Notes

Vectors in three dimensions

A vector records both a direction and a length. In three dimensions, we write

$$\mathbf v=\langle x,y,z\rangle,$$

where $x$, $y$, and $z$ are the vector's components. The symbol $\mathbf v$ is read “vector $v$.”

A point $P=(x,y,z)$ determines the position vector from the origin $O=(0,0,0)$ to $P$:

$$\overrightarrow{OP}=\langle x,y,z\rangle.$$

If

$$P=(x,y,z), \qquad Q=(a,b,c),$$

then the displacement from $P$ to $Q$ is found by subtracting the starting point from the ending point:

$$\overrightarrow{PQ} =\langle a-x,b-y,c-z\rangle.$$

This is the same as subtracting the corresponding position vectors.

Vector arithmetic

Vectors are added component by component:

$$\langle x,y,z\rangle+\langle a,b,c\rangle =\langle x+a,y+b,z+c\rangle.$$

Multiplying by a real number $\alpha$ scales every component:

$$\alpha\langle x,y,z\rangle =\langle \alpha x,\alpha y,\alpha z\rangle.$$

When $\alpha>1$, the vector becomes longer without changing direction. When $0<\alpha<1$, it becomes shorter. A negative value reverses the direction as well as changing the length.

Length, distance, and unit vectors

The norm, or magnitude, of

$$\mathbf v=\langle x,y,z\rangle$$

is

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

It is the distance from the origin to the endpoint of the vector. Scaling obeys

$$\|\alpha\mathbf v\|=|\alpha|\,\|\mathbf v\|.$$

The absolute value appears because a length cannot be negative.

The distance between two points is the norm of their displacement vector:

$$\boxed{ d(P,Q) =\|\overrightarrow{PQ}\| =\sqrt{(a-x)^2+(b-y)^2+(c-z)^2}. }$$

A unit vector has length $1$. If $\mathbf v\ne\mathbf0$, dividing by its norm keeps its direction and changes its length to $1$:

$$\boxed{\mathbf u=\frac{\mathbf v}{\|\mathbf v\|}}.$$

This process is called normalizing the vector.

The standard basis

The standard basis vectors in $\mathbb R^3$, read “$R$ three,” are

$$\mathbf i=\langle1,0,0\rangle, \qquad \mathbf j=\langle0,1,0\rangle, \qquad \mathbf k=\langle0,0,1\rangle.$$

They point along the positive coordinate directions. Every vector can be decomposed into those directions:

$$\langle x,y,z\rangle =x\mathbf i+y\mathbf j+z\mathbf k.$$

The same coordinate ideas work in $\mathbb R^2$ and, more generally, in $\mathbb R^n$.

The dot product

For

$$\mathbf v=\langle x_1,y_1,z_1\rangle, \qquad \mathbf w=\langle x_2,y_2,z_2\rangle,$$

the dot product is

$$\boxed{ \mathbf v\cdot\mathbf w =x_1x_2+y_1y_2+z_1z_2. }$$

The result is a real number, not another vector. The dot product satisfies

$$\mathbf v\cdot\mathbf v=\|\mathbf v\|^2,$$ $$\mathbf u\cdot(\mathbf v+\mathbf w) =\mathbf u\cdot\mathbf v+\mathbf u\cdot\mathbf w,$$ $$\mathbf v\cdot\mathbf w=\mathbf w\cdot\mathbf v,$$

and

$$(\alpha\mathbf v)\cdot\mathbf w =\alpha(\mathbf v\cdot\mathbf w) =\mathbf v\cdot(\alpha\mathbf w).$$

Angles between vectors

Let $\theta$ be the angle between two nonzero vectors $\mathbf v$ and $\mathbf w$. Their dot product also has the geometric form

$$\boxed{ \mathbf v\cdot\mathbf w =\|\mathbf v\|\,\|\mathbf w\|\cos\theta. }$$

Therefore,

$$\boxed{ \cos\theta =\frac{\mathbf v\cdot\mathbf w} {\|\mathbf v\|\,\|\mathbf w\|}. }$$

This formula connects coordinates to geometry: the component calculation on the left tells us the angle between the directions.

Why the angle formula works

The vectors $\mathbf v$, $\mathbf w$, and $\mathbf w-\mathbf v$ form a triangle. The law of cosines gives

$$\|\mathbf w-\mathbf v\|^2 =\|\mathbf v\|^2+\|\mathbf w\|^2 -2\|\mathbf v\|\,\|\mathbf w\|\cos\theta.$$

The dot product gives another expression for the same squared length:

$$\begin{aligned} \|\mathbf w-\mathbf v\|^2 &=(\mathbf w-\mathbf v)\cdot(\mathbf w-\mathbf v)\\ &=\|\mathbf w\|^2-2\mathbf v\cdot\mathbf w+\|\mathbf v\|^2. \end{aligned}$$

Equating these expressions and canceling the common squared norms leaves

$$\mathbf v\cdot\mathbf w =\|\mathbf v\|\,\|\mathbf w\|\cos\theta.$$

Orthogonal vectors

Two vectors are orthogonal when they meet at a right angle. For nonzero vectors, $\theta=\pi/2$, so

$$\cos\left(\frac\pi2\right)=0.$$

Thus,

$$\boxed{ \mathbf v\cdot\mathbf w=0 \quad\Longleftrightarrow\quad \mathbf v\text{ and }\mathbf w\text{ are orthogonal}. }$$

For example,

$$\begin{aligned} \langle2,2,-1\rangle\cdot\langle5,-4,2\rangle &=2(5)+2(-4)+(-1)(2)\\ &=10-8-2\\ &=0. \end{aligned}$$

Therefore, the two vectors are orthogonal.

Scalar and vector projection

A projection answers this question: how much of $\mathbf v$ points in the direction of $\mathbf w$? Assume $\mathbf w\ne\mathbf0$.

The scalar projection, or component of $\mathbf v$ along $\mathbf w$, is

$$\boxed{ \operatorname{comp}_{\mathbf w}\mathbf v =\frac{\mathbf v\cdot\mathbf w}{\|\mathbf w\|} =\|\mathbf v\|\cos\theta. }$$

It is a signed length. A positive value means the component points in the same general direction as $\mathbf w$; a negative value means it points in the opposite direction.

To turn that signed length into a vector, multiply by the unit vector in the $\mathbf w$ direction:

$$\begin{aligned} \operatorname{proj}_{\mathbf w}\mathbf v &=\operatorname{comp}_{\mathbf w}\mathbf v \left(\frac{\mathbf w}{\|\mathbf w\|}\right)\\ &=\left(\frac{\mathbf v\cdot\mathbf w}{\|\mathbf w\|}\right) \left(\frac{\mathbf w}{\|\mathbf w\|}\right). \end{aligned}$$

Therefore,

$$\boxed{ \operatorname{proj}_{\mathbf w}\mathbf v =\frac{\mathbf v\cdot\mathbf w}{\|\mathbf w\|^2}\mathbf w. }$$

If $\mathbf w$ is already a unit vector, then $\|\mathbf w\|=1$ and the formula simplifies to

$$\operatorname{proj}_{\mathbf w}\mathbf v =(\mathbf v\cdot\mathbf w)\mathbf w.$$

A projection keeps only the part in one direction

Example: projecting one vector onto another

Let

$$\mathbf v=\langle1,1,2\rangle, \qquad \mathbf w=\langle-2,3,1\rangle.$$

First compute the dot product:

$$\mathbf v\cdot\mathbf w =-2+3+2=3.$$

Then compute the squared norm of $\mathbf w$:

$$\|\mathbf w\|^2 =(-2)^2+3^2+1^2 =14.$$

Substitute into the vector-projection formula:

$$\begin{aligned} \operatorname{proj}_{\mathbf w}\mathbf v &=\frac{3}{14}\mathbf w\\ &=\frac{3}{14}\langle-2,3,1\rangle\\ &=\boxed{\left\langle-\frac37,\frac9{14},\frac3{14}\right\rangle}. \end{aligned}$$

A useful check is that the answer must be a scalar multiple of $\mathbf w$. If it is not parallel to $\mathbf w$, it cannot be the projection onto $\mathbf w$.

Teaching walkthrough

The vector as a displacement in space

Imagine moving from one location to another across a room. That displacement has two fundamental properties: how far you travel and in what direction you go. A three-dimensional vector captures both:

The numbers $x$, $y$, and $z$ are the vector's components, recording how far the vector reaches along each coordinate axis.

To find the length, or norm, of this vector, we calculate:

When $\mathbf v\ne\mathbf 0$ and we care only about direction rather than magnitude, we can rescale $\mathbf v$ to have length $1$ by dividing by its norm:

This $\mathbf u$ is a unit vector. Keeping track of pure direction this way will prove essential when we project vectors later on.

The dot product: connecting algebra to geometry

When two vectors point in different directions, we often need to measure how they relate. Given $\mathbf v = \langle x_1, y_1, z_1 \rangle$ and $\mathbf w = \langle x_2, y_2, z_2 \rangle$, their dot product is computed by multiplying matching components and adding the results:

A common point of confusion is expecting another vector as the answer. The dot product always produces a single real number (a scalar).

The reason this simple calculation is so useful is its geometric identity:

where $\theta$ is the angle between the nonzero vectors $\mathbf v$ and $\mathbf w$. By rearranging this relationship, we can determine their angle from their coordinates:

Notice what happens when two nonzero vectors are perpendicular ($\theta = \pi/2$). Because $\cos(\pi/2) = 0$, their dot product must be zero:

This provides a direct algebraic test for right angles without having to measure any angles visually.

Measuring along a direction: scalar and vector projection

Now suppose $\mathbf w\ne\mathbf 0$. We can answer a practical question: how much of a vector $\mathbf v$ points in the direction of $\mathbf w$?

Imagine dropping a perpendicular line from the tip of $\mathbf v$ straight onto the line containing $\mathbf w$. This casts a "shadow" along $\mathbf w$. We can describe this shadow in two ways:

1. Scalar projection ($\operatorname{comp}_{\mathbf w}\mathbf v$): This is the signed length of the shadow: $\operatorname{comp}_{\mathbf w}\mathbf v = \frac{\mathbf v \cdot \mathbf w}{\|\mathbf w\|}$ If $\mathbf v\ne\mathbf0$ as well and $\theta$ is the angle between the vectors, this also equals $\|\mathbf v\|\cos\theta$. It is positive if the shadow points in the same general direction as $\mathbf w$, and negative if it points in the opposite direction.

2. Vector projection ($\operatorname{proj}_{\mathbf w}\mathbf v$): This is the shadow turned into an actual vector. We take the signed length from the scalar projection and multiply it by the unit vector in the direction of $\mathbf w$: $\operatorname{proj}_{\mathbf w}\mathbf v = \left(\frac{\mathbf v \cdot \mathbf w}{\|\mathbf w\|}\right)\left(\frac{\mathbf w}{\|\mathbf w\|}\right) = \frac{\mathbf v \cdot \mathbf w}{\|\mathbf w\|^2}\mathbf w$

A reliable check on your work is that $\operatorname{proj}_{\mathbf w}\mathbf v$ must always end up as a scalar multiple of $\mathbf w$, because its entire purpose is to represent the part of $\mathbf v$ that lies directly along $\mathbf w$.