# 6.04: Direction Cosines

## Unit Vectors

Corresponding to every vector $$x$$ is a unit vector $$u_x$$ pointing in the same direction as $$x$$. The term unit vector means that the norm of the vector is 1:

$||u_x||=1$

The question is, given $$x$$, how can we find $$u_x$$? The first part of the answer is that $$u_x$$ will have to be a positive scalar multiple of $$x$$ in order to point in the same direction as $$x$$, as shown in this Figure. Thus

$u_x=αx$

But what is $$α$$? We need to choose $$α$$ so that the norm of $$u_x$$ will be 1:

$||ux||=1$

$|αx||=1$

$|α|||x||=1$

$α=\frac {1} {||x||}$

We have dropped the absolute value bars on $$α$$ because $$||x||$$ is positive. The $$α$$ that does the job is 1 over the norm of $$x$$. Now we can write formulas for $$u_x$$ in terms of $$x$$ and $$x$$ in terms of $$u_x$$:

$u_x=\frac 1 {||x||}x$

$x=||x||u_x$

So the vector $$x$$ is its direction vector $$u_x$$, scaled by its Euclidean norm.

## Unit Coordinate Vectors

There is a special set of unit vectors called the unit coordinate vectors. The unit coordinate vector $$e_i$$ is a unit vector in $$∇^n$$ that points in the positive direction of the $$i^{\mathrm{th}}$$ coordinate axis. The Figure shows the three unit coordinate vectors in $$∇^3$$.

For three-dimensional space, $$R^3$$, the unit coordinate vectors are

$e_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},e_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, e_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

More generally, in n-dimensional space, there are $$n$$ unit coordinate vectors given by

You should satisfy yourself that this definition results in vectors with a norm of l.

Exercise $$\PageIndex{1}$$

Find the norm of the vector au where $$u$$ is a unit vector.

Exercise $$\PageIndex{2}$$

Find the unit vector $$u_x$$ in the direction of

1. $$x=\begin{bmatrix} 3 \\ 4 \end{bmatrix}$$
2. $$x=\begin{bmatrix} 3 \\ 6 \\ −2 \end{bmatrix}$$
3. $$x=\begin{bmatrix} 1 \\ −1 \\ 1 \\ −1 \end{bmatrix}$$

## Direction Cosines

We often say that vectors “have magnitude and direction.” This is more or less obvious from "Linear Algebra: Vectors", where the three-dimensional vector $$x$$ has length $$\sqrt{x^2_1+x^2_2+x^2_3}$$ and points in a direction defined by the components $$x_1$$, $$x_2$$, and $$x_3$$. It is perfectly obvious from Equation $$\PageIndex{8}$$ where $$x$$ is written as $$x=||x||\,u_x$$. But perhaps there is another representation for a vector that places the notion of magnitude and direction in even clearer evidence.

The Figure below shows an arbitrary vector $$x∈R^3$$ and the three-dimensional unit coordinate vectors $$e_1,e_2,e_3$$. The inner product between the vector $$x$$ and the unit vector $$e_k$$ just reads out the $$k^\mathrm{th}$$ component of $$x$$:

$(x,e_k)=(e_k,x)=x_k$

Since this is true even in $$R^n$$, any vector $$x∈R^n$$ has the following representation in terms of unit vectors:

$x=(x,e_1)e_1+(x,e_2)e_2+⋯+(x,e_n)e_n$

Let us now generalize our notion of an angle $$θ$$ between two vectors to $$R^n$$ as follows:

$\cosθ=\frac {(x,y)} {||x||||y||}$

The celebrated Cauchy-Schwarz inequality establishes that $$−1−≤ \cos θ≤1$$. With this definition of angle, we may define the angle $$θ_k$$ that a vector makes with the unit vector $$e_k$$ to be

$\cosθ_k=\frac {(x,e_k)} {||x||||ek||}$

But the norm of $$e_k$$ is 1, so

$\cosθ_k=\frac {(x,ek)} {||x||}=\frac {x_k} {||x||}$

When this result is substituted into the representation of x in Equation $$\PageIndex{11}$$, we obtain the formula

$x=||x||\cosθ_1e_1+||x||\cosθ_2e_2+⋯+||x||\cosθ_ne_n=||x||(\cosθ_1e_1+\cosθ_2e_2+⋯+\cosθ_ne_n)$

This formula really shows that the vector $$x$$ has “magnitude” $$||x||$$ and direction $$(θ_1,θ_2,...,θ_n)$$ and that the magnitude and direction are sufficient to determine $$x$$. We call ($$\cosθ_1, \cos θ_2,..., \cos θ_n$$) the direction cosines for the vector $$x$$. In the three-dimensional case, they are illustrated in the Figure above.

Exercise $$\PageIndex{3}$$

Show that Equation $$\PageIndex{12}$$ agrees with the usual definition of an angle in two dimensions.

Exercise $$\PageIndex{4}$$

Find the cosine of the angle between $$x$$ and $$y$$ where

1. $$x=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\;y=\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$$
2. $$x=\begin{bmatrix} 1 \\ −1 \\ 1 \\ −1 \end{bmatrix}\;y=\begin{bmatrix} −1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$$
3. $$x=\begin{bmatrix} 2 \\ 1 \\ −2 \end{bmatrix}\;y=\begin{bmatrix} 4 \\ 2 \\ −4 \end{bmatrix}$$

If we compare Equation $$\PageIndex{8}$$ and Equation $$\PageIndex{15}$$ we see that the direction vector $$u_x$$ is composed of direction cosines: