## Section2.8Cross Products

The vector cross product is a mathematical 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 value. The $$\times$$ symbol is used to indicate this operation.

Cross products are used in mechanics to find the moment of a force about about a point.

The cross product is a vector multiplication process defined by

\begin{equation} \vec{A} \times \vec{B}= A \; B \sin \theta \; \hat{\vec{u}}\text{.}\label{cross-product-def}\tag{2.8.1} \end{equation}

The result is a vector mutually perpendicular to the first two with a sense determined by the right hand rule. If $$\vec{A}$$ and $$\vec{B}$$ are in the $$xy$$ plane, this is

\begin{equation} \vec{A} \times \vec{B}=(A_y B_x - A_x B_y)\ \khat\text{.}\label{cross-product-det}\tag{2.8.2} \end{equation}

The operation is not commutative, in fact

\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 $$\vec{B},$$ which is also 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.

Notice that all the terms in the cross product equation are similar to those of 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 it is advisable to use a vector determinant method discussed here.

### Subsection2.8.1Cross Product of Arbitrary Vectors

The cross product of two three-dimensional vectors can be calculated by evaluating the determinant of this $$3 \times 3$$ matrix.

\begin{equation} \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} \label{cross-product-1}\tag{2.8.3} \end{equation}

Here, the first row are the unit vectors, the second row are the components of $$\vec{A}$$ and the third row are the components of $$\vec{B}\text{.}$$

Calculating the $$3 \times 3$$ determinant can be reduced to calculating three $$2 \times 2$$ determinants using the method of cofactors, as follows

\begin{equation} \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\tag{2.8.4} \end{equation}

Finally a $$2 \times 2$$ determinant can be evaluated with the formula

\begin{equation} \begin{vmatrix} a \amp b \\ c \amp d\end{vmatrix} = a d - b c\tag{2.8.5} \end{equation}

After simplifying, the resulting formula for a three-dimensional cross product is

\begin{equation} \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\label{cross-product-2}\tag{2.8.6} \end{equation}

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

\begin{equation} \vec{A} \times \vec{B} = (A_y B_z - A_z B_y) \ihat + ( A_z B_x -A_x B_z) \jhat + (A_x B_y - A_y B_x)\khat\text{,}\label{cross-product-3}\tag{2.8.7} \end{equation}

which is mathematically equivalent to equation (2.8.6).

In two dimensions, vectors $$\vec{A}$$ and $$\vec{B}$$ have no $$z$$ components, so (2.8.3) reduces to

\begin{equation} \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{.}\label{cross-prod-2d}\tag{2.8.8} \end{equation}

This equation produces the same result as equation (2.8.1) and you may use it if it is more convenient.

###### Example2.8.2.2-D Cross Product. 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 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.

From the diagram:

\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} \end{align*}

From (2.8.8):

\begin{align*} \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*}
###### Example2.8.3.3-D Cross Product.

Find the cross product of $$\vec{A} = \langle 2,4,-1 \rangle$$ and $$\vec{B} = \langle 10, 25, 20 \rangle\text{.}$$ The components of $$\vec{A}$$ are in meters and $$\vec{B}$$ are in Newtons.

\begin{gather*} \vec{A} \times \vec{B} = \Nm{\langle 105, -50, 10 \rangle} \end{gather*}
Solution 1.

To solve, set up the augmented determinant and evaluate it by adding the left-to-right diagonals and subtracting the right-to-left diagonals. (2.8.6).

\begin{align*} \vec{A} \times \vec{B} \amp= \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ 2 \amp 4 \amp -1 \\ 10 \amp 25 \amp 20 \end{vmatrix} \begin{matrix} \ihat \amp \jhat\\ 2 \amp 4\\ 10 \amp 25 \end{matrix}\\ \amp = (4)(20)\;\ihat + (-1)(10)\;\jhat + (2)(25) \;\khat - (4)(10) \;\khat -(-1)(25) \;\ihat - (2)(20) \;\jhat\\ \amp = (80 + 25) \;\ihat + (-10 - 40) \;\jhat + (50 - 40) \;\khat\\ \amp = \Nm{\langle 105, -50, 10 \rangle} \end{align*}
Solution 2.

Calculating three-dimensional cross products by hand is tedious and error prone. Whenever you can, you should use technology to do the grunt work for you and focus on the meaning of the results. In this solution we will use an embedded Sage calculator to calculate the cross product. This same calculator can be used to do other problems.

Given:

\begin{align*} \vec{A} \amp = \m{\langle 2,4,-1 \rangle}\\ \vec{B} \amp = \N{\langle 10, 25, 20 \rangle}. \end{align*}

$$\vec{A}$$ 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?

### Subsection2.8.2Cross Product of Unit Vectors

Since unit vectors have a magnitude of one and are perpendicular to each other, the magnitude of the cross product of two perpendicular unit vectors will be one by (2.8.1). The direction is determined by the right hand rule. On the other hand, whenever you cross a unit vector with itself, the result is zero since $$\theta=0\text{.}$$

One way to apply the right hand rule is to hold your right hand flat and point your fingers in the direction of the first vector, then curl them towards the second vector. When you do, your thumb will be oriented in the direction of the cross product.

To illustrate, imagine unit vectors $$\ihat$$ and $$\jhat$$ drawn on a white board in the normal orientation — $$\ihat$$ pointing right, $$\jhat$$ pointing up. Orient your right hand with your fingers pointing to the right along $$\ihat\text{,}$$ then curl them towards $$\jhat$$ and your thumb will point out of the board and establish that the direction of $$\ihat \times \jhat=\khat\text{.}$$ Now try to cross $$-\ihat$$ with $$\jhat$$ and you will find that your thumb now points into the board.

You should be able to convince yourself that the cross products of the positive unit 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 (counterclockwise) 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 separate the negative values until after you have taken the cross product, so for example

\begin{equation*} -\jhat \times \ihat = (-1)\left ( \jhat \times \ihat \right )=(-1)(-\khat )=+\khat \text{.} \end{equation*}