Key Questions
- How is a cross product different than a dot product?
- What is a determinant?
- What defines a right-handed Cartesian coordinate system?
- How do you use the cross-product circle to find the cross product of two unit vectors?
Section 2.8 Cross 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 a point.
Instructions.
This interactive shows the relation of the cross product to the two vectors and the angle between them.
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{.}\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{.}\tag{2.8.2}
\end{equation}
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 \(\vec{B},\) 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. The magnitude of the cross product of two perpendicular unit vectors is one.
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 in Subsection 2.8.3.
Subsection 2.8.1 Direction 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 one or the other 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 you are crossing 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 right-hand rule method can also be used to find a cross product’s direction. Start with your right hand flat and fingertips pointing along the first vector you are crossing with your fingertips pointing in the direction of the arrowhead. Then rotate your hand until the second vector you are crossing is perpendicular to your fingers and imagine that it pushes your fingers into a curl around your thumb. In this position, your thumb defines the direction of the cross product.
Subsection 2.8.2 Cross Product of Unit Vectors
The Figure 2.8.4.(a) demonstrates how you can 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, we find that that 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 negatives until after you have taken the cross product, like the following.
\begin{equation*}
-\jhat \times \ihat = (-1)\left ( \jhat \times \ihat \right )=(-1)(-\khat )=+\khat
\end{equation*}
Subsection 2.8.3 Cross 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} \tag{2.8.3}
\end{equation}
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 \(\vec{B}\text{.}\)
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\text{.}\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\text{.}\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\text{.}\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{,}\tag{2.8.7}
\end{equation}
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{.}\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.
Example 2.8.7. 2D Cross Product.
The two vectors \(\vec{A}\) and \(\vec{B}\) shown lie in the \(xy\) plane. Determine the cross product \(\vec{A} \times \vec{B}\text{.}\)
Answer.
Solution 1.
Solution 2.
\begin{gather*}
\vec{A} \times \vec{B} = -\N{1,697}^2 \; \khat
\end{gather*}
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*}
In this solution we will use (2.8.8).
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}
\end{align*}
\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*}
Example 2.8.8. 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.
Answer.
Solution 1.
Solution 2.
\begin{gather*}
\vec{A} \times \vec{B} = \Nm{\langle 105, -50, 10 \rangle}
\end{gather*}
To solve, set up the augmented determinant and evaluate it by adding the left-to-right diagonals and subtracting the right-to-left diagonals using equation (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 \cdot 20)\;\ihat + (-1 \cdot 10)\;\jhat + (2 \cdot 25) \;\khat - (4 \cdot 10) \;\khat -(-1 \cdot 25) \;\ihat - (2 \cdot 20) \;\jhat \\
\amp = (80 + 25) \;\ihat + (-10 - 40) \;\jhat + (50 - 40) \;\khat \\
\amp = \Nm{\langle 105, -50, 10 \rangle}
\end{align*}
Thus, the force \(\vec{B}\) creates a three-dimensional rotational moment equal to \(\Nm{\langle 105, -50, 10 \rangle}.\)
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}\text{.}
\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?
