \vec{P}\cdot\vec{Q} = 0 \)

  • Two vectors \( \vec{P} \) and \( \vec{Q} \) are perpendicular if and only if \( \vec{P}\cdot\vec{Q} = 0 \). This occurs only if the cosine of the angle between them is 0, and a cosine of an angle is 0 only if the angle is 90°.
  • Perpendicular vectors are orthogonal to one another. The zero vector is orthogonal to all vectors since for all \( \vec{0}\cdot\vec{P} = 0 \)

\( \vec{P}\cdot\vec{Q} > 0 \)

  • Both vectors are pointing in the same direction.

\( \vec{P}\cdot\vec{Q} < 0 \)

  • Both vectors are pointing in the opposite direction.


A vector can be projected onto another vector by decomposing some vector \( \vec{P} \) into two components that are parallel and perpendicular to another vector \( \vec{Q} \).

The projection will result in a vector in the parallel direction of \( \vec{Q} \), with a magnitude equal to the projection (it may not be unit-length anymore.) \[ proj_{\vec{Q}}\vec{P} = \frac{\vec{P}\cdot\vec{Q}}{{\lVert Q \rVert}^2}\vec{Q} \]

\[ perp_{\vec{Q}}\vec{P} = \vec{P} - proj_{\vec{Q}}\vec{P} \]

The projection can also be viewed as a linear transformation and thereby a matrix-vector multiplication.

\[ proj_{\vec{Q}}\vec{P} = \frac{1}{{\lVert Q \rVert}^2} \begin{bmatrix}Q_x^2 & Q_x Q_y & Q_x Q_z \\ Q_x Q_y & Q_y^2 & Q_y Q_z \\ Q_x Q_z & Q_y Q_z & Q_z^2 \end{bmatrix}\begin{bmatrix}P_x \\ P_y \\ P_z \end{bmatrix} \]

Cross Product

The cross product applies to 3D vectors only. It is known as the vector product and it returns a new vector that is perpendicular to both of the vectors being multiplied together.

\[ \vec{P} × \vec{Q} = \langle P_y Q_z - P_z Q_y, \; P_z Q_x - P_x Q_z, \; P_a Q_y - P_y Q_x \rangle \]

The cross product also has trigonometric significance.

\[ \lVert \vec{P} × \vec{Q} \rVert = \lVert \vec{P} \rVert \lVert \vec{Q} \rVert sin \alpha \]

Although the cross product returns a vector that is perpendicular to the two vectors being multiplied, there are two possible directions. Cross products follow a pattern called the right hand rule. The following theorems demonstrate this.

\[ \vec{i} × \vec{j} = \vec{k} \qquad \quad \vec{j} × \vec{i} = -\vec{k} \] \[ \vec{j} × \vec{k} = \vec{i} \qquad \quad \vec{k} × \vec{j} = -\vec{i} \] \[ \vec{k} × \vec{i} = \vec{j} \qquad \quad \vec{i} × \vec{k} = -\vec{j} \]

Calculating the Area of an arbitrary triangle using cross product

\[ A = \frac{1}{2}{(V_2 - V_1) × (V_3 - V_1)} \]

Special Vectors

Unit Vector

\[ \text{Any} \; \vec{v} \; \text{such that} \; \lVert v \rVert = 1 \]

Zero Vector

\[ \vec{0} = \langle 0, \; 0, \; \ldots, \; 0 \rangle \]

i, j, and k basis vectors

See the discussion above on Vector Space for information about basis vectors.

The basis vectors \( \vec{i} \), \( \vec{j} \), and \( \vec{k} \) are unit vectors that point along the x-, y-, and z-axis, respectively in whatever coordinate space is contextually relevant.

\[ \vec{i} = \langle 1, \; 0, \; 0 \rangle \] \[ \vec{j} = \langle 0, \; 1, \; 0 \rangle \] \[ \vec{k} = \langle 0, \; 0, \; 1 \rangle \]