Lesson Explainer: Cross Product in 3D Mathematics

In this explainer, we will learn how to find the cross product of two vectors in space and how to use it to find the area of geometric shapes.

There are two ways to multiply vectors together. You may already be familiar with the dot product, also called scalar product. This product leads to a scalar quantity that is given by the product of the magnitudes of both vectors multiplied by the cosine of the angle between the two vectors. As for the cross product, it is a multiplication of vectors that leads to a vector.

Definition: Cross Product

The cross product of two vectors ⃑𝐴 and ⃑𝐡 is another vector given by ⃑𝐴×⃑𝐡=‖‖⃑𝐴‖‖‖‖⃑𝐡‖‖|πœƒ|⃑𝑒,sin where πœƒ is the angle between ⃑𝐴 and ⃑𝐡 and ⃑𝑒 is a unit vector perpendicular to the plane that contains ⃑𝐴 and ⃑𝐡. The direction of ⃑𝑒 is given by the so-called right-hand rule.

The right-hand rule is a mnemotechnic method using the right hand that allows us to find the direction (up or down) of ⃑𝐴×⃑𝐡 with respect to the plane defined by ⃑𝐴 and ⃑𝐡 (considered horizontal).

Taking our right hand, aligning our first finger with ⃑𝐴 and the second with ⃑𝐡, the direction of the thumbβ€”up or downβ€”indicates the direction of ⃑𝐴×⃑𝐡.

In a slightly different version, we use a half-closed right hand with the fingers showing the angle from ⃑𝐴 to ⃑𝐡, and the thumb then gives the direction of ⃑𝐴×⃑𝐡.

The direction of the vector ⃑𝐴×⃑𝐡 is also given by the movement up or down of, for instance, a bottle lid that one would turn in the same sense of rotation as when going from ⃑𝐴 to ⃑𝐡. One unscrews (opens) a bottle when turning the lid counterclockwise: the lid goes up (if the bottle is vertical), and vice versa. This is illustrated here with a nut, where the symbol βŠ™ means a vector perpendicular to the plane of the paper/screen pointing toward us (the dot in the center suggests we see the tip of the arrow), and the symbol βŠ— means a vector perpendicular to the plane of the paper/screen pointing to the other side of the plane with respect to us.

All these rules show the same thing; choose the rule that you prefer and stick to itβ€”it will be useful not only in math but also in physics.

In a three-dimensional Cartesian coordinate system, the right-handed set of unit vectors (⃑𝑖,⃑𝑗,βƒ‘π‘˜) is such that βƒ‘π‘˜=⃑𝑖×⃑𝑗, ⃑𝑖=βƒ‘π‘—Γ—βƒ‘π‘˜, and ⃑𝑗=βƒ‘π‘˜Γ—βƒ‘π‘–.

One can easily retrieve these equations using what we have shown so far. Take, for instance, βƒ‘π‘—Γ—βƒ‘π‘˜. Using the diagram above, if we align our index with ⃑𝑗 and middle finger with βƒ‘π‘˜, our thumb is pointing toward the right, that is, in the same direction as ⃑𝑖. We could also look at ⃑𝑗 and βƒ‘π‘˜ from, for example, the positive side of ⃑𝑖 and see that the shorter turn from ⃑𝑗 to βƒ‘π‘˜ is counterclockwise, which means that the resulting vector is pointing to us (we unscrew a lid). As ⃑𝑗 and βƒ‘π‘˜ are perpendicular unit vectors, the product of their magnitudes is 1 and the sine of the angle between them 90∘ is 1. So, we find that βƒ‘π‘—Γ—βƒ‘π‘˜=⃑𝑖.

From these rules, we see that ⃑𝐴×⃑𝐡 and ⃑𝐡×⃑𝐴 have opposite directions since the rotation from ⃑𝐴 to ⃑𝐡 is the opposite of that from ⃑𝐡 to ⃑𝐴. Hence, ⃑𝐴×⃑𝐡=βˆ’βƒ‘π΅Γ—βƒ‘π΄.

We say that the cross product is anticommutative.

Therefore, we find that ⃑𝑗×⃑𝑖=βˆ’βƒ‘π‘˜,βƒ‘π‘˜Γ—βƒ‘π‘—=βˆ’βƒ‘π‘–,βƒ‘π‘–Γ—βƒ‘π‘˜=βˆ’βƒ‘π‘—.and

In addition, ⃑𝐴×⃑𝐴=0 since the angle between ⃑𝐴 and ⃑𝐴 is zero and sin0=0. (Also, there is no rotation from ⃑𝐴 to ⃑𝐴, so no movement of the β€œlid,” and no plane can be defined with only one vector, so no perpendicular direction to the plane can be defined either.)

You may have learned already how to find the cross product of two vectors in the plane by calculating a 2Γ—2 determinant with components. There is a similar way with vectors in the space. We can find it by rewriting each vector in terms of its components: ⃑𝐴=𝐴⃑𝑖+𝐴⃑𝑗+π΄βƒ‘π‘˜βƒ‘π΅=𝐡⃑𝑖+𝐡⃑𝑗+π΅βƒ‘π‘˜.ο—ο˜ο™ο—ο˜ο™and

Using the distributive property of the cross product (which we will not prove here), we have ⃑𝐴×⃑𝐡=𝐴⃑𝑖+𝐴⃑𝑗+π΄βƒ‘π‘˜ο‡Γ—ο€»π΅βƒ‘π‘–+𝐡⃑𝑗+π΅βƒ‘π‘˜ο‡=π΄π΅βƒ‘π‘˜βˆ’π΄π΅βƒ‘π‘—βˆ’π΄π΅βƒ‘π‘˜+𝐴𝐡⃑𝑖+π΄π΅βƒ‘π‘—βˆ’π΄π΅βƒ‘π‘–,ο—ο˜ο™ο—ο˜ο™ο—ο˜ο—ο™ο˜ο—ο˜ο™ο™ο—ο™ο˜ since ⃑𝑖×⃑𝑖=⃑𝑗×⃑𝑗=βƒ‘π‘˜Γ—βƒ‘π‘˜=0.

Rearranging, we find ⃑𝐴×⃑𝐡=ο€Ήπ΄π΅βˆ’π΄π΅ο…βƒ‘π‘–βˆ’(π΄π΅βˆ’π΄π΅)⃑𝑗+ο€Ήπ΄π΅βˆ’π΄π΅ο…βƒ‘π‘˜.ο˜ο™ο™ο˜ο—ο™ο™ο—ο—ο˜ο˜ο—

We recognize here the evaluation of a 3Γ—3 determinant. Therefore, the cross product is given by ⃑𝐴×⃑𝐡=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΄π΄π΄π΅π΅π΅|||||.ο—ο˜ο™ο—ο˜ο™

Definition: Cross Product of Two Vectors in the Three-Dimensional Cartesian Coordinate System

For two 3D vectors ⃑𝐴=𝐴⃑𝑖+𝐴⃑𝑗+π΄βƒ‘π‘˜ο—ο˜ο™ and ⃑𝐡=𝐡⃑𝑖+𝐡⃑𝑗+π΅βƒ‘π‘˜ο—ο˜ο™ in the coordinate system 𝑂,⃑𝑖,⃑𝑗,βƒ‘π‘˜ο‡, the cross product of ⃑𝐴 and ⃑𝐡 is ⃑𝐴×⃑𝐡=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΄π΄π΄π΅π΅π΅|||||=ο€Ήπ΄π΅βˆ’π΄π΅ο…βƒ‘π‘–βˆ’(π΄π΅βˆ’π΄π΅)⃑𝑗+ο€Ήπ΄π΅βˆ’π΄π΅ο…βƒ‘π‘˜.ο—ο˜ο™ο—ο˜ο™ο˜ο™ο™ο˜ο—ο™ο™ο—ο—ο˜ο˜ο—

Let us apply this definition of the cross product with a first example.

Example 1: Finding the Cross Product of Two Vectors given Their Components

Let ⃑𝑉=⃑𝑖 and οƒŸπ‘Š=3⃑𝑖+2⃑𝑗+4βƒ‘π‘˜. Calculate βƒ‘π‘‰Γ—οƒŸπ‘Š.

Answer

The vectors are given here in terms of the unit vectors ⃑𝑖, ⃑𝑗, and βƒ‘π‘˜ in a Cartesian coordinate system.

Let us write them in component form: ⃑𝑉=(1,0,0)οƒŸπ‘Š=(3,2,4).and

We know that βƒ‘π‘‰Γ—οƒŸπ‘Š=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π‘‰π‘‰π‘‰π‘Šπ‘Šπ‘Š|||||=||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜100324||||=(0Γ—4βˆ’2Γ—0)βƒ‘π‘–βˆ’(1Γ—4βˆ’0Γ—3)⃑𝑗+(1Γ—2βˆ’3Γ—0)βƒ‘π‘˜=0βƒ‘π‘–βˆ’4⃑𝑗+2βƒ‘π‘˜=(0,βˆ’4,2).ο—ο˜ο™ο—ο˜ο™

In the next example, we are going to do the same, but we will just have to carry out some vector subtraction before.

Example 2: Finding the Cross Product of Two Vectors Obtained by Subtracting Vectors

If ⃑𝐴=(3,4,βˆ’4), ⃑𝐡=(2,5,βˆ’4), and ⃑𝐢=(βˆ’4,βˆ’4,2), find ο€Ίβƒ‘π΄βˆ’βƒ‘π΅ο†Γ—ο€Ίβƒ‘πΆβˆ’βƒ‘π΄ο†.

Answer

Let us first find the components of βƒ‘π΄βˆ’βƒ‘π΅ and βƒ‘πΆβˆ’βƒ‘π΄: βƒ‘π΄βˆ’βƒ‘π΅=(3βˆ’2,4βˆ’5,βˆ’4βˆ’(βˆ’4))=(1,βˆ’1,0),βƒ‘πΆβˆ’βƒ‘π΄=(βˆ’4βˆ’3,βˆ’4βˆ’4,2βˆ’(βˆ’4))=(βˆ’7,βˆ’8,6).

Applying the rule for calculating the cross product of two vectors from their components, we find that ο€Ίβƒ‘π΄βˆ’βƒ‘π΅ο†Γ—ο€Ίβƒ‘πΆβˆ’βƒ‘π΄ο†=||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜1βˆ’10βˆ’7βˆ’86||||=(βˆ’1Γ—6βˆ’(βˆ’8)Γ—0)βƒ‘π‘–βˆ’(1Γ—6βˆ’(βˆ’7)Γ—0)⃑𝑗+(1Γ—(βˆ’8)βˆ’(βˆ’7)Γ—(βˆ’1))βƒ‘π‘˜=βˆ’6βƒ‘π‘–βˆ’6βƒ‘π‘—βˆ’15βƒ‘π‘˜.

In the previous example, the easiest way to find the cross product was first to determine the components of βƒ‘π΄βˆ’βƒ‘π΅ and βƒ‘πΆβˆ’βƒ‘π΄. However, we could also have used the fact that the cross product is distributive: ο€Ίβƒ‘π΄βˆ’βƒ‘π΅ο†Γ—ο€Ίβƒ‘πΆβˆ’βƒ‘π΄ο†=βƒ‘π΄Γ—βƒ‘πΆβˆ’βƒ‘π΄Γ—βƒ‘π΄βˆ’βƒ‘π΅Γ—βƒ‘πΆ+⃑𝐡×⃑𝐴=βƒ‘π΄Γ—βƒ‘πΆβˆ’βƒ‘π΅Γ—βƒ‘πΆ+⃑𝐡×⃑𝐴, since ⃑𝐴×⃑𝐴=0 (the angle between a vector and itself is zero).

The distributivity can be easily shown using the way we calculate the cross product from the vector components: ⃑𝐴+⃑𝐡×⃑𝐢=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΄+𝐡𝐴+𝐡𝐴+𝐡𝐢𝐢𝐢|||||=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΄π΄π΄πΆπΆπΆ|||||+|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΅π΅π΅πΆπΆπΆ|||||.ο—ο—ο˜ο˜ο™ο™ο—ο˜ο™ο—ο˜ο™ο—ο˜ο™ο—ο˜ο™ο—ο˜ο™

Let us now use what we know about the direction of the cross product of two vectors to answer the following question.

Example 3: Finding the Unit Vectors Orthogonal to Two Given Vectors

Find the unit vectors that are perpendicular to both ⃑𝐴=(4,2,0) and ⃑𝐡=(4,6,βˆ’4).

Answer

We know that the cross product of two vectors is a vector perpendicular to both vectors (and in fact to any combination of those two vectors, that is, to any vector contained in the plane defined by the two vectors). Finding the unit vectors (i.e., with a magnitude of 1) collinear to this vector will give us our solutions, with one vector in the same direction as the cross product of ⃑𝐴 and ⃑𝐡 and the other with an opposite direction.

Let us first find ⃑𝐴×⃑𝐡: ⃑𝐴×⃑𝐡=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΄π΄π΄π΅π΅π΅|||||=||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜42046βˆ’4||||=(2Γ—(βˆ’4)βˆ’6Γ—0)βƒ‘π‘–βˆ’(4Γ—(βˆ’4)βˆ’4Γ—0)⃑𝑗+(4Γ—6βˆ’4Γ—2)βƒ‘π‘˜=βˆ’8⃑𝑖+16⃑𝑗+16βƒ‘π‘˜.ο—ο˜ο™ο—ο˜ο™

The unit vector in the same direction of ⃑𝐴×⃑𝐡 is ⃑𝐴×⃑𝐡‖‖⃑𝐴×⃑𝐡‖‖=βˆ’8⃑𝑖+16⃑𝑗+16βƒ‘π‘˜βˆš64+256+256=βˆ’8⃑𝑖+16⃑𝑗+16βƒ‘π‘˜24=βˆ’βƒ‘π‘–+2⃑𝑗+2βƒ‘π‘˜3=ο€Όβˆ’13,23,23.

And the unit vector in the opposite direction is simply this one multiplied by βˆ’1, that is, ο€Ό13,βˆ’23,βˆ’23.

The cross product of two vectors has geometric properties other than the fact that it is perpendicular to the plane defined by the two vectors: its magnitude is the area of the parallelogram spanned by the two vectors.

If we look at two vectors ⃑𝐴 and ⃑𝐡 as shown in the diagram, we see that ‖‖⃑𝐡‖‖|πœƒ|=π‘‚π‘Œsin, that is, the height of the parallelogram spanned by ⃑𝐴 and ⃑𝐡. Hence, ‖‖⃑𝐴‖‖‖‖⃑𝐡‖‖|πœƒ|sin is the area of the parallelogram 𝑂𝐴𝐢𝐡. We take here the absolute value of the sine of πœƒ as πœƒ is a directed angle. If we consider ⃑𝐴×⃑𝐡 here, angle πœƒ is defined as the angle from ⃑𝐴 to ⃑𝐡; it is a negative angle and its sine is negative. Or, if one uses only positive angles, πœƒ is an angle whose measure is between 180∘ and 360∘ (namely, 360βˆ’βˆ π΄π‘‚π΅βˆ˜); its sine is thus negative.

Let us use this to find an area.

Example 4: Finding the Area of a Parallelogram given Two 3D Vectors Forming Two Adjacent Sides

𝐴𝐡𝐢𝐷 is a parallelogram with 𝐴𝐡=(βˆ’1,1,3) and 𝐴𝐷=(3,4,1). Find the area of 𝐴𝐡𝐢𝐷. Give your answer to one decimal place.

Answer

If 𝐴𝐡𝐢𝐷 is a parallelogram, then 𝐴𝐡 and 𝐴𝐷 are two of its adjacent sides. Its area is then given by ‖‖𝐴𝐡×𝐴𝐷‖‖. Let us first find 𝐴𝐡×𝐴𝐷: 𝐴𝐡×𝐴𝐷=||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜βˆ’113341||||=βˆ’11⃑𝑖+10βƒ‘π‘—βˆ’7βƒ‘π‘˜.

Hence, ‖‖𝐴𝐡×𝐴𝐷‖‖=√121+100+49=√270β‰…16.41.tod.p

The area of 𝐴𝐡𝐢𝐷 is 16.4 area units to 1 decimal place.

In the figure above the previous example, the area of triangle 𝑂𝐴𝐡 is half that of 𝑂𝐴𝐢𝐡. It follows that the area of a triangle 𝐴𝐡𝐢 is equal to half the magnitude of the cross product of two of the three vectors making its sides; that is, areaof𝐴𝐡𝐢=12‖‖𝐴𝐡×𝐴𝐢‖‖=12β€–β€–οƒ π΅π΄Γ—οƒŸπ΅πΆβ€–β€–=12β€–β€–οƒŸπΆπ΅Γ—οƒ πΆπ΄β€–β€–.

The order of the vectors and their directions (𝐴𝐡 or 𝐡𝐴) will only change the angle between the two vectors. As we are interested here only in the magnitude of the cross product, it does not matter how we choose the two vectors between the three points and their order for the cross product. We have, however, chosen here to write the vectors starting at one of the triangle vertices to help visualize the parallelogram spanned by these two vectors and the triangle as half of the parallelogram.

Let us use the meaning of the cross product in a geometric context with the last example.

Example 5: Finding the Area of a Triangle given Its Vertices in 3D

Given that 𝐷=(0,βˆ’2,βˆ’8), 𝐸=(6,4,6), and 𝐹=(βˆ’4,βˆ’9,βˆ’2), determine the area of the triangle 𝐷𝐸𝐹 approximated to the nearest hundredth.

Answer

We know that the area of a triangle is equal to half the magnitude of the cross product of two of the three vectors making its sides. So, if we take, for instance, the two sides around vertex 𝐸, we have areaof𝐷𝐸𝐹=12β€–β€–οƒŸπΈπΉΓ—οƒ πΈπ·β€–β€–.

Let us first find the components of οƒŸπΈπΉ, 𝐸𝐷, and οƒŸπΈπΉΓ—οƒ πΈπ·: οƒŸπΈπΉ=(βˆ’4βˆ’6,βˆ’9βˆ’4,βˆ’2βˆ’6)=(βˆ’10,βˆ’13,βˆ’8),𝐸𝐷=(0βˆ’6,βˆ’2βˆ’4,βˆ’8βˆ’6)=(βˆ’6,βˆ’6,βˆ’14),οƒŸπΈπΉΓ—οƒ πΈπ·=||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜βˆ’10βˆ’13βˆ’8βˆ’6βˆ’6βˆ’14||||=(βˆ’13Γ—(βˆ’14)βˆ’(βˆ’6)Γ—(βˆ’8))βƒ‘π‘–βˆ’(βˆ’10Γ—(βˆ’14)βˆ’(βˆ’6)Γ—(βˆ’8))⃑𝑗+(βˆ’10Γ—(βˆ’6)βˆ’(βˆ’6)Γ—(βˆ’13))βƒ‘π‘˜=134βƒ‘π‘–βˆ’92βƒ‘π‘—βˆ’18βƒ‘π‘˜.

Therefore, β€–β€–οƒŸπΈπΉΓ—οƒ πΈπ·β€–β€–=√134+92+18, and areaofareaunitstod.p𝐷𝐸𝐹=12β€–β€–οƒŸπΈπΉΓ—οƒ πΈπ·β€–β€–=12√134+92+18β‰…81.772.

Key Points

  • The cross product of two vectors ⃑𝐴 and ⃑𝐡 is another vector given by ⃑𝐴×⃑𝐡=‖‖⃑𝐴‖‖‖‖⃑𝐡‖‖|πœƒ|⃑𝑒,sin where πœƒ is the angle between ⃑𝐴 and ⃑𝐡 and ⃑𝑒 is a unit vector perpendicular to the plane that contains ⃑𝐴 and ⃑𝐡, where the direction of ⃑𝑒 is given by the so-called right-hand rule.
  • For two 3D vectors ⃑𝐴=𝐴⃑𝑖+𝐴⃑𝑗+π΄βƒ‘π‘˜ο—ο˜ο™ and ⃑𝐡=𝐡⃑𝑖+𝐡⃑𝑗+π΅βƒ‘π‘˜ο—ο˜ο™ in the coordinate system 𝑂,⃑𝑖,⃑𝑗,βƒ‘π‘˜ο‡, the cross product of ⃑𝐴 and ⃑𝐡 is ⃑𝐴×⃑𝐡=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΄π΄π΄π΅π΅π΅|||||=ο€Ήπ΄π΅βˆ’π΄π΅ο…βƒ‘π‘–βˆ’(π΄π΅βˆ’π΄π΅)⃑𝑗+ο€Ήπ΄π΅βˆ’π΄π΅ο…βƒ‘π‘˜.ο—ο˜ο™ο—ο˜ο™ο˜ο™ο™ο˜ο—ο™ο™ο—ο—ο˜ο˜ο—
  • The cross product is anticommutative: ⃑𝐴×⃑𝐡=βˆ’βƒ‘π΅Γ—βƒ‘π΄.
  • The cross product is distributive: ⃑𝐴+⃑𝐡×⃑𝐢=⃑𝐴×⃑𝐢+⃑𝐡×⃑𝐢.
  • The cross product of two collinear vectors is zero, and so ⃑𝐴×⃑𝐴=0.
  • The area of the parallelogram spanned by ⃑𝐴 and ⃑𝐡 is given by ‖‖⃑𝐴×⃑𝐡‖‖. It follows that the area of the triangle with ⃑𝐴 and ⃑𝐡 defining two of its sides is given by 12‖‖⃑𝐴×⃑𝐡‖‖.

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