When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:

[latex]\begin{vmatrix}a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3\end{vmatrix}=d \quad\begin{vmatrix}b_1&b_2&b_3 \\a_1&a_2&a_3 \\ c_1&c_2&c_3\end{vmatrix}=-d \quad \begin{vmatrix}b_1&b_2&b_3 \\c_1&c_2&c_3 \\ a_1&a_2&a_3\end{vmatrix}=d \quad\begin{vmatrix}c_1&c_2&c_3 \\b_1&b_2&b_3 \\ a_1&a_2&a_3\end{vmatrix}=-d[/latex].

Verifying this fact is straightforward, but rather messy. 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]

Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let [latex]{\bf{u}}=u_1{\bf{i}}+u_2{\bf{j}}+u_3{\bf{k}}[/latex], [latex]{\bf{v}}=v_1{\bf{i}}+v_2{\bf{j}}+v_3{\bf{k}}[/latex], and [latex]{\bf{w}}=w_1{\bf{i}}+w_2{\bf{j}}+w_3{\bf{k}}[/latex]. Applying Theorem: Calculating a Triple Scalar Product, we have

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

We can obtain the determinant for calculating [latex]{\bf{u}}\cdot({\bf{w}}\times{\bf{v}})[/latex] by switching the bottom two rows of [latex]{\bf{u}}\cdot({\bf{v}}\times{\bf{w}})[/latex]. Therefore, [latex]{\bf{u}}\cdot({\bf{v}}\times{\bf{w}})=-{\bf{u}}\cdot({\bf{w}}\times{\bf{v}})[/latex].

Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:

[latex]\begin{aligned} {\bf{u}}\cdot({\bf{v}}\times{\bf{w}})&=-{\bf{u}}\cdot({\bf{w}}\times{\bf{v}}) \\ {\bf{u}}\cdot({\bf{v}}\times{\bf{w}})&={\bf{v}}\cdot({\bf{w}}\times{\bf{u}})={\bf{w}}\cdot({\bf{u}}\times{\bf{v}}). \end{aligned}[/latex]

Let [latex]\textbf u[/latex] and [latex]\textbf v[/latex] be two vectors in standard position. If [latex]\textbf u[/latex] and [latex]\textbf v[/latex] are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in Theorem: Area of a Parallelogram that the area of this parallelogram is [latex]||{\bf{u}}\times{\bf{v}}||[/latex]. Now suppose we add a third vector [latex]\textbf w[/latex] that does not lie in the same plane as [latex]\textbf u[/latex] and [latex]\textbf v[/latex] but still shares the same initial point. Then these vectors form three edges of a parallelepiped, a three-dimensional prism with six faces that are each parallelograms, as shown in Figure 3. The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of [latex]\textbf u[/latex], [latex]\textbf v[/latex], and [latex]\textbf w[/latex] provides a simple method for calculating the volume of the parallelepiped defined by these vectors.

Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.

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{aligned} v&=||\text{proj}_{{\bf{v}}\times{\bf{w}}}{\bf{u}}|| \ ||{\bf{v}}\times{\bf{w}}|| \\ &=\left|\frac{{\bf{u}}\cdot({\bf{v}}\times{\bf{w}})}{||{\bf{v}}\times{\bf{w}}||}\right| \||{\bf{v}}\times{\bf{w}}|| \\ &=|{\bf{u}}\cdot({\bf{v}}\times{\bf{w}})|. \end{aligned}[/latex]

[latex]_\blacksquare[/latex]

Example: calculating the volume of a parallelepiped

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).

Figure 4. A parallelepiped with adjacent edges [latex]\textbf u[/latex], [latex]\textbf v[/latex], and [latex]\textbf w[/latex]

Show Solution

We have

[latex]\begin{aligned} {\bf{u}}\cdot({\bf{v}}\times{\bf{w}})&=\begin{vmatrix}-1&-2&1 \\4&3&2 \\0&-5&-2\end{vmatrix}=(-1)\begin{vmatrix}3&2 \\-5&-2\end{vmatrix}+2\begin{vmatrix}4&2 \\ 0&-2\end{vmatrix}+\begin{vmatrix} 4&3 \\ 0&-5\end{vmatrix} \\ &=(-1)(-6+10)+2(-8-0)+(-20-0) \\ &=-4-16-20 \\ &=-40. \end{aligned}[/latex]

Thus, the volume of the parallelepiped is [latex]|-40|=40[/latex] units³.

Watch the following video to see the worked solution to the above Try IT.

Applications of the Cross Product

The cross product appears in many practical applications in mathematics, physics, and engineering. Let’s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields (Introduction to Vector Calculus).

We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.

Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.

DEFINITION

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Torque, [latex]\tau[/latex] (the Greek letter tau), measures the tendency of a force to produce rotation about an axis of rotation. Let [latex]\textbf r[/latex] be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector [latex]\textbf F[/latex] represent the force. Then torque is equal to the cross product of [latex]\textbf r[/latex] and [latex]\textbf F[/latex]:

[latex]\tau={\bf{r}}\times{\bf{F}}[/latex].

See Figure 5.

Figure 5. Torque measures how a force causes an object to rotate.

Think about using a wrench to tighten a bolt. The torque [latex]\tau[/latex] applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in the context of rotation, whereas work typically involves motion along a line.

Example: evaluating torque

A bolt is tightened by applying a force of [latex]6[/latex] N to a [latex]0.15[/latex]-m wrench (Figure 6). The angle between the wrench and the force vector is [latex]40^{\small\circ}[/latex]. Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.

Figure 6. Torque describes the twisting action of the wrench.

Show Solution

Substitute the given information into the equation defining torque:

[latex]||\tau||=||{\bf{r}}\times{\bf{F}}||=||{\bf{r}}|| \ ||{\bf{F}}||\sin\theta=(0.15\text{ m})(6\text{ N})\sin{40^{\small\circ}}\approx0.58\text{N}\cdot\text{m}[/latex].

try it

Calculate the force required to produce [latex]15\text{N}\cdot\text{m}[/latex] torque at an angle of [latex]30^{\small\circ}[/latex] from a [latex]150[/latex]-cm rod.

Show Solution

[latex]20[/latex] N.

Watch the following video to see the worked solution to the above Try IT.

You can view the transcript for “CP 2.42” here (opens in new window).