Unit Vectors

The concept of a unit vector is very simple, but very important and foundational in the field of Linear Algebra.

A unit vector is simply a vector whose length equals $1$ .

The use of a vector with length $1$ is that you don’t care about the length - you only care about the direction it’s pointing to. The unit vector isolates the direction from its magnitude (length);

This could be useful because maybe you want to scale a vector to have some specific length. To do this you could first find the unit vector which points in the same direction as your given vector. Then you could multiply the unit vector with a scalar (scalar multiplication) corresponding to your desired length.

To find the unit vector $\vec{u}$ of a vector $\vec{v}$ , you just divide $\vec{v}$ by its length.

That is

\vec{u} = \dfrac{\vec{v}}{\lVert \vec{v} \rVert}.

If $\vec{u}$ is found as shown above, then

\lVert \vec{u} \rVert = 1.

NOTE: You cannot find the unit vector which points in the same direction as a zero vector. For one, it would not make any sense, since the zero vector has no direction. But most importantly, the length of a zero vector is $0$ , and the formula above would require you to divide by $0$ in that case, which is undefined.

NOTE: Division by a scalar is simply multiplying by the multiplicative inverse. What that means is that dividing by some scalar $s$ is the same as multiplying by $\frac{1}{s}$ . It is basically scalar multiplication.

Example

For example, let us find the unit vector which points in the same direction as the vector

\vec{v} = \begin{bmatrix} 3 \\ 3 \\ 3 \\ 3 \end{bmatrix}.

We divide by the length to find the unit vector $\vec{u}$ :

\begin{align*} \vec{u} &= \dfrac{\vec{v}}{\lVert \vec{v} \rVert} \\[1em] &= \dfrac{\vec{v}}{\sqrt{3^2 + 3^2 + 3^2 + 3^2}} \\[1em] &= \dfrac{\vec{v}}{\sqrt{36}} \\[1em] &= \dfrac{\vec{v}}{6} \\[1em] &= \begin{bmatrix} \frac{3}{6} \\[0.5em] \frac{3}{6} \\[0.5em] \frac{3}{6} \\[0.5em] \frac{3}{6} \\[0.5em] \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{2} \\[0.5em] \frac{1}{2} \\[0.5em] \frac{1}{2} \\[0.5em] \frac{1}{2} \\[0.5em] \end{bmatrix}. \end{align*}

If you now compute $\lVert \vec{u} \rVert$ , you will find that it will equal 1:

\begin{align*} \lVert \vec{u} \rVert &= \sqrt{ \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 } \\[1em] &= \sqrt{ \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} } \\[1em] &= \sqrt{1} = 1 \end{align*}