# Engineering Statics: Open and Interactive

## Section2.8Cross Products

The vector cross product is a multipliation operation applied to two vectors which produces a third mutually perpendicular vector as a result. It’s sometimes called the vector product to emphasize this and to distinguish it from the dot product which produces a scalar result. The $$\times$$ symbol is used to indicate this operation.
Cross products are used in mechanics to find the moment of a force about a point.
The cross product is a vector multiplication process defined by
$$\vec{A} \times \vec{B}= A \; B \sin \theta \; \hat{\vec{u}}\text{.}\tag{2.8.1}$$
The result is a vector mutually perpendicular to both with a sense determined by the right-hand rule. If $$\vec{A}$$ and $$\vec{B}$$ are in the $$xy$$ plane, this is
$$\vec{A} \times \vec{B}=(A_y B_x - A_x B_y)\ \khat\text{.}\tag{2.8.2}$$
The operation is not commutative, in fact reversing the order introduces a negative sign.
\begin{equation*} \vec{A}\times \vec{B}= - \vec{B} \times \vec{A}\text{.} \end{equation*}
The magnitude of the cross product is the product of the perpendicular component of $$\vec{A}$$ with the magnitude of $$\vec{B},\text{.}$$ This is the area of the parallelogram formed by vectors $$\vec{A}$$ and $$\vec{B}\text{.}$$ The magnitude of the cross product is zero if $$\vec{A}$$ and $$\vec{B}$$ are parallel, and it is maximum when they are perpendicular. The magnitude of the cross product of two perpendicular unit vectors is one.
Notice that the cross product equation are similar to the dot product, except that $$\sin$$ is used rather than $$\cos$$ and the product includes a unit vector $$\hat{\vec{u}}$$ making the result a vector. This unit vector $$\hat{\vec{u}}$$ is simple to find in a two-dimensional problem as it will always be perpendicular to the page, but for three-dimensional cross products a vector determinant is used, as discussed in Subsection 2.8.3.

### Subsection2.8.1Direction of the Vector Cross Product

The direction of a cross product is determined by the right-hand rule. There are two ways to apply the right-hand rule, the three-finger method, and the point-and-curl method. You don’t need both, but you will need to master at least one to find the direction of cross products.
The three-finger method uses the fact that your extended index finger, middle finger, and thumb are all roughly mutually perpendicular. If you align your index finger with the first vector and your middle finger with the second, then your thumb will point in the direction of the cross product. Alternately, if you align your thumb with the first vector and your index finger with the second, your middle finger will point in the direction of the cross-product.
The point-and-curl method involves placing your right hand flat with your fingertips pointing in the direction of the the first vector. Then rotate your hand until the second vector is can curl your fingers around your thumb. In this position, your thumb defines the direction of the cross product.

### Subsection2.8.2Cross Product of Unit Vectors

The Figure 2.8.4.(a) demonstrates how you apply these techniques to find the cross product of $$\ihat \times \jhat\text{.}$$ Assuming the $$x$$ axis points right and the $$y$$ axis points up, the cross product points in the positive $$z$$ direction. Recalling that the magnitude of the cross product of two peperpedicular unit vectors is one, we conclude that
\begin{equation*} \ihat\times\jhat = \khat\text{.} \end{equation*}
Similarly, the cross products of the other pairs of vectors are:
\begin{align*} \ihat \times \ihat \amp =0 \amp \ihat \times \jhat \amp = \khat \amp \ihat \times \khat \amp = -\jhat\\ \jhat \times \ihat \amp = -\khat \amp \jhat \times \jhat \amp = 0 \amp \jhat \times \khat \amp = \ihat \\ \khat \times \ihat \amp = \jhat \amp \khat \times \jhat \amp =-\ihat \amp \khat \times \khat \amp =0 \end{align*}
An alternate way to remember this is to use the cross-product circle shown. For example when you cross $$\ihat$$ with $$\jhat$$ you are going in the positive (counter-clockwise) direction around the blue inner circle and thus the answer is $$+\khat\text{.}$$ But when you cross $$\jhat$$ into $$\ihat$$ you go in the negative (clockwise) direction around the circle and thus get a $$-\khat\text{.}$$ Remember that the order of cross products matter. If you put the vectors in the wrong order you will introduce a sign error.
If you have any negative unit vectors it is easiest to pull out the negative signs before you take the cross product, like the following.
\begin{equation*} -\jhat \times \ihat = (-1)\left ( \jhat \times \ihat \right )=(-1)(-\khat )=+\khat \end{equation*}

### Subsection2.8.3Cross Product of Arbitrary Vectors

The cross product of two arbitrary three-dimensional vectors can be calculated by evaluating the determinant of this $$3 \times 3$$ matrix.
$$\vec{A} \times \vec{B} = \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ A_x \amp A_y \amp A_z \\ B_x \amp B_y \amp B_z \end{vmatrix}\tag{2.8.3}$$
Here, the first row contains the unit vectors, the second row contains the components of $$\vec{A}\text{,}$$ and the third row, the components of $$\vec{B}\text{.}$$ The determinant of this $$3 \times 3$$ matrix is evaluated using the method of cofactors, as follows
$$\vec{A} \times \vec{B} = +\begin{vmatrix} A_y \amp A_z \\ B_y \amp B_z \end{vmatrix} \ \ihat -\begin{vmatrix} A_x \amp A_z \\ B_x \amp B_z \end{vmatrix}\ \jhat +\begin{vmatrix} A_x \amp A_y \\ B_x \amp B_y \end{vmatrix}\ \khat\text{.}\tag{2.8.4}$$
Each term contains a $$2 \times 2$$ determinant which is evaluated with the formula
$$\begin{vmatrix} a \amp b \\ c \amp d\end{vmatrix} = a d - b c\text{.}\tag{2.8.5}$$
After simplifying, the resulting formula for a three-dimensional cross product is
$$\vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\ \ihat - ( A_x B_z - A_z B_x)\ \jhat + (A_x B_y - A_y B_x)\ \khat\text{.}\tag{2.8.6}$$
In practice, the easiest way to remember this equation is to use the augmented determinant below, where the first two columns have been copied and placed after the determinant. The cross product is then calculated by adding the product of the red diagonals and subtracting the product of blue diagonals. The result is identical to (2.8.6).
In two dimensions, vectors $$\vec{A}$$ and $$\vec{B}$$ have no $$z$$ components, so (2.8.6) reduces to
$$\vec{A} \times \vec{B} = \begin{vmatrix}\ihat \amp \jhat \amp \khat \\ A_x \amp A_y \amp 0 \\ B_x \amp B_y \amp 0 \end{vmatrix} = (A_x B_y - A_y B_x)\ \khat\text{.}\tag{2.8.7}$$
This equation produces the same result as equation (2.8.1) and you may use it if it is more convenient.

#### Example2.8.7.2D Cross Product.

The two vectors $$\vec{A}$$ and $$\vec{B}$$ shown lie in the $$xy$$ plane. Determine the cross product $$\vec{A} \times \vec{B}\text{.}$$
\begin{gather*} \vec{A} \times \vec{B} = -\N{1,697}^2 \ \khat \end{gather*}
Solution 1.
In this solution we will apply equation (2.8.1).
\begin{align*} \vec{A} \times \vec{B} \amp = A \; B \sin \theta \;\hat{\vec{u}} \end{align*}
The direction of the cross product is determined by applying the right-hand rule. With the right hand, rotating $$\vec{A}$$ towards $$\vec{B}$$ we find that our thumb points into the $$xy$$ plane, so the direction of $$\hat{\vec{u}}$$ is $$-\khat\text{.}$$
\begin{align*} \vec{A} \times \vec{B} \amp = (\N{60}) (\N{40}) \sin \ang{45} (-\khat)\\ \amp = \N{1,697}^2 \; (-\khat)\\ \amp = - \N{1,697}^2 \; \khat \end{align*}
Solution 2.
In this solution we will use (2.8.7).
First, establish a coordinate system with the origin $$P$$ and with the $$x$$ axis aligned with $$\vec{A}\text{,}$$ then find the rectangular components and apply the cross product equation.
\begin{align*} A_x \amp= \N{60} \amp A_y \amp= \N{0}\\ B_x \amp= \N{40}\cos \ang{45} \amp B_y \amp= -\N{40} \sin \ang{45}\\ \amp = \N{28.28} \amp \amp= \N{-28.28}\\ \\ \vec{A} \times \vec{B} \amp = (A_x B_y - A_y B_x) \khat\\ \amp = \N{(60)(-28.28) - (0)(28.28)}^2 \khat\\ \amp = \N{-1697}^2 \; \khat \end{align*}

#### Example2.8.8.3D Cross Product.

Find the cross product of $$\vec{A} = \m{\langle 2,4,-1 \rangle}$$ and $$\vec{B} = \N{\langle 10, 25, 20 \rangle}\text{.}$$
Here, we are crossing a distance$$\vec{A}$$ and with a force$$\vec{B}.$$ This calculation is equivalent to finding the moment about a point $$P$$ caused by force $$\vec{B}$$ acting distance $$\vec{A}$$ from $$P\text{.}$$ You will learn about moments in Chapter 4.
Thus, the force $$\vec{B}$$ creates a three-dimensional rotational moment equal to $$\Nm{\langle 105, -50, 10 \rangle}.$$
$$\vec{A}$$ and $$\vec{B}$$ are defined in the first two lines, and A.cross_product(B) is the expression to be evaluated. Click Evaluate to see the result. You’ll have to work out the correct units for yourself.
Try changing the third line to B.cross_product(A). What changes?