Cross Product and Applications: Learn It 4

The Triple Scalar Product

Since the cross product of two vectors produces a vector, we can take the dot product of this result with a third vector. This combination—the dot product of a vector with the cross product of two other vectors—is called the triple scalar product because the final result is a scalar.

triple scalar product

For vectors [latex]\mathbf{u}[/latex], [latex]\mathbf{v}[/latex], and [latex]\mathbf{w}[/latex]:

[latex]\text{Triple scalar product }= \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})[/latex]

Just as we used determinants to simplify cross product calculations, we can use them for triple scalar products.

 triple scalar product formula

For vectors [latex]\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k}[/latex], [latex]\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}[/latex], and [latex]\mathbf{w} = w_1\mathbf{i} + w_2\mathbf{j} + w_3\mathbf{k}[/latex]:

[latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix}u_1&u_2&u_3 \\ v_1&v_2&v_3 \\w_1&w_2&w_3 \end{vmatrix}[/latex]

Proof

The calculation is straightforward.

[latex]\large{\begin{aligned} {\bf{u}}\cdot({\bf{v}}\times{\bf{w}})&=\langle u_1,u_2,u_3\rangle\cdot\langle v_2w_3-v_3w_2,-v_1w_3+v_3w_1,v_1w_2-v_2w_1\rangle \\ &=u_1(v_2w_3-v_3w_2)+u_2(-v_1w_3+v_3w_1)+u_3(v_1w_2-v_2w_1) \\ &=u_1(v_2w_3-v_3w_2)-u_2(v_1w_3-v_3w_1)+u_3(v_1w_2-v_2w_1) \\ &=\begin{vmatrix}u_1&u_2&u_3 \\ v_1&v_2&v_3 \\w_1&w_2&w_3 \end{vmatrix}. \end{aligned}}[/latex]
[latex]_\blacksquare[/latex]
 Instead of computing [latex]\mathbf{v} \times \mathbf{w}[/latex] first and then taking the dot product with [latex]\mathbf{u}[/latex], you can simply evaluate a [latex]3 \times 3[/latex] determinant.
Let [latex]{\bf{u}}=\langle1,3,5\rangle[/latex], [latex]{\bf{v}}=\langle2,-1,0\rangle[/latex], and [latex]{\bf{w}}=\langle-3,0,-1\rangle[/latex]. Calculate the triple scalar product [latex]{\bf{u}}\cdot({\bf{v}}\times{\bf{w}})[/latex].

Order Matters

When you rearrange vectors in the triple scalar product, you’re essentially reordering rows in the determinant matrix. Each time you swap two rows, the determinant changes sign.

Let’s take a look at this with an example:

[latex]\begin{aligned} \begin{vmatrix}1&2&1 \\ -2&0&3 \\4&1&-1\end{vmatrix} &=\begin{vmatrix}0&3 \\1&-1\end{vmatrix}-2\begin{vmatrix}-2&3 \\4&-1\end{vmatrix}+\begin{vmatrix}-2&0 \\4&1\end{vmatrix} \\ &=(0-3)-2(2-12)+(-2-0)=-3+20-2=15. \end{aligned}[/latex]

Switching the top two rows we have

[latex]\begin{aligned} \begin{vmatrix}-2&0&3 \\1&2&1\\ 4&1&-1\end{vmatrix} &=-2\begin{vmatrix}2&1 \\1&-1\end{vmatrix}+3\begin{vmatrix}1&2 \\4&1\end{vmatrix} \\ &=-2(-2-1)+3(1-8)=6-21=-15. \end{aligned}[/latex]

Notice the sign change!

This leads to important identities for triple scalar products.

triple scalar product identities

  • [latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = -\mathbf{u} \cdot (\mathbf{w} \times \mathbf{v})[/latex] (switching cross product order)
  • [latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) = \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})[/latex] (cyclic permutations)

Volume of Parallelepipeds

The triple scalar product has a powerful geometric interpretation. When three non-coplanar vectors [latex]\mathbf{u}[/latex], [latex]\mathbf{v}[/latex], and [latex]\mathbf{w}[/latex] share the same initial point, they form the edges of a parallelepiped—a 3D prism where all faces are parallelograms.

theorem: volume of a parallelepiped

The volume of a parallelepiped with adjacent edges given by vectors [latex]\mathbf{u}[/latex], [latex]\mathbf{v}[/latex], and [latex]\mathbf{w}[/latex] is:

[latex]V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})|[/latex]

We use absolute value because volume is always positive.

This figure is a parallelepided, a three dimensional parallelogram. Three of the sides are represented with vectors. The base has vectors v and w. The vertical side has vector u. All three vectors have the same initial point. A perpendicular vector is drawn from this common point. It is labeled “proj sub (v x w) u.”
Figure 3. The height of the parallelepiped is given by [latex]||\text{proj}_{{\bf{v}}\times{\bf{w}}}{\bf{u}}||[/latex].

 

Proof

The area of the base of the parallelepiped is given by [latex]||{\bf{v}}\times{\bf{w}}||[/latex]. The height of the figure is given by [latex]||\text{proj}_{{\bf{v}}\times{\bf{w}}}{\bf{u}}||[/latex]. The volume of the parallelepiped is the product of the height and the area of the base, so we have

[latex]\begin{array}{rcl} V & = & \left\|\operatorname{proj}_{\mathbf{v}\times\mathbf{w}}\mathbf{u}\right\|\left\|\mathbf{v}\times\mathbf{w}\right\| \\[6pt] & = & \left|\dfrac{\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})}{\left\|\mathbf{v}\times\mathbf{w}\right\|}\right|\left\|\mathbf{v}\times\mathbf{w}\right\| \\[6pt] & = & \left|\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})\right|. \end{array}[/latex]

[latex]_\blacksquare[/latex]

Let [latex]{\bf{u}}=\langle-1,-2,1\rangle[/latex], [latex]{\bf{v}}=\langle4,3,2\rangle[/latex], and [latex]{\bf{w}}=\langle0,-5,-2\rangle[/latex]. Find the volume of the parallelepiped with adjacent edges [latex]\textbf u[/latex], [latex]\textbf v[/latex], and [latex]\textbf w[/latex] (Figure 4)."This