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The Cross Product
Calculate the cross product of two given vectors.
The Cross Product
This section only expects students to have familiarity with the notation and operations of vectors from and only when working with the triple product do we use the dot product in this section. As stated earlier, we suggest at least a week on the combined topics in , , and .
This section emphasizes the cross product as a tool for finding a third vector orthogonal to the two input vectors that will create a right handed coordinate system. We introduce the ideas of determinants of two by two and three by three matrices without motivation or context (other than as a useful computational tool).
In our discussion of applications, the idea of area for the associated parallelogram is described and used. This idea will be used several times in when we are approximating surface area for curved surfaces and setting up surface integrals.
We encourage all users to look at to see an activity that encourages students to think about the types of objects being used in the many vector operations. This is a key meta skill for student to improve on before courses like linear algebra and differential equations which can be notationally difficult for students.
Introduction
The last two sections have introduced some basic algebraic operations and useful geometric interpretations for vector addition, scalar multiplication, and the dot product. In this section, we will define and explore the cross product, another algebraic operation which measures useful geometric information related to pairs of vectors in \(\R^3\).
Remember that we use a right-handed coordinate system, as described in . In particular, recall that the vectors \(\vi\), \(\vj\), and \(\vk\) are oriented as shown below in . We would like to think of a right-handed coordinate system in terms of vectors that are not just \(\vi\), \(\vj\), and \(\vk\).
We will call an ordered list of vectors \(\{\vu, \vv, \vw\}\) a right-handed coordinate system (or right-handed orientation) provided that putting the fingers of your right hand in the direction of \(\vu\) and curling your fingers in the direction of \(\vv\) results in the thumb of your right hand pointing in the direction of \(\vw\). Remember that to get this orientation correct, you may need to rotate your right hand to get your fingers to curl in the correct direction, which will then give you the proper direction for the third vector \(\vw\).
We would like to create a product of two vectors, which we will call the cross product, that creates a right handed coordinate system with the input vectors as the first two vectors in our coordinate system. We will denote the cross product of \(\vu\) and \(\vv\) by \(\vu \times \vv\). If \(\vu\) and \(\vv\) are not parallel, we want the cross product of \(\vu\) with \(\vv\) to make \(\{\vu, \vv, \vu\times\vv\}\) a right-handed coordinate system. Additionally, we would like the cross product to work well with linear combinations of vectors. Specifically, we want the following algebraic property to be satisfied: \[\begin{aligned}\end{aligned}\] An important difference to note here is that the cross product of two vectors will be a vector, whereas the dot product of two vectors gives a scalar.
In other sources, you may see the dot product referred to as the scalar product of vectors and the cross product referred to as the vector product. These terms comes from the result of the dot product being a scalar and the result of the cross product being a vector.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Computing the cross product
As we have seen in Preview Activity, the cross product \(\vu\times\vv\) will be defined for two vectors \(\vu\) and \(\vv\) in \(\R^3\) and produces another vector in \(\R^3\). Using the right-hand rule, we saw that \[\begin{aligned}\vi\times\vj \amp = \vk \amp \vi\times\vk \amp= -\vj \amp \vj\times\vk \amp= \vi \\ \vj\times\vi \amp = -\vk \amp \vk\times\vi \amp= \vj \amp \vk\times\vj \amp= -\vi\end{aligned}\]
We also want the cross product to distribute over linear combinations so that \((k \vv+\vu) \times \vw = k(\vv \times \vw) + (\vu \times \vw)\). Using the ideas directly above, we can compute the cross product in terms of the components of general vectors. Applying these ideas of right-handed coordinate systems and the cross product working with linear combinations of vectors gives us the following: \[\begin{aligned}\vu\times\vv =\mathstrut \amp (u_1\vi + u_2\vj + u_3\vk) \times (v_1\vi + v_2\vj + v_3\vk) \\ =\mathstrut \amp u_1\vi\times (v_1\vi + v_2\vj + v_3\vk)+u_2\vj\times (v_1\vi + v_2\vj + v_3\vk) \\ \mathstrut \amp +u_3\vk\times (v_1\vi + v_2\vj + v_3\vk) \\ =\mathstrut \amp u_1v_1\vi\times\vi + u_1v_2\vi\times\vj + u_1v_3\vi\times\vk + u_2v_1\vj\times\vi + u_2v_2\vj\times\vj \\ \mathstrut \amp + u_2v_3\vj\times\vk +u_3v_1\vk\times\vi + u_3v_2\vk\times\vj + u_3v_3\vk\times\vk \\ =\mathstrut \amp u_1v_2\vk - u_1v_3\vj - u_2v_1\vk + u_2v_3\vi +u_3v_1\vj - u_3v_2\vi \\ =\mathstrut \amp (u_2v_3-u_3v_2)\vi - (u_1v_3-u_3v_1)\vj + (u_1v_2-u_2v_1)\vk.\end{aligned}\]
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.
As presented, the expression in is complicated, and the variety of subscripts and minus signs may appear difficult to remember. However, we can rewrite this definition using determinants, which allows important structure to emerge. The determinant of a \(2\times2\) matrix is \[\begin{aligned}\end{aligned}\]
It follows that we can thus rewrite Equation in the form \[\begin{aligned}\end{aligned}\]
For those familiar with the determinant of a \(3\times3\) matrix, we write the mnemonic as \[\begin{aligned}\end{aligned}\]
The next activity gives you an opportunity to practice computing cross products and to start investigating its algebraic properties.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
The Vector Nature of the Cross Product
The output of the cross product is a vector and thus has both magnitude and direction. In this section, we will split up the cross product's geometric meaning into these two fundamental properties of vectors. We will begin by first mentioning that the direction of the cross product (as defined in equation ) satisfies the claim need from the introduction to this section.
We now look at the magnitude of \(\vu\times\vv\) in terms of the vector attributes (magnitude and direction) of the input vectors \(\vu\) and \(\vv\).
Note that stated above implies \(\vu\times\vv = \vzero\) if \(\vu\) and \(\vv\) are parallel. If \(\vu\) and \(\vv\) are parallel, then \(\theta=0\) or \(\pi\). Thus, equation tells us that \(\sin(\theta)=0\) implies \(\vu\times\vv = \vzero\).
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Applications of the Cross Product
Equation is also related to the parallelogram formed by two vectors \(\vu\) and \(\vv\), as shown in Figure. We say that \(\vw\) is the outward direction of the parallelogram formed by \(\vu\) and \(\vv\) provided that \(\vw\) is perpendicular (in \(\R^3\)) to the parallelogram and \(\{\vu,\vv,\vw\}\) has a right-handed orientation. You should now convince yourself that in the outward direction is coming directly out of the page (or screen). Remember that the order of our vectors matters for orientation purposes, so the outward direction for the parallelogram formed by \(\vv\) and \(\vu\) would be into the page.
The next activity gives you a chance to practice thinking about how to use the cross product to answer questions about parallelograms.
Activity
Find the area of the parallelogram formed by the vectors \(\vu = \langle 1,3, -2\rangle\) and \(\vv=\langle 3,0,1\rangle\).
Find the area of the parallelogram in \(\R^3\) whose vertices are \((1,0,1)\), \((0,0,1)\), \((2,1,0)\), and \((1,1,0)\).
There is yet one more geometric implication we may draw from . Suppose \(\vu\), \(\vv\), and \(\vw\) are vectors in \(\R^3\) that are not coplanar. (That is, \(\vw\) is not a linear combination of \(\vu\) and \(\vv\).) In this case, \(\vu\), \(\vv\), and \(\vw\) define a three-dimensional solid called a parallelepiped and depicted in Figure.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Comparing the dot and cross products
In this subsection, we will summarize the last few sections by comparing and contrasting several properties of the dot and cross products.
\(\vu\cdot\vv\) is a scalar, while \(\vu\times\vv\) is a vector.
\(\vu\cdot\vv = \vv\cdot\vu\), while \(\vu\times\vv = -\vv\times\vu\)
\(\vu\cdot\vv = \Vert\vu\Vert\Vert\vv\Vert\cos(\theta)\), while \(\Vert\vu\times\vv\Vert = \Vert\vu\Vert\Vert\vv\Vert\sin(\theta)\).
\(\vu\cdot\vv = 0\) if \(\vu\) and \(\vv\) are perpendicular, while \(\vu\times\vv = \vzero\) if \(\vu\) and \(\vv\) are parallel.
Object Types and Vector Notation
In this final subsection, we practice making sense of different expressions involving vectors by describing what kind object the result will be. It is easy to get caught up in all of the manipulations of symbols and new operations and write an expression that looks reasonable but is not defined. For example, \((\vu \cdot \vv) \times \vw\) looks like a nice and reasonable calculation involving vectors (and looks a lot like a scalar triple product!). However, \((\vu \cdot \vv) \times \vw\) is undefined because the quantity in the parentheses is a scalar and you can't take the cross product of a scalar and a vector.
Activity
In this activity, we are focused on the type of objects being used and whether the expression makes sense to do at all. We are not going to worry about interpreting or understanding what is being measured by these expressions.
For each of the expressions below, state whether the result is a scalar, a vector, or undefined. You should write a sentence or two about each to explain your reasoning. (Assume that vectors are nonzero and not orthogonal.)
- \(\frac{\vv}{\vw}\)
- \(\frac{\vv}{\vw\cdot\vv}\)
- \((\vu\times\vw)+\vv\)
- \(k(\vu\cdot\vw)+c\vv\)
- \(k(\vu\cdot\vv)\cdot\vw\)
- \(\frac{\vv}{\vecmag{\vw}}\)
- \((k+\vu)\times \vv\)
- \(k+(\vu\times \vv)\)
- \(\vv+(\vu\cdot \vw)\)
- \(k+\vecmag{\vu\times \vw}\)
Use the operations of dot product and vector subtraction to write an expression involving \(\vu\), \(\vv\), and \(\vw\) that evaluates to a scalar. You can use other operations if you want.
Use the operations of cross product, scalar multiplication, and vector addition to write an expression involving \(\vu\), \(\vv\), and \(\vw\) that evaluates to a vector. You can use other operations if you want.
Use the operations of dot product and vector addition to write an expression involving \(\vu\), \(\vv\), and \(\vw\) that is undefined. You can use other operations if you want.
The Cross Product and Its Properties
The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let \(\text{u}=〈{u}_{1},{u}_{2},{u}_{3}〉\) and \(\text{v}=〈{v}_{1},{v}_{2},{v}_{3}〉\) be nonzero vectors. We want to find a vector \(\text{w}=〈{w}_{1},{w}_{2},{w}_{3}〉\) orthogonal to both \(\text{u}\) and \(\text{v}\)—that is, we want to find \(\text{w}\) such that \(\text{u}\cdot \text{w}=0\) and \(\text{v}\cdot \text{w}=0.\) Therefore, \({w}_{1},\) \({w}_{2},\) and \({w}_{3}\) must satisfy
\[\begin{array}{lll}{u}_{1}{w}_{1}+{u}_{2}{w}_{2}+{u}_{3}{w}_{3} & = & 0 \\ {v}_{1}{w}_{1}+{v}_{2}{w}_{2}+{v}_{3}{w}_{3} & = & 0.\end{array}\]If we multiply the top equation by \({v}_{3}\) and the bottom equation by \({u}_{3}\) and subtract, we can eliminate the variable \({w}_{3},\) which gives
\[({u}_{1}{v}_{3}-{v}_{1}{u}_{3}){w}_{1}+({u}_{2}{v}_{3}-{v}_{2}{u}_{3}){w}_{2}=0.\]If we select
\[\begin{array}{lll}{w}_{1} & = & {u}_{2}{v}_{3}-{u}_{3}{v}_{2} \\ {w}_{2} & = & \text{-}({u}_{1}{v}_{3}-{u}_{3}{v}_{1}),\end{array}\]we get a possible solution vector. Substituting these values back into the original equations gives
\[{w}_{3}={u}_{1}{v}_{2}-{u}_{2}{v}_{1}.\]That is, vector
\[\text{w}=〈{u}_{2}{v}_{3}-{u}_{3}{v}_{2},\text{-}({u}_{1}{v}_{3}-{u}_{3}{v}_{1}),{u}_{1}{v}_{2}-{u}_{2}{v}_{1}〉\]is orthogonal to both \(\text{u}\) and \(\text{v},\) which leads us to define the following operation, called the cross product.
From the way we have developed \(\text{u}\ \times \ \text{v},\) it should be clear that the cross product is orthogonal to both \(\text{u}\) and \(\text{v}.\) However, it never hurts to check. To show that \(\text{u}\ \times \ \text{v}\) is orthogonal to \(\text{u},\) we calculate the dot product of \(\text{u}\) and \(\text{u}\ \times \ \text{v}.\)
\[\text{i}\ \times \ \text{i}=\text{j}\ \times \ \text{j}=\text{k}\ \times \ \text{k}=0.\]Condensed — the full section is in OpenStax Calculus Volume 3.
Determinants and the Cross Product
Using to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using determinant notation.
A \(2\ \times \ 2\) determinant is defined by
\[|\begin{array}{ll}{a}_{1} & {a}_{2} \\ {b}_{1} & {b}_{2}\end{array}|={a}_{1}{b}_{2}-{b}_{1}{a}_{2}.\]For example,
\[|\begin{array}{ll}3 & -2 \\ 5 & 1\end{array}|=3(1)-5(-2)=3+10=13.\]A \(3\ \times \ 3\) determinant is defined in terms of \(2\ \times \ 2\) determinants as follows:
\[|\begin{array}{lll}{a}_{1} & {a}_{2} & {a}_{3} \\ {b}_{1} & {b}_{2} & {b}_{3} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{array}|={a}_{1}|\begin{array}{ll}{b}_{2} & {b}_{3} \\ {c}_{2} & {c}_{3}\end{array}|-{a}_{2}|\begin{array}{ll}{b}_{1} & {b}_{3} \\ {c}_{1} & {c}_{3}\end{array}|+{a}_{3}|\begin{array}{ll}{b}_{1} & {b}_{2} \\ {c}_{1} & {c}_{2}\end{array}|.\]is referred to as the expansion of the determinant along the first row. Notice that the multipliers of each of the \(2\ \times \ 2\) determinants on the right side of this expression are the entries in the first row of the \(3\ \times \ 3\) determinant. Furthermore, each of the \(2\ \times \ 2\) determinants contains the entries from the \(3\ \times \ 3\) determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, \({a}_{1}\) is the multiplier, and the \(2\ \times \ 2\) determinant contains the entries that remain if you cross out the first row and first column of the \(3\ \times \ 3\) determinant. Similarly, for the second term, the multiplier is \({a}_{2},\) and the \(2\ \times \ 2\) determinant contains the entries that remain if you cross out the first row and second column of the \(3\ \times \ 3\) determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.
Example
Try it.
Evaluate the determinant \(|\begin{array}{lll}2 & 5 & -1 \\ -1 & 1 & 3 \\ -2 & 3 & 4\end{array}|.\)
Solution
We have
\[\begin{array}{llllllllllllll}|\begin{array}{lll}2 & 5 & -1 \\ -1 & 1 & 3 \\ -2 & 3 & 4\end{array}| & =2|\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}|-5|\begin{array}{ll}-1 & 3 \\ -2 & 4\end{array}|-1|\begin{array}{ll}-1 & 1 \\ -2 & 3\end{array}| \\ & =2(4-9)-5(-4+6)-1(-3+2) \\ & =2(-5)-5(2)-1(-1)=-10-10+1 \\ & =-19.\end{array}\]Condensed — the full section is in OpenStax Calculus Volume 3.
Using the Cross Product
The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped. The following examples illustrate these calculations.
Example
Try it.
Let \(\text{a}=〈5,2,-1〉\) and \(\text{b}=〈0,-1,4〉.\) Find a unit vector orthogonal to both \(\text{a}\) and \(\text{b}.\)
Solution
The cross product \(\text{a}\ \times \ \text{b}\) is orthogonal to both vectors \(\text{a}\) and \(\text{b}.\) We can calculate it with a determinant:
\[\begin{array}{llllllllllllll}\text{a}\ \times \ \text{b} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ 5 & 2 & -1 \\ 0 & -1 & 4\end{array}|=|\begin{array}{ll}2 & -1 \\ -1 & 4\end{array}|\text{i}-|\begin{array}{ll}5 & -1 \\ 0 & 4\end{array}|\text{j}+|\begin{array}{ll}5 & 2 \\ 0 & -1\end{array}|\text{k} \\ & =(8-1)\text{i}-(20-0)\text{j}+(-5-0)\text{k} \\ & =7\text{i}-20\text{j}-5\text{k}.\end{array}\]Normalize this vector to find a unit vector in the same direction:
\[‖\text{a}\ \times \ \text{b}‖=\sqrt{{(7)}^{2}+{(-20)}^{2}+{(-5)}^{2}}=\sqrt{474}.\]Thus, \(〈\frac{7}{\sqrt{474}},\frac{-20}{\sqrt{474}},\frac{-5}{\sqrt{474}}〉\) is a unit vector orthogonal to \(\text{a}\) and \(\text{b}.\)
To use the cross product for calculating areas, we state and prove the following theorem.
Condensed — the full section is in OpenStax Calculus Volume 3.
The Triple Scalar Product
Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.
The area of the base of the parallelepiped is given by \(‖\text{v}\ \times \ \text{w}‖.\) The height of the figure is given by \(‖{\text{proj}}_{\text{v\times w}}\text{u}‖.\) The volume of the parallelepiped is the product of the height and the area of the base, so we have
\[\begin{array}{ll}V & =‖{\text{proj}}_{\text{v}\ \times \ \text{w}}\text{u}‖‖\text{v}\ \times \ \text{w}‖ \\ & =|\frac{\text{u}\cdot (\text{v}\ \times \ \text{w})}{‖\text{v}\ \times \ \text{w}‖}|‖\text{v}\ \times \ \text{w}‖ \\ & =|\text{u}\cdot (\text{v}\ \times \ \text{w})|.\end{array}\]□
Example
Try it.
Let \(\text{u}=〈-1,-2,1〉,\text{v}=〈4,3,2〉,\ \text{and}\ \text{w}=〈0,-5,-2〉.\) Find the volume of the parallelepiped with adjacent edges \(\text{u},\text{v},\ \text{and}\ \text{w}\) ().
Solution
We have
\[\begin{array}{llllllllllllll}\text{u}\cdot (\text{v}\ \times \ \text{w}) & =|\begin{array}{lll}-1 & -2 & 1 \\ 4 & 3 & 2 \\ 0 & -5 & -2\end{array}|=(-1)|\begin{array}{ll}3 & 2 \\ -5 & -2\end{array}|+2|\begin{array}{ll}4 & 2 \\ 0 & -2\end{array}|+|\begin{array}{ll}4 & 3 \\ 0 & -5\end{array}| \\ & =(-1)(-6+10)+2(-8-0)+(-20-0) \\ & =-4-16-20 \\ & =-40.\end{array}\]Thus, the volume of the parallelepiped is \(|-40|=40\) units3.
Condensed — the full section is in OpenStax Calculus Volume 3.
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).
Example
Try it.
Use the triple scalar product to show that vectors \(\text{u}=〈2,0,5〉,\text{v}=〈2,2,4〉,\ \text{and}\ \text{w}=〈1,-1,3〉\) are coplanar—that is, show that these vectors lie in the same plane.
Solution
Start by calculating the triple scalar product to find the volume of the parallelepiped defined by \(\text{u},\text{v},\ \text{and}\ \text{w}\text{:}\)
\[\begin{array}{llllllll}\text{u}\cdot (\text{v}\ \times \ \text{w}) & =|\begin{array}{lll}2 & 0 & 5 \\ 2 & 2 & 4 \\ 1 & -1 & 3\end{array}| \\ & =[2(2)(3)+(0)(4)(1)+5(2)(-1)]-[5(2)(1)+(2)(4)(-1)+(0)(2)(3)] \\ & =2-2 \\ & =0.\end{array}\]The volume of the parallelepiped is \(0\) units3, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.
Example
Try it.
Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points \(P=(9,-3,-2),Q=(1,3,0),\) and \(R=(-2,5,0).\)
Solution
The plane must contain vectors \(\overset{\to}{PQ}\) and \(\overset{\to}{QR}\text{:}\)
\[\begin{array}{l}\overset{\to}{PQ}=〈1-9,3-(-3),0-(-2)〉=〈-8,6,2〉 \\ \overset{\to}{QR}=〈-2-1,5-3,0-0〉=〈-3,2,0〉.\end{array}\]The cross product \(\overset{\to}{PQ}\ \times \ \overset{\to}{QR}\) produces a vector orthogonal to both \(\overset{\to}{PQ}\) and \(\overset{\to}{QR}.\) Therefore, the cross product is orthogonal to the plane that contains these two vectors:
\[\begin{array}{llllllll}\overset{\to}{PQ}\ \times \ \overset{\to}{QR} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ -8 & 6 & 2 \\ -3 & 2 & 0\end{array}| \\ & =0\text{i}-6\text{j}-16\text{k}-(-18\text{k}+4\text{i}+0\text{j}) \\ & =-4\text{i}-6\text{j}+2\text{k}.\end{array}\]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.
Condensed — the full section is in OpenStax Calculus Volume 3.
Key Concepts
- The cross product \(\text{u}\ \times \ \text{v}\) of two vectors \(\text{u}=〈{u}_{1},{u}_{2},{u}_{3}〉\) and \(\text{v}=〈{v}_{1},{v}_{2},{v}_{3}〉\) is a vector orthogonal to both \(\text{u}\) and \(\text{v}.\) Its length is given by \(‖\text{u}\ \times \ \text{v}‖=‖\text{u}‖\cdot ‖\text{v}‖\cdot \text{sin}\ \theta ,\) where \(\theta\) is the angle between \(\text{u}\) and \(\text{v}.\) Its direction is given by the right-hand rule.
- The algebraic formula for calculating the cross product of two vectors,
\(\text{u}=〈{u}_{1},{u}_{2},{u}_{3}〉\ \text{and}\ \text{v}=〈{v}_{1},{v}_{2},{v}_{3}〉,\) is
\(\text{u}\ \times \ \text{v}=({u}_{2}{v}_{3}-{u}_{3}{v}_{2})\text{i}-({u}_{1}{v}_{3}-{u}_{3}{v}_{1})\text{j}+({u}_{1}{v}_{2}-{u}_{2}{v}_{1})\text{k}.\) - The cross product satisfies the following properties for vectors \(\text{u},\text{v},\ \text{and}\ \text{w},\) and scalar \(c\text{:}\)
- \(\text{u}\ \times \ \text{v}=\text{-}(\text{v}\ \times \ \text{u})\)
- \(\text{u}\ \times \ (\text{v}+\text{w})=\text{u}\ \times \ \text{v}+\text{u}\ \times \ \text{w}\)
- \(c(\text{u}\ \times \ \text{v})=(c\text{u})\ \times \ \text{v}=\text{u}\ \times \ (c\text{v})\)
- \(\text{u}\ \times \ 0=0\ \times \ \text{u}=0\)
- \(\text{v}\ \times \ \text{v}=0\)
- \(\text{u}\cdot (\text{v}\ \times \ \text{w})=(\text{u}\ \times \ \text{v})\cdot \text{w}\)
- The cross product of vectors \(\text{u}=〈{u}_{1},{u}_{2},{u}_{3}〉\) and \(\text{v}=〈{v}_{1},{v}_{2},{v}_{3}〉\) is the determinant \(|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ {u}_{1} & {u}_{2} & {u}_{3} \\ {v}_{1} & {v}_{2} & {v}_{3}\end{array}|.\)
- If vectors \(\text{u}\) and \(\text{v}\) form adjacent sides of a parallelogram, then the area of the parallelogram is given by \(‖\text{u}\ \times \ \text{v}‖.\)
- The triple scalar product of vectors \(\text{u},\) \(\text{v},\) and \(\text{w}\) is \(\text{u}\cdot (\text{v}\ \times \ \text{w}).\)
- The volume of a parallelepiped with adjacent edges given by vectors \(\text{u},\text{v},\ \text{and}\ \text{w}\) is \(V=|\text{u}\cdot (\text{v}\ \times \ \text{w})|.\)
- If the triple scalar product of vectors \(\text{u},\text{v},\ \text{and}\ \text{w}\) is zero, then the vectors are coplanar. The converse is also true: If the vectors are coplanar, then their triple scalar product is zero.
- The cross product can be used to identify a vector orthogonal to two given vectors or to a plane.
- Torque \(\tau\) measures the tendency of a force to produce rotation about an axis of rotation. If force \(\text{F}\) is acting at a distance \(\text{r}\) from the axis, then torque is equal to the cross product of \(\text{r}\) and \(\text{F}\text{:}\) \(\tau =\text{r}\ \times \ \text{F}.\)
Key Equations
| The cross product of two vectors in terms of the unit vectors | \(\text{u}\ \times \ \text{v}=({u}_{2}{v}_{3}-{u}_{3}{v}_{2})\text{i}-({u}_{1}{v}_{3}-{u}_{3}{v}_{1})\text{j}+({u}_{1}{v}_{2}-{u}_{2}{v}_{1})\text{k}\) |
The Cross Product
For the following exercises, the vectors \(\text{u}\) and \(\text{v}\) are given.
- Find the cross product \(\text{u}\ \times \ \text{v}\) of the vectors \(\text{u}\) and \(\text{v}.\) Express the answer in component form.
- Sketch the vectors \(\text{u},\text{v},\) and \(\text{u}\ \times \ \text{v}.\)
In the following exercises, vectors \(\text{u}\) and \(\text{v}\) are given. Find unit vector \(\text{w}\) in the direction of the cross product vector \(\text{u}\ \times \ \text{v}.\) Express your answer using standard unit vectors.
For the following exercises, the vectors \(\text{u}\) and \(\text{v}\) are given. Use determinant notation to find vector \(\text{w}\) orthogonal to vectors \(\text{u}\) and \(\text{v}.\)
In the following exercises, vectors \(\text{u},\text{v},\text{and}\ \text{w}\) are given.
- Find the triple scalar product \(\text{u}\cdot (\text{v}\ \times \ \text{w}).\)
- Find the volume of the parallelepiped with the adjacent edges \(\text{u},\text{v},\text{and}\ \text{w}.\)
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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In , we stated Because the cross product works well with linear combinations, \(\vu\times\vv\) is perpendicular to any linear combination of \(\vu\) and \(\vv\). We want to verify this algebraically here. Specifically, show that \(\vu\times\vv\) will be orthogonal to any vector of the form \(a\vu+b\vv\).
Jawaabta muuji
\((\vu\times\vv)\cdot(a\vu+b\vv)=a[(\vu\times\vv)\cdot\vu] + b[(\vu\times\vv)\cdot\vv] = a [0]+b[0]\)
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Let \(\text{p}=〈-1,2,5〉\ \text{and}\ \text{q}=〈4,0,-3〉\) (). Find \(\text{p}\ \times \ \text{q}.\)
Jawaabta muuji
Substitute the components of the vectors into :
\[\begin{array}{ll}\text{p}\ \times \ \text{q} & =⟨-1,2,5⟩\ \times \ ⟨4,0,-3⟩ \\ & =⟨{p}_{2}{q}_{3}-{p}_{3}{q}_{2},-\left({p}_{1}{q}_{3}-{p}_{3}{q}_{1}\right),{p}_{1}{q}_{2}-{p}_{2}{q}_{1}⟩ \\ & =⟨{p}_{2}{q}_{3}-{p}_{3}{q}_{2},{p}_{3}{q}_{1}-{p}_{1}{q}_{3},{p}_{1}{q}_{2}-{p}_{2}{q}_{1}⟩ \\ & =⟨2(-3)-5(0),5\left(4\right)-(-1)(-3),(-1)0-2(4)⟩ \\ & =⟨-6,17,-8⟩.\end{array}\] -
Find \(\text{p}\ \times \ \text{q}\) for \(\text{p}=〈5,1,2〉\) and \(\text{q}=〈-2,0,1〉.\) Express the answer using standard unit vectors.
Jawaabta muuji
\(\text{i}-9\text{j}+2\text{k}\)
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Let \(\text{u}=〈0,2,1〉\) and \(\text{v}=〈3,-1,0〉.\) Calculate \(\text{u}\ \times \ \text{v}\) and \(\text{v}\ \times \ \text{u}\) and graph them.
Jawaabta muuji
We have
\[\begin{array}{lll}\text{u}\ \times \ \text{v} & = & 〈(0+1),\text{-}(0-3),(0-6)〉=〈1,3,-6〉 \\ \text{v}\ \times \ \text{u} & = & 〈(-1-0),\text{-}(3-0),(6-0)〉=〈-1,-3,6〉.\end{array}\]We see that, in this case, \(\text{u}\ \times \ \text{v}=\text{-}(\text{v}\ \times \ \text{u})\) (). We prove this in general later in this section.
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Suppose vectors \(\text{u}\) and \(\text{v}\) lie in the xy-plane (the z-component of each vector is zero). Now suppose the x- and y-components of \(\text{u}\) and the y-component of \(\text{v}\) are all positive, whereas the x-component of \(\text{v}\) is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does \(\text{u}\ \times \ \text{v}\) point?
Jawaabta muuji
Up (the positive z-direction)
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Find \(\text{i}\ \times \ (\text{j}\ \times \ \text{k}).\)
Jawaabta muuji
We know that \(\text{j}\ \times \ \text{k}=\text{i}.\) Therefore, \(\text{i}\ \times \ (\text{j}\ \times \ \text{k})=\text{i}\ \times \ \text{i}=0.\)
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Find \((\text{i}\ \times \ \text{j})\ \times \ (\text{k}\ \times \ \text{i}).\)
Jawaabta muuji
\(\text{-}\text{i}\)
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Use the cross product properties to calculate \((2\text{i}\ \times \ 3\text{j})\ \times \ \text{j}.\)
Jawaabta muuji
\[\begin{array}{ll}(2\text{i}\ \times \ 3\text{j})\ \times \ \text{j} & =2(\text{i}\ \times \ 3\text{j})\ \times \ \text{j} \\ & =2(3)(\text{i}\ \times \ \text{j})\ \times \ \text{j} \\ & =(6\text{k})\ \times \ \text{j} \\ & =6(\text{k}\ \times \ \text{j}) \\ & =6(\text{-}\text{i})=-6\text{i}.\end{array}\] -
Use the properties of the cross product to calculate \((\text{i}\ \times \ \text{k})\ \times \ (\text{k}\ \times \ \text{j}).\)
Jawaabta muuji
\(\text{-}\text{k}\)
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Use to find the magnitude of the cross product of \(\text{u}=〈0,4,0〉\) and \(\text{v}=〈0,0,-3〉.\)
Jawaabta muuji
We have
\[\begin{array}{ll}‖\text{u}\ \times \ \text{v}‖ & =‖\text{u}‖\cdot ‖\text{v}‖\cdot \text{sin}\ \theta \\ & =\sqrt{{0}^{2}+{4}^{2}+{0}^{2}}\cdot \sqrt{{0}^{2}+{0}^{2}+{(-3)}^{2}}\cdot \text{sin}\ \frac{\pi }{2} \\ & =4(3)(1)=12.\end{array}\] -
Use to find the magnitude of \(\text{u}\ \times \ \text{v},\) where \(\text{u}=〈-8,0,0〉\) and \(\text{v}=〈0,2,0〉.\)
Jawaabta muuji
\(16\)
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Evaluate the determinant \(|\begin{array}{lll}2 & 5 & -1 \\ -1 & 1 & 3 \\ -2 & 3 & 4\end{array}|.\)
Jawaabta muuji
We have
\[\begin{array}{llllllllllllll}|\begin{array}{lll}2 & 5 & -1 \\ -1 & 1 & 3 \\ -2 & 3 & 4\end{array}| & =2|\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}|-5|\begin{array}{ll}-1 & 3 \\ -2 & 4\end{array}|-1|\begin{array}{ll}-1 & 1 \\ -2 & 3\end{array}| \\ & =2(4-9)-5(-4+6)-1(-3+2) \\ & =2(-5)-5(2)-1(-1)=-10-10+1 \\ & =-19.\end{array}\] -
Evaluate the determinant \(|\begin{array}{lll}1 & -2 & -1 \\ 3 & 2 & -3 \\ 1 & 5 & 4\end{array}|.\)
Jawaabta muuji
\(40\)
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Let \(\text{p}=〈-1,2,5〉\) and \(\text{q}=〈4,0,-3〉.\) Find \(\text{p}\ \times \ \text{q}.\)
Jawaabta muuji
We set up our determinant by putting the standard unit vectors across the first row, the components of \(\text{u}\) in the second row, and the components of \(\text{v}\) in the third row. Then, we have
\[\begin{array}{llllllllllllll}\text{p}\ \times \ \text{q} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ -1 & 2 & 5 \\ 4 & 0 & -3\end{array}|=|\begin{array}{ll}2 & 5 \\ 0 & -3\end{array}|\text{i}-|\begin{array}{ll}-1 & 5 \\ 4 & -3\end{array}|\text{j}+|\begin{array}{ll}-1 & 2 \\ 4 & 0\end{array}|\text{k} \\ & =(-6-0)\text{i}-(3-20)\text{j}+(0-8)\text{k} \\ & =-6\text{i}+17\text{j}-8\text{k}.\end{array}\]Notice that this answer confirms the calculation of the cross product in .
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Use determinant notation to find \(\text{a}\ \times \ \text{b},\) where \(\text{a}=〈8,2,3〉\) and \(\text{b}=〈-1,0,4〉.\)
Jawaabta muuji
\(8\text{i}-35\text{j}+2\text{k}\)
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Let \(\text{a}=〈5,2,-1〉\) and \(\text{b}=〈0,-1,4〉.\) Find a unit vector orthogonal to both \(\text{a}\) and \(\text{b}.\)
Jawaabta muuji
The cross product \(\text{a}\ \times \ \text{b}\) is orthogonal to both vectors \(\text{a}\) and \(\text{b}.\) We can calculate it with a determinant:
\[\begin{array}{llllllllllllll}\text{a}\ \times \ \text{b} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ 5 & 2 & -1 \\ 0 & -1 & 4\end{array}|=|\begin{array}{ll}2 & -1 \\ -1 & 4\end{array}|\text{i}-|\begin{array}{ll}5 & -1 \\ 0 & 4\end{array}|\text{j}+|\begin{array}{ll}5 & 2 \\ 0 & -1\end{array}|\text{k} \\ & =(8-1)\text{i}-(20-0)\text{j}+(-5-0)\text{k} \\ & =7\text{i}-20\text{j}-5\text{k}.\end{array}\]Normalize this vector to find a unit vector in the same direction:
\[‖\text{a}\ \times \ \text{b}‖=\sqrt{{(7)}^{2}+{(-20)}^{2}+{(-5)}^{2}}=\sqrt{474}.\]Thus, \(〈\frac{7}{\sqrt{474}},\frac{-20}{\sqrt{474}},\frac{-5}{\sqrt{474}}〉\) is a unit vector orthogonal to \(\text{a}\) and \(\text{b}.\)
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Find a unit vector orthogonal to both \(\text{a}\) and \(\text{b},\) where \(\text{a}=〈4,0,3〉\) and \(\text{b}=〈1,1,4〉.\)
Jawaabta muuji
\(〈\frac{-3}{\sqrt{194}},\frac{-13}{\sqrt{194}},\frac{4}{\sqrt{194}}〉\)
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Let \(P=(1,0,0),Q=(0,1,0),\ \text{and}\ R=(0,0,1)\) be the vertices of a triangle (). Find its area.
Jawaabta muuji
We have \(\overset{\to}{PQ}=〈0-1,1-0,0-0〉=〈-1,1,0〉\) and \(\overset{\to}{PR}=〈0-1,0-0,1-0〉=〈-1,0,1〉.\) The area of the parallelogram with adjacent sides \(\overset{\to}{PQ}\) and \(\overset{\to}{PR}\) is given by \(‖\overset{\to}{PQ}\ \times \ \overset{\to}{PR}‖\text{:}\)
\[\begin{array}{lllllllll}\overset{\to}{PQ}\ \times \ \overset{\to}{PR} & = & |\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}|=(1-0)\text{i}-(-1-0)\text{j}+(0-(-1))\text{k}=\text{i}+\text{j}+\text{k} \\ ‖\overset{\to}{PQ}\ \times \ \overset{\to}{PR}‖ & = & ‖〈1,1,1〉‖=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=\sqrt{3}.\end{array}\]The area of \(\text{\Delta }PQR\) is half the area of the parallelogram, or \(\sqrt{3}\text{/}2.\)
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Find the area of the parallelogram \(PQRS\) with vertices \(P(1,1,0),Q(7,1,0),R(9,4,2),\) and \(S(3,4,2).\)
Jawaabta muuji
\(6\sqrt{13}\)
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Let \(\text{u}=〈1,3,5〉,\text{v}=〈2,-1,0〉\ \text{and}\ \text{w}=〈-3,0,-1〉.\) Calculate the triple scalar product \(\text{u}\cdot (\text{v}\ \times \ \text{w}).\)
Jawaabta muuji
Apply directly:
\[\begin{array}{llllllll}\text{u}\cdot (\text{v}\ \times \ \text{w}) & =|\begin{array}{lll}1 & 3 & 5 \\ 2 & -1 & 0 \\ -3 & 0 & -1\end{array}| \\ & =1|\begin{array}{ll}-1 & 0 \\ 0 & -1\end{array}|-3|\begin{array}{ll}2 & 0 \\ -3 & -1\end{array}|+5|\begin{array}{ll}2 & -1 \\ -3 & 0\end{array}| \\ & =(1-0)-3(-2-0)+5(0-3) \\ & =1+6-15=-8.\end{array}\] -
Calculate the triple scalar product \(\text{a}\cdot (\text{b}\ \times \ \text{c}),\) where \(\text{a}=〈2,-4,1〉,\) \(\text{b}=〈0,3,-1〉,\) and \(\text{c}=〈5,-3,3〉.\)
Jawaabta muuji
\(17\)
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Let \(\text{u}=〈-1,-2,1〉,\text{v}=〈4,3,2〉,\ \text{and}\ \text{w}=〈0,-5,-2〉.\) Find the volume of the parallelepiped with adjacent edges \(\text{u},\text{v},\ \text{and}\ \text{w}\) ().
Jawaabta muuji
We have
\[\begin{array}{llllllllllllll}\text{u}\cdot (\text{v}\ \times \ \text{w}) & =|\begin{array}{lll}-1 & -2 & 1 \\ 4 & 3 & 2 \\ 0 & -5 & -2\end{array}|=(-1)|\begin{array}{ll}3 & 2 \\ -5 & -2\end{array}|+2|\begin{array}{ll}4 & 2 \\ 0 & -2\end{array}|+|\begin{array}{ll}4 & 3 \\ 0 & -5\end{array}| \\ & =(-1)(-6+10)+2(-8-0)+(-20-0) \\ & =-4-16-20 \\ & =-40.\end{array}\]Thus, the volume of the parallelepiped is \(|-40|=40\) units3.
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Find the volume of the parallelepiped formed by the vectors \(\text{a}=3\text{i}+4\text{j}-\text{k},\) \(\text{b}=2\text{i}-\text{j}-\text{k},\) and \(\text{c}=3\text{j}+\text{k}.\)
Jawaabta muuji
\(8\) units3
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Use the triple scalar product to show that vectors \(\text{u}=〈2,0,5〉,\text{v}=〈2,2,4〉,\ \text{and}\ \text{w}=〈1,-1,3〉\) are coplanar—that is, show that these vectors lie in the same plane.
Jawaabta muuji
Start by calculating the triple scalar product to find the volume of the parallelepiped defined by \(\text{u},\text{v},\ \text{and}\ \text{w}\text{:}\)
\[\begin{array}{llllllll}\text{u}\cdot (\text{v}\ \times \ \text{w}) & =|\begin{array}{lll}2 & 0 & 5 \\ 2 & 2 & 4 \\ 1 & -1 & 3\end{array}| \\ & =[2(2)(3)+(0)(4)(1)+5(2)(-1)]-[5(2)(1)+(2)(4)(-1)+(0)(2)(3)] \\ & =2-2 \\ & =0.\end{array}\]The volume of the parallelepiped is \(0\) units3, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.
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Are the vectors \(\text{a}=\text{i}+\text{j}-\text{k},\) \(\text{b}=\text{i}-\text{j}+\text{k},\) and \(\text{c}=\text{i}+\text{j}+\text{k}\) coplanar?
Jawaabta muuji
No, the triple scalar product is \(-4\ne 0,\) so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.
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Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points \(P=(9,-3,-2),Q=(1,3,0),\) and \(R=(-2,5,0).\)
Jawaabta muuji
The plane must contain vectors \(\overset{\to}{PQ}\) and \(\overset{\to}{QR}\text{:}\)
\[\begin{array}{l}\overset{\to}{PQ}=〈1-9,3-(-3),0-(-2)〉=〈-8,6,2〉 \\ \overset{\to}{QR}=〈-2-1,5-3,0-0〉=〈-3,2,0〉.\end{array}\]The cross product \(\overset{\to}{PQ}\ \times \ \overset{\to}{QR}\) produces a vector orthogonal to both \(\overset{\to}{PQ}\) and \(\overset{\to}{QR}.\) Therefore, the cross product is orthogonal to the plane that contains these two vectors:
\[\begin{array}{llllllll}\overset{\to}{PQ}\ \times \ \overset{\to}{QR} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ -8 & 6 & 2 \\ -3 & 2 & 0\end{array}| \\ & =0\text{i}-6\text{j}-16\text{k}-(-18\text{k}+4\text{i}+0\text{j}) \\ & =-4\text{i}-6\text{j}+2\text{k}.\end{array}\] -
A bolt is tightened by applying a force of \(6\) N to a 0.15-m wrench (). The angle between the wrench and the force vector is \(40\text{^{\circ}}.\) Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.
Jawaabta muuji
Substitute the given information into the equation defining torque:
\[‖\tau ‖=‖\text{r}\ \times \ \text{F}‖=‖\text{r}‖‖\text{F}‖\text{sin}\ \theta =(0.15\ \text{m})(6\ \text{N})\text{sin}\ 40\text{^{\circ}}\approx 0.58\ \text{N}\cdot \text{m}.\] -
Calculate the force required to produce \(15\ \text{N}\cdot \text{m}\) torque at an angle of \(30º\) from a 150-cm rod.
Jawaabta muuji
\(20\) N
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\(\text{u}=〈2,0,0〉,\) \(\text{v}=〈2,2,0〉\)
Jawaabta muuji
a. \(\text{u}\ \times \ \text{v}=〈0,0,4〉;\)
b. -
\(\text{u}=〈3,2,-1〉,\) \(\text{v}=〈1,1,0〉\)
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\(\text{u}=2\text{i}+3\text{j},\) \(\text{v}=\text{j}+2\text{k}\)
Jawaabta muuji
a. \(\text{u}\ \times \ \text{v}=〈6,-4,2〉;\)
b. -
\(\text{u}=2\text{j}+3\text{k},\) \(\text{v}=3\text{i}+\text{k}\)
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Simplify \((\text{i}\ \times \ \text{i}-2\text{i}\ \times \ \text{j}-4\text{i}\ \times \ \text{k}+3\text{j}\ \times \ \text{k})\ \times \ \text{i}.\)
Jawaabta muuji
\(-2\text{j}-4\text{k}\)
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Simplify \(\text{j}\ \times \ (\text{k}\ \times \ \text{j}+2\text{j}\ \times \ \text{i}-3\text{j}\ \times \ \text{j}+5\text{i}\ \times \ \text{k}).\)
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\(\text{u}=〈3,-1,2〉,\) \(\text{v}=〈-2,0,1〉\)
Jawaabta muuji
\(\text{w}=-\frac{1}{3\sqrt{6}}\text{i}-\frac{7}{3\sqrt{6}}\text{j}-\frac{2}{3\sqrt{6}}\text{k}\)
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\(\text{u}=〈2,6,1〉,\) \(\text{v}=〈3,0,1〉\)
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\(\text{u}=\overset{\to}{AB},\) \(\text{v}=\overset{\to}{AC},\) where \(A(1,0,1),\) \(B(1,-1,3),\) and \(C(0,0,5)\)
Jawaabta muuji
\(\text{w}=-\frac{4}{\sqrt{21}}\text{i}-\frac{2}{\sqrt{21}}\text{j}-\frac{1}{\sqrt{21}}\text{k}\)
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\(\text{u}=\overset{\to}{OP},\) \(\text{v}=\overset{\to}{PQ},\) where \(P(-1,1,0)\) and \(Q(0,2,1)\)
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Determine the real number \(\alpha\) such that \(\text{u}\ \times \ \text{v}\) and \(\text{i}\) are orthogonal, where \(\text{u}=3\text{i}+\text{j}-5\text{k}\) and \(\text{v}=4\text{i}-2\text{j}+\alpha \text{k}.\)
Jawaabta muuji
\(\alpha =10\)
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Show that \(\text{u}\ \times \ \text{v}\) and \(2\text{i}-14\text{j}+2\text{k}\) cannot be orthogonal for any \(\alpha\) real number, where \(\text{u}=\text{i}+7\text{j}-\text{k}\) and \(\text{v}=\alpha \text{i}+5\text{j}+\text{k}.\)
Symbols used here
Ratio of a circle's circumference to its diameter, 3.14159…
The usual name for an angle.
Ratios of sides in a right triangle; coordinates on the unit circle.
Chance of A; chance of A given that B happened.
i² = −1.
1/360 of a full turn. 180° = π radians.
Inequalities that allow equality; < and > exclude it.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Antiderivative (indefinite) or signed area from a to b (definite).
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: The Cross Product
- Calculate the cross product of two given vectors.
- Use determinants to calculate a cross product.
- Find a vector orthogonal to two given vectors.
- Determine areas and volumes by using the cross product.
- Calculate the torque of a given force and position vector.
- The cross product
- The algebraic formula for calculating the cross product of two vectors,
- The cross product satisfies the following properties for vectors
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Ku day inaad ku
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
In ka badan Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems