The Cross Product of Vectors in Three-Dimensional Space

# The Cross Product of Vectors in Three-Dimensional Space

Recall that for any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ that the dot product $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n$ produces a scalar. We will now look at another type of vector product known as the Cross Product. We emphasize that the cross product between two vectors is only defined in $\mathbb{R}^3$.

 Definition: If $\vec{u}, \vec{v} \in \mathbb{R}^3$ then the Cross Product of $\vec{u}$ and $\vec{v}$ denoted $\vec{u} \times \vec{v}$ is a unique vector that is perpendicular to both $\vec{u}$ and $\vec{v}$ and forms a righthand system with $\vec{u}$, $\vec{v}$, and $\vec{u} \times \vec{v}$ respectively. The cross product $\vec{u} \times \vec{v}$ can be calculated as $\vec{u} \times \vec{v} = (u_2v_3 - u_3v_2)\vec{i} + (u_3v_1 - u_1v_3)\vec{j} + (u_1v_2 - u_2v_1) \vec{k}$. Alternatively $\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3 \end{vmatrix}$ when evaluated by cofactor expansion along the first row.

Using the latter definition of the cross product of two vectors in terms of a 3 x 3 Determinant:

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\begin{align} \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3 \end{vmatrix} = \begin{vmatrix} u_2 & u_3\\ v_2 & v_3 \end{vmatrix} \vec{i} - \begin{vmatrix} u_1 & u_3\\ v_1 & v_3 \end{vmatrix} \vec{j} + \begin{vmatrix} u_1 & u_2\\ v_1 & v_2 \end{vmatrix} \vec{k} \end{align}

Thus, this "determinant" form of the cross product can be rather useful instead of memorizing the longer formula for the cross product. We will now look at some properties of the cross product and prove a few of them. The rest will be left for the reader to prove.

 Theorem 1: If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ and $k$ is a scalar then the following properties about the cross products hold: a) $\vec{u} \times \vec{u} = \vec{0}$. b) $\vec{u} \cdot (\vec{u} \times \vec{v}) = 0$ and $\vec{v} \times (\vec{u} \times \vec{v}) = \vec{0}$. c) $\vec{u} \times \vec{v} = - \vec{v} \times \vec{u}$. (The Anticommutative Property) d) $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$. e) $\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}$ and $(\vec{u} + \vec{v}) \times \vec{w} = \vec{u} \times \vec{w} + \vec{v} \times \vec{w}$. (The Distributive Properties). f) $\vec{u} \cdot ( \vec{v} \times \vec{w} ) = (\vec{u} \times \vec{v}) \cdot \vec{w}$. g) $(k \vec{u}) \times \vec{v} = \vec{u} \times (k \vec{v}) = k (\vec{u} \times \vec{v})$.
• Proof of a)
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\begin{align} \vec{u} \times \vec{u} = \begin{vmatrix} u_2 & u_3\\ u_2 & u_3\end{vmatrix} \vec{i} - \begin{vmatrix} u_1 & u_3\\ u_1 & u_3\end{vmatrix} \vec{j} + \begin{vmatrix} u_1 & u_2\\ u_1 & u_2\end{vmatrix} \vec{k} \\ \vec{u} \times \vec{u}= (u_2u_3 - u_2u_3)\vec{i} - (u_1u_3 - u_1u_3)\vec{j} + (u_1u_2 - u_1u_2)\vec{k} \\ \vec{u} \times \vec{u}= 0\vec{i} + 0\vec{j} + 0\vec{k} \\ \vec{u} \times \vec{u}= \vec{0} \end{align}
• Proof of c) Consider the following geometric argument:
• Notice that $\vec{u} \times \vec{v}$ is the vector perpendicular to $\vec{u}$ and $\vec{v}$ that forms a right-hand system with $\vec{u}$, $\vec{v}$, and $\vec{u} \times \vec{v}$ respectively. Now notice that $\vec{v} \times \vec{u}$ is the vector perpendicular to $\vec{u}$ and $\vec{v}$ still, however instead $\vec{v} \times \vec{u}$ forms a right-hand system with $\vec{v}$, $\vec{u}$, and $\vec{v} \times \vec{u}$ respectively.
• Therefore $\vec{u} \times \vec{v} \| \vec{v} \times \vec{u}$ and these vectors are pointing in opposite directions. The only thing to show further is that their magnitudes are the same, which we will leave to the reader, and so $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$. $\blacksquare$
 Theorem 2: If $\vec{u}, \vec{v} \in \mathbb{R}^3$ and $0 ≤ \theta ≤ \pi$ is the angle between $\vec{u}$ and $\vec{v}$ then $\| \vec{u} \times \vec{v} \| = \| \vec{u} \| \| \vec{v} \| \sin \theta$.
• Proof:
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