Objectives
How and when is the cross product of two vectors defined?
What geometric information does the cross product provide?
Section1.4The Cross Product
The last two sections have introduced some basic algebraic operations on vectorsaddition, scalar multiplication, and the dot productwith useful geometric interpretations. In this section, we will meet a final algebraic operation, the cross product, which again conveys important geometric information.
To begin, we must emphasize that the cross product is only defined for vectors \(\vu\) and \(\vv\) in \(\R^3\text{.}\) Also, remember that we use a right-hand coordinate system, as described in Section1.1. In particular, recall that the vectors \(\vi\text{,}\) \(\vj\text{,}\) and \(\vk\) are oriented as shown below in Figure1.4.1. Earlier, we noticed that if we point the index finger of our right hand in the direction of \(\vi\) and our middle finger in the direction of \(\vj\text{,}\) then our thumb points in the direction of \(\vk\text{.}\)
Exploration1.4.1
The cross product of two vectors, \(\vu\) and \(\vv\text{,}\) will itself be a vector denoted \(\vu\times\vv\text{.}\) The direction of \(\vu\times\vv\) is determined by the right-hand rule: if we point the index finger of our right hand in the direction of \(\vu\) and our middle finger in the direction of \(\vv\text{,}\) then our thumb points in the direction of \(\vu\times\vv\text{.}\)
We begin by defining the cross products using the vectors \(\vi\text{,}\) \(\vj\text{,}\) and \(\vk\text{.}\) Referring to Figure1.4.1, explain why \(\vi\text{,}\) \(\vj\text{,}\) \(\vk\) in that order form a right-hand system. We then define \(\vi \times \vj\) to be \(\vk\) that is \(\vi\times\vj = \vk\text{.}\)
Now explain why \(\vi\text{,}\) \(\vk\text{,}\) and \(-\vj\) in that order form a right-hand system. We then define \(\vi \times \vk\) to be \(-\vj\) that is \(\vi\times\vk=-\vj\text{.}\)
Continuing in this way, complete the missing entries in Table1.4.2.
\(\vi\times\vj = \vk\) \(\vi\times\vk = -\vj\) \(\vj\times\vk =\) \(\vj\times\vi =\) \(\vk\times\vi =\) \(\vk\times\vj =\) Table1.4.2 Up to this point, the products you have seen, such as the product of real numbers and the dot product of vectors, have been commutative, meaning that the product does not depend on the order of the terms. For instance, \(2\cdot5 = 5\cdot 2\text{.}\) The table above suggests, however, that the cross product is anti-commutative: for any vectors \(\vu\) and \(\vv\) in \(\R^3\text{,}\) \(\vu\times\vv = -\vv\times\vu\text{.}\) If we consider the case when \(\vu=\vv\text{,}\) this shows that \(\vv\times\vv = -(\vv\times\vv)\text{.}\) What does this tell us about \(\vv\times\vv\text{;}\) in particular, what vector is unchanged by scalar multiplication by \(-1\text{?}\)
-
It is not difficult to show that the cross product interacts with scalar multiplication and vector addition as one would expect: that is
\begin{align*} (c\vu) \times \vv =\mathstrut \amp c(\vu \times \vv) \\ (\vu + \vv) \times\vw =\mathstrut \amp (\vu\times\vw) + (\vv\times\vw) \end{align*}We can combine these properties to make cross product calculations a bit easier. For example,
\begin{align*} (2\vi + \vj) \times \vk =\mathstrut \amp (2\vi \times \vk) + (\vj \times \vk) \\ =\mathstrut \amp 2(\vi \times \vk) + (\vj \times \vk) \\ =\mathstrut \amp -2 \vj + \vi. \end{align*}Using these properties along with Table1.4.2, find the cross product \(\vu\times\vv\) if \(\vu = 2\vi + 3\vj\) and \(\vv = -\vi + \vk\text{.}\)
Verify that the cross product \(\vu\times\vv\) you just found in part (e) is orthogonal to both \(\vu\) and \(\vv\text{.}\)
-
Consider the vectors \(\vu\) and \(\vv\) in the \(xy\)-plane as shown below in Figure1.4.3.
Figure1.4.3Two vectors in the \(xy\)-plane Explain why \(\vu = |\vu|\vi\) and \(\vv = |\vv|\cos(\theta) \vi + |\vv|\sin(\theta) \vj\text{.}\) Then compute the length of \(|\vu\times\vv|\text{.}\)
Subsection1.4.1Computing the cross product
As we have seen in Preview Activity1.4.1, the cross product \(\vu\times\vv\) is defined for two vectors \(\vu\) and \(\vv\) in \(\R^3\) and produces another vector in \(\R^3\text{.}\) Using the right-hand rule, we saw that
| \(\vi\times\vj = \vk\) |
| \(\vj\times\vi = -\vk\) |
| \(\vi\times\vk = -\vj\) |
| \(\vk\times\vi = \vj\) |
| \(\vj\times\vk = \vi\) |
| \(\vk\times\vj = -\vi\) |
Investigation1.4.2
If, in addition, we assume the cross product behaves like we think a product should (e.g., the cross product distributes over vector addition), we can compute the cross product in terms of the components of general vectors to find a formula for the cross product. Doing so we see that
\begin{align*}
\vu\times\vv =\mathstrut \amp (u_1\vi + u_2\vj + u_3\vk) \times (v_1\vi + v_2\vj + v_3\vk)\\
=\mathstrut \amp u_1\vi\times (v_1\vi + v_2\vj + v_3\vk)+u_2\vj\times (v_1\vi + v_2\vj + v_3\vk) \\
\mathstrut \amp +u_3\vk\times (v_1\vi + v_2\vj + v_3\vk)\\
=\mathstrut \amp u_1v_1\vi\times\vi + u_1v_2\vi\times\vj + u_1v_3\vi\times\vk + u_2v_1\vj\times\vi + u_2v_2\vj\times\vj \\
\mathstrut \amp + u_2v_3\vj\times\vk +u_3v_1\vk\times\vi + u_3v_2\vk\times\vj + u_3v_3\vk\times\vk\\
=\mathstrut \amp u_1v_2\vk - u_1v_3\vj - u_2v_1\vk + u_2v_3\vi +u_3v_1\vj -
u_3v_2\vi\\
=\mathstrut \amp (u_2v_3-u_3v_2)\vi - (u_1v_3-u_3v_1)\vj + (u_1v_2-u_2v_1)\vk.
\end{align*}
(Like the dot product, the cross product arises in physical applications, e.g., torque, but it is more convenient mathematically to begin from an algebraic perspective.)
The previous calculations lead us to define the cross product of vectors in \(\R^3\) as follows. The cross product \(\vu \times \vv\) of vectors \(\vu = u_1\vi + u_2\vj + u_3\vk\) and \(\vv = v_1\vi + v_2\vj + v_3\vk\) in \(\R^3\) is the vector
Definition1.4.4
\begin{equation}
(u_2v_3-u_3v_2)\vi - (u_1v_3-u_3v_1)\vj + (u_1v_2-u_2v_1)\vk.\label{E_9_4_cross_def}\tag{1.4.1}
\end{equation}
At first, this may look intimidating and difficult to remember. However, if we rewrite the expression in Equation(1.4.1) using determinants, important structure emerges. The determinant of a \(2\times2\) matrix is
\begin{equation*}
\left|
\begin{array}{cc}
a \amp b \\
c \amp d
\end{array}
\right|
=ad - bc.
\end{equation*}
It follows that we can thus rewrite Equation(1.4.1) in the following form.
Formula for the cross product
\begin{equation*}
\vu\times\vv =
\left|
\begin{array}{cc}
u_2 \amp u_3 \\
v_2 \amp v_3
\end{array}
\right| \vi
-
\left|
\begin{array}{cc}
u_1 \amp u_3 \\
v_1 \amp v_3
\end{array}
\right| \vj
+
\left|
\begin{array}{cc}
u_1 \amp u_2 \\
v_1 \amp v_2
\end{array}
\right| \vk.
\end{equation*}
For those familiar with the determinant of a \(3\times3\) matrix, we write the mnemonic as
\begin{equation*}
\vu\times\vv =
\left|
\begin{array}{ccc}
\vi \amp \vj \amp \vk \\
u_1 \amp u_2 \amp u_3 \\
v_1 \amp v_2 \amp v_3
\end{array}
\right|.
\end{equation*}
Example1.4.5
Suppose \(\vu = \langle 0, 1, 3\rangle\) and \(\vv = \langle 2, -1, 0\rangle\text{.}\) Use the formula (1.4.1) for the following.
Find the cross product \(\vu\times\vv\text{.}\)
Evaluate the dot products \(\vu\cdot(\vu\times\vv)\) and \(\vv\cdot(\vu\times\vv)\text{.}\) What does this tell you about the geometric relationship among \(\vu\text{,}\) \(\vv\text{,}\) and \(\vu\times\vv\text{?}\)
Find the cross product \(\vv\times \vi\text{.}\)
Multiplication of real numbers is associative, which means, for instance, that \((2\cdot 5)\cdot 3 = 2\cdot(5\cdot 3)\text{.}\) Is it true that the cross product of vectors is associative? For instance, is it true that \((\vu\times\vv)\times\vi = \vu\times(\vv\times\vi)\text{?}\)
Find the cross product \(\vu\times\vu\text{.}\)
Solution
- \begin{align*} \vu\times\vv \amp = \langle 0,1,3 \rangle \times \langle 2,-1,0 \rangle \\ \amp = \left| \begin{array}{ccc} \vi \amp \vj \amp \vk \\ 0 \amp 1 \amp 3 \\ 2 \amp -1 \amp 0 \end{array} \right|\\ \amp = \left| \begin{array}{cc} 1 \amp 3 \\ -1 \amp 0 \end{array} \right| \vi - \left| \begin{array}{cc} 0 \amp 3 \\ 2 \amp 0 \end{array} \right| \vj + \left| \begin{array}{cc} 0 \amp 1 \\ 2 \amp -1 \end{array} \right| \vk \\ \amp = (1\cdot 0 - 3 \cdot -1)\vi - (0\cdot 0 - 3 \cdot 2) \vj + (0 \cdot -1 - 1 \cdot 2) \vk\\ \amp = \langle 3, 6, -2 \rangle \end{align*}
-
\begin{align*} \vu \cdot (\vu \times \vv) \amp = \langle 0,1,3 \rangle \cdot \langle 3,6,-2 \rangle = 0 + 6 - 6 = 0 \\ \vv \cdot (\vu \times \vv) \amp = \langle 2,-1,0 \rangle \cdot \langle 3,6,-2 \rangle = 6 - 6 + 0 = 0 \end{align*}
Recall that two vectors are perpendicular if and only if their dot product is \(0\text{.}\) So we have verified two geometric facts: the vectors \(\vu\) and \(\vu \times \vv\) are perpendicular, and the vectors \(\vv\) and \(\vu \times \vv\) are perpendicular.
- \begin{equation*} \vv \times \vi = \left| \begin{array}{ccc} \vi \amp \vj \amp \vk \\ 2 \amp -1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array} \right| = - \vi - 2 \vj + 0 \vk = \langle -1, -2, 0 \rangle \end{equation*}
You can check that \((\vu \times \vv) \times \vi = \langle 6,-3,0 \rangle\) and that \(\vu \times (\vv \times \vi) = \langle 6,-3,1 \rangle\text{.}\) These two vectors are not equal, so the cross product is not associative in general.
-
\begin{equation*} \vu \times \vu = \left| \begin{array}{ccc} \vi \amp \vj \amp \vk \\ 0 \amp 1 \amp 3 \\ 0 \amp 1 \amp 3 \end{array} \right| = (3-3)\vi - (0 - 0)\vj + (0 - 0) \vk = \langle 0,0,0 \rangle. \end{equation*}
Notice that the cross product of any vector with itself will always be the zero vector.
The cross product satisfies the following properties, some of which were illustrated in Preview Activity1.4.1 and may be easily verified from the definition (1.4.1).
Properties of the cross product
Let \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) be vectors in \(\R^3\text{,}\) and let \(c\) be a scalar. Then
\(\vu \times \vv = -(\vv \times \vu)\)
\((\vu +\vv)\times \vw = (\vu \times \vw) + (\vv \times \vw)\)
\((c\vu)\times \vw = c(\vu \times \vw) = \vu \times (c\vw)\)
\(\vu\times\vv = \vzero\) if \(\vu\) and \(\vv\) are parallel.
-
The cross product is not associative; that is, in general
\begin{equation*} (\vu\times\vv)\times\vw \neq \vu\times(\vv\times\vw). \end{equation*}
Just as we found for the dot product, the cross product provides us with useful geometric information. In particular, both the length and direction of the cross product \(\vu\times\vv\) encode information about the geometric relationship between \(\vu\) and \(\vv\text{.}\)
Subsection1.4.2The Length of \(\vu\times\vv\)
We may ask whether the length \(|\vu\times\vv|\) has any relationship to the lengths of \(\vu\) and \(\vv\text{.}\) To investigate, we will compute the square of the length \(|\vu\times\vv|^2\) and denote by \(\theta\) the angle between \(\vu\) and \(\vv\text{,}\) as in Section1.4.6.
Investigation1.4.3
We start with \(|\vu \times \vv|^2\text{,}\) expand it out using the definition of cross product, and do a lot of algebra:
\begin{align*}
|\vu\times\vv|^2=\mathstrut \amp (u_2v_3-u_3v_2)^2 + (u_1v_3-u_3v_1)^2 +
(u_1v_2-u_2v_1)^2\\
=\mathstrut \amp u_2^2 v_3^2- 2u_2u_3v_2v_3 +u_3^2v_2^2 +
u_1^2 v_3^2 -2u_1u_3v_1v_3 +u_3^2v_1^2 \\
\mathstrut \amp + u_1^2 v_2^2 -2u_1u_2v_1v_2 +u_2^2v_1^2\\
=\mathstrut \amp u_1^2(v_2^2+v_3^2) + u_2^2(v_1^2+v_3^2) + u_3^2(v_1^2+v_2^2) \\
\mathstrut \amp - 2(u_1u_2v_1v_2 + u_1u_3v_1v_3 + u_2u_3v_2v_3)\\
=\mathstrut \amp u_1^2(v_1^2+v_2^2+v_3^2) + u_2^2(v_1^2+v_2^2+v_3^2) +
u_3^2(v_1^2+v_2^2+v_3^2) \\
\mathstrut \amp -
(u_1^2v_1^2 + u_2^2v_2^2 + u_3^2v_3^2 +
2(u_1u_2v_1v_2 + u_1u_3v_1v_3 + u_2u_3v_2v_3))\\
=\mathstrut \amp (u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)-(u_1v_1+u_2v_2+u_3v_3)^2\\
=\mathstrut \amp |\vu|^2|\vv|^2-(\vu\cdot\vv)^2\\
=\mathstrut \amp |\vu|^2|\vv|^2(1-\cos^2(\theta))\\
=\mathstrut \amp |\vu|^2|\vv|^2\sin^2(\theta).
\end{align*}
Therefore, we have found \(|\vu\times\vv|^2 = |\vu|^2|\vv|^2\sin^2(\theta)\text{.}\) Taking square-roots we obtain the identity below.
The length of the cross product, algebraic version
\begin{equation}
|\vu\times\vv| = |\vu||\vv|\sin(\theta).\label{E_9_4_cross_length}\tag{1.4.2}
\end{equation}
Note that Item4 in the list of properties above stated above says that \(\vu\times\vv = \vzero\) if \(\vu\) and \(\vv\) are parallel. This is reflected in Equation(1.4.2) since \(\sin(\theta)=0\) if \(\vu\) and \(\vv\) are parallel, which implies that \(\vu\times\vv = \vzero\text{.}\)
Equation (1.4.2) also has a geometric implication. Consider the parallelogram formed by two vectors \(\vu\) and \(\vv\text{,}\) as shown in Figure1.4.6.
Remember that the area of a parallelogram is the product of its base and height. As shown in the figure, we may consider the base of the parallelogram to be \(|\vu|\) and the height to be \(|\vv|\sin(\theta)\text{.}\) This means that the area of the parallelogram formed by \(\vu\) and \(\vv\) is
\begin{equation*}
|\vu||\vv|\sin(\theta) = |\vu\times\vv|.
\end{equation*}
This leads to the following interesting fact.
The length of the cross product, geometric version
The length, \(|\vu \times \vv|\text{,}\) of the cross product of vectors \(\vu\) and \(\vv\) is the area of the parallelogram determined by \(\vu\) and \(\vv\text{.}\)
Note also that if \(\vu = u_1\vi + u_2\vj + 0\vk\) and \(\vv = v_1\vi + v_2\vj + 0\vk\) are vectors in the \(xy\)-plane, then Equation(1.4.1) shows that the area of the parallelogram determined by \(\vu\) and \(\vv\) is \(|\vu \times \vv| = |u_1v_2-u_2v_1|\) is the absolute value of the \(2 \times 2\) determinant \(\left| \begin{array}{cc} u_1 \amp u_2 \\ v_1 \amp v_2 \end{array} \right|.\) So the absolute value of a determinant of a \(2 \times 2 \) matrix is also the area of a parallelogram.
Example1.4.7
Find the area of the parallelogram formed by the vectors \(\vu = \langle 1,3, -2\rangle\) and \(\vv=\langle 3,0,1\rangle\text{.}\)
Find the area of the parallelogram in \(\R^3\) whose vertices are \((1,0,1)\text{,}\) \((0,0,1)\text{,}\) \((2,1,0)\text{,}\) and \((1,1,0)\text{.}\) (Hint: It might be helpful to draw a picture to see how the vertices are arranged so you can determine which vectors you might use.)
Solution
-
The area of this parallelogram is the length of the cross product:
\begin{align*} \vu \times \vv \amp = \left| \begin{array}{ccc} \vi \amp \vj \amp \vk \\ 1 \amp 3 \amp -2 \\ 3 \amp 0 \amp 1 \end{array} \right| \\ \amp = (3 - 0) \vi - (1 - -6)\vj + (0 - 9) \vk\\ \amp = \langle 3, -6, -9 \rangle \end{align*}The area of the parallelogram is the length of this vector, which is
\begin{equation*} | \langle 3, -6, -9 \rangle | = \sqrt{3^2 + (-6)^2 + (-9)^2} = \sqrt{9 + 36 + 81} = \sqrt{146} \end{equation*} -
Write \(\vu\) for the vector from \((1,0,1)\) to \((0,0,1)\text{,}\) which in coordinate form is \(\langle -1, 0, 0 \rangle\text{.}\) Write \(\vv\) for the vector from \((1,0,1)\) to \((2,1,0)\text{,}\) so \(\vv = \langle 1,1,-1 \rangle\text{.}\) Then the parallelogram is the one determined by \(\vu\) and \(\vv\text{,}\) so its area is the length of the cross product of \(\vu\) and \(\vv\text{.}\)
\begin{align*} |\vu \times \vv| \amp = \left| \begin{array}{ccc} \vi \amp \vj \amp \vk \\ -1 \amp 0 \amp 0 \\ 1 \amp 1 \amp -1 \end{array} \right| \\ \amp = |\langle 0, -1, -1 \rangle| \\ \amp = \sqrt 2 \end{align*}
Subsection1.4.3The Direction of \(\vu\times\vv\)
Now that we understand the length of \(\vu\times\vv\text{,}\) we will investigate its direction. Remember from Preview Activity1.4.1 that cross products involving the vectors \(\vi\text{,}\) \(\vj\text{,}\) and \(\vk\) resulted in vectors that are orthogonal to the two terms. We will see that this holds more generally.
Investigation1.4.4
To verify this we compute \(\vu\cdot(\vu\times\vv)\text{,}\) and see that
\begin{align*}
\vu\cdot(\vu\times\vv)=\mathstrut \amp u_1(u_2v_3-u_3v_2) - u_2(u_1v_3-u_3v_1) +
u_3(u_1v_2-u_2v_1)\\
=\mathstrut \amp u_1u_2v_3 - u_1u_3v_2 - u_2u_1v_3+u_2u_3v_1 + u_3u_1v_2 -
u_3u_2v_1\\
=\mathstrut \amp 0
\end{align*}
To summarize, we have \(\vu\cdot(\vu\times\vv) = 0\text{,}\) which implies that \(\vu\) is orthogonal to \(\vu\times\vv\text{.}\) In the same way, we can show that \(\vv\) is orthogonal to \(\vu\times\vv\text{.}\) The net effect is that \(\vu\times\vv\) is a vector that is perpendicular to both \(\vu\) and \(\vv\text{,}\) and hence \(\vu\times\vv\) is perpendicular to the plane determined by \(\vu\) and \(\vv\text{.}\) Moreover, the direction of \(\vu\times\vv\) is determined by applying the right-hand rule to \(\vu\) and \(\vv\text{,}\) as we saw in Preview Activity1.4.1. In light of our earlier work that showed \(|\vu||\vv|\sin(\theta) = |\vu\times\vv|.\text{,}\) we may now express \(\vu \times \vv\) in the following different way.
The cross product as normal vector
Suppose that \(\vu\) and \(\vv\) are not parallel and that \(\vn\) is the unit vector perpendicular to the plane containing \(\vu\) and \(\vv\) determined by the right-hand rule. Then
\begin{equation*}
\vu\times\vv = |\vu||\vv|\sin(\theta) ~\vn.
\end{equation*}
There is yet one more geometric implication we may draw from this result. Suppose \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) are vectors in \(\R^3\) that are not coplanar and that form a three-dimension parallelepiped as shown in Figure1.4.8.
The volume of the parallelepiped is determined by multiplying \(A\text{,}\) the area of the base, by the height \(h\text{.}\) As we have just seen, the area of the base is \(|\vu\times\vv|\text{.}\) Moreover, the height \(h=|\vw|\cos(\alpha)\) where \(\alpha\) is the angle between \(\vw\) and the vector \(\vn\text{,}\) which is orthogonal to the plane formed by \(\vu\) and \(\vv\text{.}\) Since \(\vn\) is parallel to \(\vu\times\vv\text{,}\) the angle between \(\vw\) and \(\vu\times\vv\) is also \(\alpha\text{.}\) This shows that
\begin{equation*}
|(\vu\times\vv)\cdot\vw| = |\vu\times\vv||\vw|\cos(\alpha) = Ah,
\end{equation*}
and therefore
The cross product and the volume of a parallelepiped
The volume of the parallelepiped determined by \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) is \(|(\vu\times\vv)\cdot\vw|\text{.}\)
As a dot product of two vectors, the quantity \((\vu \times \vv) \cdot \vw\) is a scalar and is called the triple scalar product.
Example1.4.9
Suppose \(\vu = \langle 3, 5, -1\rangle\) and \(\vv = \langle 2, -2, 1\rangle\text{.}\)
Find two unit vectors orthogonal to both \(\vu\) and \(\vv\text{.}\)
Find the volume of the parallelepiped formed by the vectors \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw = \langle 3,3,1\rangle\text{.}\)
Given the vectors \(\vu\) and \(\vv\) shown below in Figure1.4.10, sketch the cross products \(\vu\times\vv\) and \(\vv\times\vu\text{.}\)
Figure1.4.10Vectors \(\vu\) and \(\vv\)
Solution
-
First we compute the cross product.
\begin{align*} \vu \times \vv \amp = \left| \begin{array}{ccc} \vi \amp \vj \amp \vk \\ 3 \amp 5 \amp -1 \\ 2 \amp -2 \amp 1 \end{array} \right| \\ \amp = (5 - 2)\vi - (3 - - 2)\vj + (-6 - 10)\vk \\ \amp = \langle 3, -1, -16 \rangle \end{align*}This is not a unit vector, but if we scale it by a factor of the reciprocal of its length, we will get a unit vector that points in the same direction:
\begin{align*} \frac{\langle 3, -1, -16 \rangle }{| \langle 3, -1, -16 \rangle |} \amp = \frac{\langle 3, -1, -16 \rangle }{\sqrt{3^2 + (-1)^2 + (-16)^2}}\\ \amp = \frac{1}{266} \langle 3, -1, -16 \rangle \end{align*} -
We could compute e.g. \(|(\vu \times \vw) \cdot \vv|\) or \(|(\vv \times \vw) \cdot \vu|\text{,}\) but since we have already computed \(\vu \times \vv\) it will be easiest to do the following.
\begin{equation*} |(\vu \times \vv) \times \vw| = |\langle 3, -1, -16 \rangle \cdot \langle 3,3,1 \rangle| = |-10| = 10 \end{equation*}
Subsection1.4.4Torque is measured by a cross product
We have seen that the cross product enables us to produce a vector perpendicular to two given vectors, to measure the area of a parallelogram, and to measure the volume of a parallelepiped. Besides these geometric applications, the cross product also enables us to describe a physical quantity called torque.
Suppose that we would like to turn a bolt using a wrench as shown in Figure1.4.11. If a force \(\vF\) is applied to the wrench and \(\vr\) is the vector from the position on the wrench at which the force is applied to center of the bolt, we define the torque, \(\tau\text{,}\) to be
\begin{equation*}
\tau=\vF\times\vr.
\end{equation*}
When a force is applied to an object, Newton's Second Law tells us that the force is equal to the rate of change of the object's linear momentum. Similarly, the torque applied to an object is equal to the rate of change of the object's angular momentum. In other words, torque will cause the bolt to rotate.
In many industrial applications, bolts are required to be tightened using a specified torque. Of course, the magnitude of the torque is \(|\tau| =|\vF\times\vr|=|\vF||\vr||\sin(\theta)\text{.}\) Thus, to produce a larger torque, we can increase either \(|\vF|\) or \(|\vr|\text{,}\) which you may know if you have ever removed lug nuts when changing a flat tire. The ancient Greek mathematician Archimedes said: Give me a lever long enough and a fulcrum on which to place it, and I shall move the world.
A modern spin on this statement is: Allow me to make \(|\vr|\) large enough, and I shall produce a torque large enough to move the world.
Subsection1.4.5Comparing the dot and cross products
Finally, it is worthwhile to compare and contrast the dot and cross products.
\(\vu\cdot\vv\) is a scalar, while \(\vu\times\vv\) is a vector.
\(\vu\cdot\vv = \vv\cdot\vu\text{,}\) while \(\vu\times\vv = -\vv\times\vu\)
\(\vu\cdot\vv = |\vu||\vv|\cos(\theta)\text{,}\) while \(|\vu\times\vv| = |\vu||\vv|\sin(\theta)\text{.}\)
\(\vu\cdot\vv = 0\) if \(\vu\) and \(\vv\) are perpendicular, while \(\vu\times\vv = \vzero\) if \(\vu\) and \(\vv\) are parallel.
Subsection1.4.6Summary
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The cross product is defined only for vectors in \(\R^3\text{.}\) The cross product of vectors \(\vu = u_1 \vi + u_2 \vj + u_3 \vk\) and \(\vv = v_1 \vi + v_2 \vj + v_3 \vk\) in \(\R^3\) is the vector
\begin{equation*} \vu \times \vv = (u_2v_3-u_3v_2) \vi - (u_1v_3 - u_3v_1) \vj + (u_1v_2 - u_2v_1) \vk. \end{equation*} -
Geometrically, the cross product is
\begin{equation*} \vu \times \vv = |\vu| \ |\vv| \ \sin(\theta) \ \vn, \end{equation*}where \(\theta\) is the angle between \(\vu\) and \(\vv\) and \(\vn\) is a unit vector perpendicular to both \(\vu\) and \(\vv\) as determined by the right-hand rule.
The cross product of vectors \(\vu\) and \(\vv\) is a vector perpendicular to both \(\vu\) and \(\vv\text{.}\)
The magnitude \(|\vu \times \vv|\) of the cross product of the vectors \(\vu\) and \(\vv\) gives the area of the parallelogram determined by \(\vu\) and \(\vv\text{.}\) Also, the scalar triple product \(|(\vu \times \vv) \cdot \vw|\) gives the volume of the parallelepiped determined by \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\text{.}\)
Subsection1.4.7Exercises
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Let \(\vu = 2\vi + \vj\) and \(\vv = \vi + 2\vj\) be vectors in \(\R^3\text{.}\)
Without doing any computations, find a unit vector that is orthogonal to both \(\vu\) and \(\vv\text{.}\) What does this tell you about the formula for \(\vu \times \vv\text{?}\)
Using the properties of the cross product and what you know about cross products involving the fundamental vectors \(\vi\) and \(\vj\text{,}\) compute \(\vu \times \vv\text{.}\)
Next, use the determinant version of Equation(1.4.1) to compute \(\vu \times \vv\text{.}\) Write one sentence that compares your results in (a), (b), and (c).
Find the area of the parallelogram determined by \(\vu\) and \(\vv\text{.}\)
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Let \(\vx = \langle 1, 1, 1 \rangle\) and \(\vy = \langle 0, 3, -2 \rangle\text{.}\)
Are \(\vx\) and \(\vy\) orthogonal? Are \(\vx\) and \(\vy\) parallel? Clearly explain how you know, using appropriate vector products.
Find a unit vector that is orthogonal to both \(\vx\) and \(\vy\text{.}\)
Express \(\vy\) as the sum of two vectors: one parallel to \(\vx\text{,}\) the other orthogonal to \(\vx\text{.}\)
Determine the area of the parallelogram formed by \(\vx\) and \(\vy\text{.}\)
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Consider the triangle in \(\R^3\) formed by \(P(3, 2, -1)\text{,}\) \(Q(1, -2, 4)\text{,}\) and \(R(4, 4, 0)\text{.}\)
Find \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\text{.}\)
Observe that the area of \(\triangle PQR\) is half of the area of the parallelogram formed by \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\text{.}\) Hence find the area of \(\triangle PQR\text{.}\)
Find a unit vector that is orthogonal to the plane that contains points \(P\text{,}\) \(Q\text{,}\) and \(R\text{.}\)
Determine the measure of \(\angle PQR\text{.}\)
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One of the properties of the cross product is that \((\vu+\vv) \times \vw = (\vu \times \vw) + (\vv \times \vw)\text{.}\) That is, the cross product distributes over vector addition on the right. Here we investigate whether the cross product distributes over vector addition on the left.
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Let \(\vu = \langle 1,2,-1 \rangle\text{,}\) \(\vv = \langle 4,-3,6 \rangle\text{,}\) and \(\vv = \langle 4,7,2 \rangle\text{.}\) Calculate
\begin{equation*} \vu \times (\vv + \vw) \ \ \text{ and } \ \ (\vu \times \vv) + (\vu \times \vw). \end{equation*}What do you notice?
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Use the properties of the cross product to show that in general
\begin{equation*} \vx \times (\vy + \vz) = (\vx \times \vy) + (\vx \times \vz) \end{equation*}for any vectors \(\vx\text{,}\) \(\vy\text{,}\) and \(\vz\) in \(\R^3\text{.}\)
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Let \(\vu = \langle u_1, u_2, u_3 \rangle\text{,}\) \(\vv = \langle v_1, v_2, v_3 \rangle\text{,}\) and \(\vw = \langle w_1, w_2, w_3 \rangle\) be vectors in \(\R^3\text{.}\) In this exercise we investigate properties of the triple scalar product \((\vu \times \vv) \cdot \vw\text{.}\)
Show that \((\vu \times \vv) \cdot \vw = \left|\begin{array}{ccc} u_1 \amp u_2 \amp u_3 \\ v_1 \amp v_2 \amp v_3 \\ w_1 \amp w_2 \amp w_3 \end{array} \right| \text{.}\)
Show that \(\left|\begin{array}{ccc} u_1 \amp u_2 \amp u_3 \\ v_1 \amp v_2 \amp v_3 \\ w_1 \amp w_2 \amp w_3 \end{array} \right| = -\left|\begin{array}{ccc} v_1 \amp v_2 \amp v_3 \\ u_1 \amp u_2 \amp u_3 \\ w_1 \amp w_2 \amp w_3 \end{array} \right|\text{.}\) Conclude that interchanging the first two rows in a \(3 \times 3\) matrix changes the sign of the determinant. In general (although we won't show it here), interchanging any two rows in a \(3 \times 3\) matrix changes the sign of the determinant.
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Use the results of parts (a) and (b) to argue that
\begin{equation*} (\vu \times \vv) \cdot \vw = (\vw \times \vu) \cdot \vv = (\vv \times \vw) \cdot \vu. \end{equation*} -
Now suppose that \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) do not lie in a plane when they eminate from a common initial point.
Given that the parallepiped determined by \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) must have positive volume, what can we say about \((\vu \times \vv) \cdot \vw\text{?}\)
Now suppose that \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) all lie in the same plane. What value must \((\vu \times \vv) \cdot \vw\) have? Why?
Explain how (i.) and (ii.) show that if \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) all eminate from the same initial point, then \(\vu\text{,}\) \(\vv\text{,}\) and \(\vw\) lie in the same plane if and only if \((\vu \times \vv) \cdot \vw = 0\text{.}\) This provides a straightforward computational method for determining when three vectors are co-planar.