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Vectors

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Prerequisites

ArithmeticLogic
GeometryAlgebra
TrigonometryFunctions
Vectors

Notation and components

Vectors may be written in several equivalent forms:

$$ \mathbf{v} = \langle 3, -2 \rangle $$
$$ \mathbf{v} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} $$
$$ \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} $$

In 3D:

$$ \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} $$

where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the standard basis vectors:

$$ \mathbf{i} = \langle 1, 0, 0 \rangle,\quad \mathbf{j} = \langle 0, 1, 0 \rangle,\quad \mathbf{k} = \langle 0, 0, 1 \rangle $$

Component form from two points

If a vector goes from point $A(x_1, y_1)$ to point $B(x_2, y_2)$, then

$$ \overrightarrow{AB} = \langle x_2 - x_1,\ y_2 - y_1 \rangle $$

In 3D:

$$ \overrightarrow{AB} = \langle x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1 \rangle $$

This is one of the most common ways vectors appear in geometry problems.

Magnitude and unit vectors

The magnitude or length of $\mathbf{v} = \langle v_1, v_2, \dots, v_n \rangle$ is

$$ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2} $$

In 2D:

$$ \|\mathbf{v}\| = \sqrt{x^2 + y^2} $$

In 3D:

$$ \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} $$

Unit vectors

A unit vector has length 1. The unit vector in the direction of $\mathbf{v}$ is

$$ \hat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|} $$

provided $\mathbf{v} \neq \mathbf{0}$.

Example:

If $\mathbf{v} = \langle 3, 4 \rangle$, then

$$ \|\mathbf{v}\| = 5,\qquad \hat{\mathbf{v}} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle $$

Direction in 2D

If $\mathbf{v} = \langle x, y \rangle$, then the direction angle $\theta$ satisfies

$$ \tan \theta = \frac{y}{x} $$

with the correct quadrant chosen from the signs of $x$ and $y$.

Basic vector operations

Addition and subtraction

For $\mathbf{u} = \langle u_1, u_2, \dots \rangle$ and $\mathbf{v} = \langle v_1, v_2, \dots \rangle$:

$$ \mathbf{u} + \mathbf{v} = \langle u_1 + v_1,\ u_2 + v_2,\ \dots \rangle $$
$$ \mathbf{u} - \mathbf{v} = \langle u_1 - v_1,\ u_2 - v_2,\ \dots \rangle $$

Geometrically, addition follows the tip-to-tail or parallelogram rule.

Scalar multiplication

For scalar $c$:

$$ c\mathbf{v} = \langle cv_1, cv_2, \dots, cv_n \rangle $$

Effects:

  • If $|c| > 1$, the vector stretches

  • If $0 < |c| < 1$, the vector shrinks

  • If $c < 0$, the direction reverses

Zero vector

The zero vector is

$$ \mathbf{0} = \langle 0, 0, \dots, 0 \rangle $$

It has zero magnitude and no direction.

Geometrically, the zero vector represents no displacement: the starting point and ending point are the same. It is not an arrow pointing in a special direction; its length is zero, so there is no orientation to measure. For example, if a particle moves from $A$ to $B$ and then returns to $A$, its net displacement is $\mathbf{0}$.

The zero vector is the additive identity because adding it changes nothing:

$$ \mathbf{v} + \mathbf{0} = \mathbf{v} $$

It is also the result when a vector is subtracted from itself:

$$ \mathbf{v} - \mathbf{v} = \mathbf{0} $$

Two details are especially important:

  • The zero vector is perpendicular to every vector in the algebraic sense that $\mathbf{0} \cdot \mathbf{v} = 0$, but the usual angle formula does not give it an angle because it requires division by $\lVert\mathbf{0}\rVert$.

  • It cannot be normalized. The expression $\mathbf{0}/\lVert\mathbf{0}\rVert$ divides by zero, so the zero vector has no associated unit vector.

Dot product

The dot product of two vectors in $\mathbb{R}^n$ is

$$ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \cdots + u_nv_n $$

In 2D:

$$ \langle a, b \rangle \cdot \langle c, d \rangle = ac + bd $$

In 3D:

$$ \langle a, b, c \rangle \cdot \langle d, e, f \rangle = ad + be + cf $$

Angle between vectors

The dot product also satisfies

$$ \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \, \|\mathbf{v}\| \cos \theta $$

where $\theta$ is the angle between the vectors.

So

$$ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \, \|\mathbf{v}\|} $$

This is often the fastest way to find an angle.

Orthogonality

Two nonzero vectors are orthogonal if their dot product is zero:

$$ \mathbf{u} \cdot \mathbf{v} = 0 $$

This means the vectors are perpendicular.

Projection idea from the dot product

The scalar component of $\mathbf{v}$ along $\mathbf{u}$ is

$$ \operatorname{comp}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|} $$

The vector projection is covered in the next section.

Cross product

The cross product is defined for vectors in $\mathbb{R}^3$.

For

$$ \mathbf{u} = \langle u_1, u_2, u_3 \rangle,\qquad \mathbf{v} = \langle v_1, v_2, v_3 \rangle $$

the cross product is

$$ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} $$

which expands to

$$ \mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1 \rangle $$

Geometric meaning

  • $\mathbf{u} \times \mathbf{v}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$

  • Its magnitude is

$$ \|\mathbf{u} \times \mathbf{v}\| = \|\mathbf{u}\| \, \|\mathbf{v}\| \sin \theta $$

  • It gives the area of the parallelogram spanned by the two vectors

So the triangle area is

$$ \frac{1}{2}\|\mathbf{u} \times \mathbf{v}\| $$

Orientation

The cross product is not commutative:

$$ \mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u}) $$

Use the right-hand rule to determine direction.

Vector projections

The projection of $\mathbf{v}$ onto $\mathbf{u}$ is the component of $\mathbf{v}$ that points along $\mathbf{u}$.

Scalar projection

$$ \operatorname{comp}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|} $$

Vector projection

$$ \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} $$

This formula is useful in decomposition problems and least-squares geometry.

Orthogonal decomposition

Any vector can be split into parallel and perpendicular parts relative to $\mathbf{u}$:

$$ \mathbf{v} = \operatorname{proj}_{\mathbf{u}} \mathbf{v} + \mathbf{v}_\perp $$

where

$$ \mathbf{v}_\perp = \mathbf{v} - \operatorname{proj}_{\mathbf{u}} \mathbf{v} $$

and $\mathbf{v}_\perp \cdot \mathbf{u} = 0$.

Lines and planes

Vectors provide compact equations for geometric objects.

Line in 2D or 3D

A line through point $\mathbf{r}_0$ with direction vector $\mathbf{d}$ has vector equation

$$ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d} $$

where $t$ is a parameter.

In components:

$$ x = x_0 + at,\qquad y = y_0 + bt,\qquad z = z_0 + ct $$

if $\mathbf{d} = \langle a, b, c \rangle$.

Plane in 3D

A plane through point $\mathbf{r}_0$ with normal vector $\mathbf{n}$ satisfies

$$ \mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0 $$

If $\mathbf{n} = \langle A, B, C \rangle$ and $\mathbf{r}_0 = (x_0, y_0, z_0)$, then

$$ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 $$

This is the standard point-normal form of a plane.

Distance from a point to a plane

For plane

$$ Ax + By + Cz + D = 0 $$

and point $(x_0, y_0, z_0)$, the distance is

$$ \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} $$

Common workflows

Find the vector from one point to another

  1. Identify the initial point $A$ and terminal point $B$

  2. Subtract coordinates: $B - A$

  3. Simplify the components

Example:

If $A = (1, -3)$ and $B = (5, 2)$, then

$$ \overrightarrow{AB} = \langle 4, 5 \rangle $$

Find an angle between vectors

  1. Compute $\mathbf{u} \cdot \mathbf{v}$

  2. Compute $\|\mathbf{u}\|$ and $\|\mathbf{v}\|$

  3. Use

$$ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \, \|\mathbf{v}\|} $$

  1. Apply inverse cosine

Test perpendicularity or parallelism

  • Perpendicular: $\mathbf{u} \cdot \mathbf{v} = 0$

  • Parallel: one vector is a scalar multiple of the other

Build a line or plane

  • Line: point plus direction vector

  • Plane: point plus normal vector

If the problem gives a geometric picture, convert it to vectors first. The algebra is usually simpler afterward.

Formula summary

Core formulas

$$ \mathbf{u} + \mathbf{v} = \langle u_1 + v_1, \dots, u_n + v_n \rangle $$
$$ c\mathbf{v} = \langle cv_1, \dots, cv_n \rangle $$
$$ \|\mathbf{v}\| = \sqrt{v_1^2 + \cdots + v_n^2} $$
$$ \mathbf{u} \cdot \mathbf{v} = \sum_{i=1}^n u_i v_i $$
$$ \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \, \|\mathbf{v}\| \cos \theta $$
$$ \mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1 \rangle $$
$$ \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} $$
$$ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d} $$
$$ \mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0 $$

Pitfalls

Sign mistakes in components

When subtracting coordinates, always use

$$ \overrightarrow{AB} = B - A $$

not the other way around unless the direction is intentionally reversed.

Confusing dot product with multiplication

The dot product produces a scalar, not a vector.

Confusing dot and cross products

  • Dot product measures alignment and angle

  • Cross product measures perpendicularity and area in 3D

Forgetting absolute value in distance formulas

Distance is always nonnegative. Use absolute value when needed.

Using the wrong dimension

  • Dot product works in any dimension

  • Cross product, in the standard form, is a 3D operation

Losing the direction of a vector

A vector is not just its length. Reversing a vector changes its direction even if the magnitude stays the same.

Final checks

Before moving on in a vector problem, verify:

  • The components point the correct way

  • The dimensions match the operation

  • Units are consistent if the vectors are physical quantities

  • The final answer has the right geometric meaning

Sources