Scalar Triple Products

If we have three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ in $\mathbb{R}^3$, then the scalar quantity $\vec{u} \cdot (\vec{v} \times \vec{w})$ appears frequently in other areas of mathematics such as Calculus and has a special name which we define below.

Definition: For any three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, the Scalar Triple Product is denoted as the dot product between $\vec{u}$ and $\vec{v} \times \vec{w}$, or rather $\vec{u} \cdot (\vec{v} \times \vec{w})$.

The following theorem will give us a method for computing the scalar triple product between three vectors in $\mathbb{R}^3$.

Theorem 1: If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, then the scalar triple product can be computed as $\vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_1 & u_2 & u_3\\ v_1 & v_2& v_3\\ w_1 & w_2 & w_3 \end{vmatrix}$.

• Proof: We note that the cross product of two vectors $\vec{v} \times \vec{w} = \begin{bmatrix} v_2 & v_3\\ w_2 & w_3 \end{bmatrix} \vec{i} - \begin{bmatrix} v_1 & v_3\\ w_1 & w_3 \end{bmatrix} \vec{j} + \begin{bmatrix} v_1 & v_2\\ w_1 & w_2 \end{bmatrix} \vec{k}$. Multiplying this by $\vec{u}$ we obtain that:

(1)

\begin{align} \vec{u} \cdot (\vec{v} \times \vec{w}) = (u_1, u_2, u_3) \cdot \left ( \begin{bmatrix} v_2 & v_3\\ w_2 & w_3 \end{bmatrix} \vec{i} - \begin{bmatrix} v_1 & v_3\\ w_1 & w_3 \end{bmatrix} \vec{j} + \begin{bmatrix} v_1 & v_2\\ w_1 & w_2 \end{bmatrix} \vec{k} \right ) \\ \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{bmatrix} v_2 & v_3\\ w_2 & w_3 \end{bmatrix} u_1 - \begin{bmatrix} v_1 & v_3\\ w_1 & w_3 \end{bmatrix} u_2 + \begin{bmatrix} v_1 & v_2\\ w_1 & w_2 \end{bmatrix} u_3 \\ \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_1 & u_2 & u_3\\ v_1 & v_2& v_3\\ w_1 & w_2 & w_3 \end{vmatrix} \\ \blacksquare \end{align}

Note: We note that when we compute the scalar triple product between $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ that the result will be a scalar and NOT a vector!

As another remark, we should note that the order of $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ matters when it comes to computing certain scalar triple products. That is, $\vec{u} \cdot (\vec{v} \times \vec{w})$ does not necessarily equal $(\vec{u} \times \vec{v}) \cdot \vec{w}$. However, for any vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, then the following scalar triple products are in fact equal:

(2)

\begin{align} \vec{u} \cdot (\vec{v} \times \vec{w}) = \vec{w} \cdot (\vec{u} \times \vec{v}) = \vec{v} \cdot (\vec{w} \times \vec{u}) \end{align}

An easy way to remember this equivalence is with the following diagram where clockwise movements of the dot and cross represent equal scalar triples.

Example 1

Find the scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ where $\vec{u} = (1, 2, 3)$, $\vec{v} = (0, 2, 0)$ and $\vec{w} = (1, 4, 1)$.

Applying our formula tells us that the scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ arises when we evaluate the following determinant $\begin{vmatrix}1 & 2 & 3\\ 0 & 2 & 0\\ 1 & 4 & 1\end{vmatrix}$. This is best done by cofactor expansion along row 2 where we get our answer to be $-4$.