Vector Cross Product

# Cross Product

The cross product is another form of vector multiplication. Unlike the dot product, the cross product results in a vector instead of a scalar. Furthermore, the cross product is defined only in $\mathbb{R}^3$.

 Definition: Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, the Cross Product denoted $\vec{u} \times \vec{v}$ results in a new vector that is perpendicular to both $\vec{u}$ and $\vec{v}$. The formula to calculate the cross product is $\vec{u} \times \vec{v} = (u_{2}v_{3} - u_{3}v_{2}, -u_{1}v_{3} + u_{3}v_{1}, u_{1}v_{2} - u_{2}v_{1})$.

For example, consider the vectors $\vec{u} = (1, 2, 3)$ and $\vec{v} = (2, 3, 4)$. To calculate the cross product $\vec{u} \times \vec{v}$, all we need to do is apply the formula to get $\vec{u} \times \vec{v} = (-1, 2, -1)$.

Of course, this formula is rather difficult to remember, so we will use somewhat of a cheat as a memory device. First set up a $2 \times 3$ matrix $A$ where the first row represents components from $\vec{u}$ and the second row represents components from $\vec{v}$, that is:

(1)
\begin{align} A = \begin{bmatrix} u_{1} & u_{2} & u_{3}\\ v_{1 }& v_{2} & v_{3} \end{bmatrix} \end{align}
• The first component of the cross product will be the determinant of the $2 \times 2$ matrix that results from deleting the first column of $A$, that is $\begin{vmatrix} u_2 & u_3\\ v_2 & v_3\end{vmatrix} = u_2v_3 - u_3v_2$.
• The second component of the cross product will be the negative of the determinant of the $2 \times 2$ matrix that results from deleting the second column of $A$, that is $-\begin{vmatrix} u_1 & u_3\\ v_1 & v_3\end{vmatrix} = -u_1v_3 + u_3v_1$.
• The third component of the cross product will be the determinant of the $2 \times 2$ matrix that results from deleting the third column of $A$, that is $\begin{vmatrix} u_1 & u_2\\ v_1 & v_2\end{vmatrix} = u_1v_2 - u_2v_1$.

Thus we derive the cross product of two vectors to be $\vec{u} \times \vec{v} = \left ( \begin{vmatrix} u_2 & u_3\\ v_2 & v_3\end{vmatrix}, -\begin{vmatrix} u_1 & u_3\\ v_1 & v_3\end{vmatrix} ,\begin{vmatrix} u_1 & u_2\\ v_1 & = v_2\end{vmatrix} \right )= (u_{2}v_{3} - u_{3}v_{2}, -u_{1}v_{3} + u_{3}v_{1}, u_{1}v_{2} - u_{2}v_{1})$.

## Example 1

Find the cross product of $\vec{u} = (2, 3)$ and $\vec{v} = (3, 4)$.

Since the cross product is only defined in 3-space, there is no solution since $\vec{u}, \vec{v} \in \mathbb{R}^2$.

## Example 2

Find the cross product given the vectors $\vec{u} = (2, 3, 1)$ and $\vec{v} = (3, -2, 1)$.

All we need to do is apply out formula as follows:

(2)
\begin{align} \vec{u} \times \vec{v} = ( - [-2], - + , [-2] - ) \\ \vec{u} \times \vec{v} = (4, 1, -13) \end{align}

# Properties of the Cross Product

 Theorem 1: Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, the dot products $\vec{u} \cdot (\vec{u} \times \vec{v}) = 0$ and $\vec{v} \cdot (\vec{u} \times \vec{v}) = 0$.
• Proof: Let $\vec{u}, \vec{v} \in \mathbb{R}^3$ and expand the formulas for the dot and cross product to get that:
(3)
\begin{align} \quad \vec{u} \cdot (\vec{u} \times \vec{v}) = (u_{1}, u_{2}, u_{3}) \cdot (u_{2}v_{3} - u_{3}v_{2}, -u_{1}v_{3} + u_{3}v_{1}, u_{1}v_{2} - u_{2}v_{1}) \\ \quad \vec{u} \cdot (\vec{u} \times \vec{v}) = u_{1}[u_{2}v_{3} - u_{3}v_{2}] - u_{2}[u_{1}v_{3} + u_{3}v_{1}] + u_{3}[u_{1}v_{2} - u_{2}v_{1}] \\ \quad \vec{u} \cdot (\vec{u} \times \vec{v}) = u_{1}u_{2}v_{3} - u_{1}u_{3}v_{2} - u_{2}u_{1}v_{3} - u_{2}u_{3}v_{1} + u_{3}u_{1}v_{2} - u_{3}u_{2}v_{1} \\ \quad \vec{u} \cdot (\vec{u} \times \vec{v}) = u_{1}u_{2}v_{3} - u_{1}u_{3}v_{2} - u_{1}u_{2}v_{3} - u_{2}u_{3}v_{1} + u_{1}u_{3}v_{2} - u_{2}u_{3}v_{1} \\ \quad \vec{u} \cdot (\vec{u} \times \vec{v}) = 0 \\ \blacksquare \end{align}

Note that the same proof can be applied for showing that $\vec{v} \cdot (\vec{u} \times \vec{v}) = 0$.

 Theorem 2: For three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, the cross product $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$.
• Proof: Consider the following $2 \times 3$ matrix: $\begin{bmatrix}u_{1} & u_{2} & u_{3} \\ u_{2}v_{3} - u_{3}v_{2}& -v_{1}w_{3} + v_{3}w_{1} & v_{1}w_{2} - v_{2}w_{1} \end{bmatrix}$. If we take the cross product with this matrix we obtain that:
(4)
\begin{align} \quad \vec{u} \times (\vec{v} \times \vec{w}) = (u_{2}[v_{1}w_{2} - v_{2}w_{1}] + u_{3}[v_{1}w_{3} + v_{3}w_{1}], u_{3}[u_{2}v_{3} - u_{3}v_{2}] - u_{1}[v_{1}w_{2} - v_{2}w_{1}] , -u_{1}[v_{1}w_{3} + v_{3}w_{1}] - u_{2}[u_{2}v_{3} - u_{3}v_{2}]) \\ \quad \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w} \\ \blacksquare \end{align}

Note that we did not expand this out in its fully entirety, however, feel free to verify.

One important aspect of the cross product that we should touch upon is that the associative property does NOT hold. If we have three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, often times $\vec{u} \times (\vec{v} \times \vec{w}) ≠ (\vec{u} \times \vec{v}) \times \vec{w}$.

We can easily see this with unit vectors. We note that:

(5)
\begin{align} \vec{i} \times (\vec{j} \times \vec{j}) = \vec{i} + \vec{0} = \vec{0} \end{align}

Furthermore, if we rearrange the vectors:

(6)
\begin{align} (\vec{i} \times \vec{j}) \times \vec{j} = \vec{k} \times \vec{j} = -\vec{i} \end{align}

Therefore there is clearly no associativity.

Similarly, the commutative property for the cross product of vectors does not always hold, that is $\vec{u} \times \vec{v} \neq \vec{v} \times \vec{u}$. This can easily be proven with the determinant method.

 Theorem 3 (Anticommutativity of the Cross Product): Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, the following cross products are equal $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$.

We will not prove Theorem 3, but it can be done using determinants.

 Theorem 4 (Distributivity of the Cross Product): Given three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, the distributive property holds such that $\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}$ and $(\vec{u} + \vec{v}) \times \vec{w} = \vec{u} \times \vec{w} + \vec{v} \times \vec{w}$.

Once again we will omit the proof of this theorem.

 Theorem 5: Given two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$, and some scalar $k$, it follows that $k(\vec{u} \times \vec{v}) = (k\vec{u}) \times \vec{v} = \vec{u} \times (k\vec{v})$.

Like the last two theorems, we will omit the proof of this theorem as it is rather straight-forward using determinants.

 Theorem 6: Given the two vectors $\vec{u}, \vec{0} \in \mathbb{R}^3$, it follows that $\vec{u} \times \vec{0} = \vec{0} \times \vec{u} = \vec{0}$.
• Proof: The cross product creates a vector that is perpendicular to both the vectors cross product multiplied together. However, the zero vector has no length or direction. Hence, there is no vector that is perpendicular to both some vector $\vec{u}$ and the zero vector $\vec{0}$. For convention, we say the result is the zero vector, as it can be assigned any direction because it has no magnitude. $\blacksquare$

Like the other proofs, theorem 6 can also be shown using determinants.

 Theorem 7: Given a vector $\vec{u} \in \mathbb{R}^3$, it follows that $\vec{u} \times \vec{u} = \vec{0}$.
• Proof: This proof is also intuitive. The zero vector can be assigned any direction despite having no magnitude. Hence, it is the only vector that is perpendicular to itself from the cross product. $\blacksquare$