The Scalar Triple Product of Vectors in Three-Dimensional Space

The Scalar Triple Product of Vectors in Three-Dimensional Space

 Definition: If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ then the Scalar Triple Product of $\vec{u}, \vec{v}, \vec{w}$ (in that order) is equal to the following determinant, $\vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix} = u_1 \begin{vmatrix} v_2 & v_3\\ w_2 & w_3 \end{vmatrix} - u_2 \begin{vmatrix} v_1 & v_3\\ w_1 & w_3 \end{vmatrix} + u_3 \begin{vmatrix} v_1 & v_2\\ w_1 & w_2 \end{vmatrix}$.

Geometrically, the absolute value of the scalar triple product, $\biggr \rvert \vec{u} \cdot ( \vec{v} \times \vec{w} ) \biggr \rvert$ is equal to the volume of the parallelepiped formed from the vectors $\vec{u}, \vec{v}, \vec{w}$ as depicted in the following image. Furthermore, it is important to note that $\vec{u} \cdot (\vec{v} \times \vec{w}) = \vec{v} \cdot (\vec{w} \times \vec{u}) = \vec{w} \cdot (\vec{u} \times \vec{v})$, which can be proven by noting the following determinants result in a two row interchanges which is equivalent to multiplying the determinant $\vec{u} \cdot (\vec{v} \times \vec{w})$ by $(-1)(-1) = 1$.

We're about to look more into lines, planes, and surfaces in general in $\mathbb{R}^3$. The following definition will be important coming up.

 Definition: Three vectors $\vec{u}, \vec{v}, \vec{w}$ are said to be Coplanar if all three vectors lie on the same plane.

For example, the vectors $(0, 1, 2), (0, 2, 3), (0, -1, 3) \in \mathbb{R}^3$ are coplanar as they all lie on the $yz$-plane. The following theorem gives us a nice way of checking if three vectors in $\mathbb{R}^3$ are coplanar.

 Theorem 1: If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ then the set of vectors $\{ \vec{u}, \vec{v}, \vec{w} \}$ are coplanar if and only if the scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w}) = 0$.
• Proof: $\Rightarrow$ Suppose that $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ are coplanar. Then these vectors lie on a plane and thus one of the dimensions of the corresponding parallelepiped formed by these vectors has zero length, and so $\vec{u} \cdot (\vec{v} \times \vec{w}) = 0$.
• $\Leftarrow$ Suppose that $\vec{u} \cdot (\vec{v} \times \vec{w}) = 0$. Then the volume spanned by $\vec{u}, \vec{v}, \vec{w}$ is zero which implies that these vectors must line on the some similar plane, and thus they are coplanar. $\blacksquare$

Example 1

Let $\vec{u} = (1, 0, 1)$, $\vec{v} = (1, 1, 1)$, $\vec{w} = (2, 0, -1)$. Compute the scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$.

Applying the formula from above we get that:

(1)
\begin{align} \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} 1 & 1\\ 0 & -1 \end{vmatrix} + \begin{vmatrix} 1 & 1\\ 2 & 0 \end{vmatrix} = -1 -2 = -3 \end{align}