Standard Unit Vectors

# Standard Unit Vectors

 Definition: The vectors $\vec{i} = (1, 0, 0)$, $\vec{j} = (0, 1, 0)$, and $\vec{k} = (0, 0, 1)$ are said to be Standard Unit Vectors, that is a vector that runs on either the $x$, $y$ or $z$-axis and has a magnitude of 1, that is $\mid \mid \vec{i} \mid \mid = \mid \mid \vec{j} \mid \mid= \mid \mid \vec{k} \mid \mid = 1$.

The following image depicts the three standard unit vectors on an $xyz$-plane:

Standard unit vectors are helpful as we can in fact represent any vector in terms of them. Consider a general vector $\vec{u} = (u_1, u_2, u_3)$. We can represent the vector $\vec{u}$ in the following way:

(1)
\begin{align} \vec{u} = u_1\vec{i} + u_2\vec{j} + u_3\vec{k} \end{align}

# Expressing Vectors with Standard Unit Vectors

Let's let some vector u exist in 3-Space. We know that u = (u1, u2, u3). We can thus express vector u in terms of standard unit vectors by the dot product property. For example:

(2)
\begin{align} \quad \vec{u} = (u_{1}, u_{2}, u_{3}) = u_{1}\vec{i} + u_{2}\vec{j} + u_{3}\vec{k} = u_{1} \cdot (1, 0, 0) + u_{2} \cdot (0, 1, 0) + u_{3} \cdot (0, 0, 1) \end{align}

## Example 1

Express the vector $\vec{u} = (2, -4, 5)$ in terms of standard unit vectors.

By simply putting what we know into the formula, we can write $\vec{u}$ rather easily in terms of standard unit vectors, that is $\vec{u} = 2\vec{i} -4\vec{j} + 5\vec{k}$.

# Representing The Cross Product in Standard Unit Vectors

One way to represent the cross product of two vectors $\vec{u} \times \vec{v}$ is with determinants and unit vectors. Firstly, create a $3 \times 3$ matrix such that the entries of the first row are the unit vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$. Make the entries of the second row be the components of $\vec{u}$, and make the entries of the third row be the components of $\vec{v}$ as shown: $\begin{bmatrix}\vec{i} & \vec{j} & \vec{k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{bmatrix}$.

The determinant of this matrix is equal to the cross product $\vec{u} \times \vec{v}$. This can easily be seen by expanding across the first row in terms of minors/cofactors:

(3)
\begin{align} \vec{u} \times \vec{v} = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{vmatrix} = \vec{i} \begin{vmatrix} u_{2} & u_{3} \\ v_{2} & v_{2} \end{vmatrix} - \vec{j} \begin{vmatrix} u_{1} & u_{3} \\ v_{1} & v_{3} \end{vmatrix} + \vec{k} \begin{vmatrix} u_{1} & u_{2} \\ v_{1} & v_{2} \end{vmatrix} \end{align}