Lesson Explainer: Cross Product in 2D | Nagwa Lesson Explainer: Cross Product in 2D | Nagwa

Lesson Explainer: Cross Product in 2D Mathematics • Third Year of Secondary School

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In this explainer, we will learn how to find the cross product of two vectors in the coordinate plane.

There are two ways to multiply vectors together. You may already be familiar with the dot product, also called the 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 a vector perpendicular to the plane that contains ⃑𝐴 and ⃑𝐡 and whose magnitude is given by ‖‖⃑𝐴×⃑𝐡‖‖=‖‖⃑𝐴‖‖‖‖⃑𝐡‖‖|πœƒ|,sin where πœƒ is the angle between ⃑𝐴 and ⃑𝐡.

From the definition of the cross product, we find that the cross product of two parallel (or collinear) vectors is zero as the sine of the angle between them (0 or 180∘) is zero. Note that no plane can be defined by two collinear vectors, so it is consistent that ⃑𝐴×⃑𝐡=0 if ⃑𝐴 and ⃑𝐡 are collinear.

From the definition above, it follows that the cross product of any two noncollinear vectors in the coordinate plane with ⃑𝑖 and ⃑𝑗 as units vectors is parallel to βƒ‘π‘˜, where βƒ‘π‘˜ is a unit vector perpendicular to the plane containing ⃑𝑖 and ⃑𝑗, as shown in the diagram.

Let us consider the two vectors ⃑𝐴 and ⃑𝐡 in the coordinate plane.

Vector ⃑𝐴 makes an angle πœƒοŠ§ with ⃑𝑖 and vector ⃑𝐡 makes an angle πœƒοŠ¨. The angle πœƒ between ⃑𝐴 and ⃑𝐡 is therefore πœƒβˆ’πœƒοŠ¨οŠ§, so we have sinsinπœƒ=(πœƒβˆ’πœƒ). Using the subtraction trigonometric identity sinsincoscossin(πœƒβˆ’πœƒ)=πœƒπœƒβˆ’πœƒπœƒοŠ¨οŠ§οŠ¨οŠ§οŠ¨οŠ§, we find that sinsincoscossinπœƒ=πœƒπœƒβˆ’πœƒπœƒ.

Let us make a matrix with the components in terms of πœƒοŠ§ and πœƒοŠ¨ of ⃑𝐴 in the first row and those of ⃑𝐡 in the second row ο„β€–β€–βƒ‘π΄β€–β€–πœƒβ€–β€–βƒ‘π΄β€–β€–πœƒβ€–β€–βƒ‘π΅β€–β€–πœƒβ€–β€–βƒ‘π΅β€–β€–πœƒοcossincossin and calculate its determinant. Recall that the determinant of a 2Γ—2 matrix is given by |||π‘Žπ‘π‘π‘‘|||=π‘Žπ‘‘βˆ’π‘π‘.

Hence, we find that |||||β€–β€–βƒ‘π΄β€–β€–πœƒβ€–β€–βƒ‘π΄β€–β€–πœƒβ€–β€–βƒ‘π΅β€–β€–πœƒβ€–β€–βƒ‘π΅β€–β€–πœƒ|||||=β€–β€–βƒ‘π΄β€–β€–β€–β€–βƒ‘π΅β€–β€–πœƒπœƒβˆ’β€–β€–βƒ‘π΄β€–β€–β€–β€–βƒ‘π΅β€–β€–πœƒπœƒ=‖‖⃑𝐴‖‖‖‖⃑𝐡‖‖(πœƒπœƒβˆ’πœƒπœƒ)=β€–β€–βƒ‘π΄β€–β€–β€–β€–βƒ‘π΅β€–β€–πœƒ,cossincossincossinsincoscossinsincossin since sincoscossinsinπœƒπœƒβˆ’πœƒπœƒ=πœƒοŠ¨οŠ§οŠ¨οŠ§.

If we combine this with the fact that ⃑𝐴×⃑𝐡 is a vector parallel to βƒ‘π‘˜, we can write the following definition of the cross product of two vectors in the coordinate plane with ⃑𝑖 and ⃑𝑗 as unit vectors.

Definition: Cross Product of Two Vectors in the Coordinate Plane

For two vectors ⃑𝐴=𝐴⃑𝑖+π΄βƒ‘π‘—ο—ο˜ and ⃑𝐡=𝐡⃑𝑖+π΅βƒ‘π‘—ο—ο˜ in the coordinate plane with ⃑𝑖 and ⃑𝑗 as unit vectors, the cross product of ⃑𝐴 and ⃑𝐡 is ⃑𝐴×⃑𝐡=|||𝐴𝐴𝐡𝐡|||βƒ‘π‘˜=ο€Ήπ΄π΅βˆ’π΅π΄ο…βƒ‘π‘˜=ο€»β€–β€–βƒ‘π΄β€–β€–β€–β€–βƒ‘π΅β€–β€–πœƒο‡βƒ‘π‘˜,ο—ο˜ο—ο˜ο—ο˜ο—ο˜sin where πœƒ is the angle between ⃑𝐴 and ⃑𝐡, and the vectors ⃑𝑖, ⃑𝑗, and βƒ‘π‘˜ are the fundamental unit vectors along the π‘₯-, 𝑦-, and 𝑧-axes, respectively, as shown in the diagram.

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

Example 1: Finding a Missing Component Given the Cross Product of Two 2D Vectors

If ⃑𝐴=3βƒ‘π‘–βˆ’5⃑𝑗, ⃑𝐡=π‘šβƒ‘π‘–+5⃑𝑗, and ⃑𝐴×⃑𝐡=50βƒ‘π‘˜, find the value of π‘š.

Answer

As ⃑𝐴=3βƒ‘π‘–βˆ’5⃑𝑗, we have 𝐴=3 and 𝐴=βˆ’5; and as ⃑𝐡=π‘šβƒ‘π‘–+5⃑𝑗, we have 𝐡=π‘šο— and 𝐡=5.

We furthermore know that ⃑𝐴×⃑𝐡=|||𝐴𝐴𝐡𝐡|||βƒ‘π‘˜=50βƒ‘π‘˜.ο—ο˜ο—ο˜

Hence ||3βˆ’5π‘š5||=503Γ—5βˆ’(βˆ’5)π‘š=5015+5π‘š=505π‘š=50βˆ’15π‘š=7.

Let us further practice calculating the cross product of two vectors with a further question involving vector addition.

Example 2: Calculating the Cross Product of Two 2D Vectors

Given that ⃑𝐴=7⃑𝑖+2⃑𝑗, ⃑𝐡=βˆ’βƒ‘π‘–+2⃑𝑗, and ⃑𝐢=6⃑𝑖+6⃑𝑗, determine ⃑𝐢+⃑𝐴×⃑𝐡.

Answer

Let us first work out ⃑𝐢+⃑𝐴: ⃑𝐢+⃑𝐴=7⃑𝑖+2⃑𝑗+6⃑𝑖+6⃑𝑗=13⃑𝑖+8⃑𝑗.

From this, we deduce that the π‘₯- and 𝑦-components of ⃑𝐢+⃑𝐴 are (𝐢+𝐴)=13 and (𝐢+𝐴)=8. Also, from ⃑𝐡=βˆ’βƒ‘π‘–+2⃑𝑗, we know that 𝐡=βˆ’1 and 𝐡=2.

We can now calculate ⃑𝐢+⃑𝐴×⃑𝐡 as ⃑𝐢+⃑𝐴×⃑𝐡=|||(𝐢+𝐴)(𝐢+𝐴)𝐡𝐡|||βƒ‘π‘˜=||138βˆ’12||βƒ‘π‘˜=(13Γ—2βˆ’(βˆ’1)Γ—8)βƒ‘π‘˜=34βƒ‘π‘˜.ο—ο˜ο—ο˜

With the previous example, we can wonder whether the cross product is distributive, that is, do we have ⃑𝐢+⃑𝐴×⃑𝐡=⃑𝐢×⃑𝐡+⃑𝐴×⃑𝐡?

It can be found out easily when looking more closely at the determinant we used to calculate ⃑𝐢+⃑𝐴×⃑𝐡: |||(𝐢+𝐴)(𝐢+𝐴)𝐡𝐡|||=(𝐢+𝐴)π΅βˆ’π΅(𝐢+𝐴).ο—ο˜ο—ο˜ο—ο˜ο—ο˜

The rule for vector addition is (𝐢+𝐴)=𝐢+𝐴 and (𝐢+𝐴)=𝐢+𝐴; therefore, |||(𝐢+𝐴)(𝐢+𝐴)𝐡𝐡|||=𝐢𝐡+π΄π΅βˆ’π΅πΆβˆ’π΅π΄=πΆπ΅βˆ’π΅πΆ+π΄π΅βˆ’π΅π΄=|||𝐢𝐢𝐡𝐡|||+|||𝐴𝐴𝐡𝐡|||.ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜ο—ο˜

It follows that ⃑𝐢+⃑𝐴×⃑𝐡=⃑𝐢×⃑𝐡+⃑𝐴×⃑𝐡; the cross product is therefore distributive.

Property: Distributivity of the Cross Product

The cross product is distributive: ⃑𝐢+⃑𝐴×⃑𝐡=⃑𝐢×⃑𝐡+⃑𝐴×⃑𝐡.

We will now use our knowledge of how cross products are calculated to find an unknown vector given the results of its cross product with two known vectors.

Example 3: Finding a Vector Given its Cross Product with Two Known Vectors

If ⃑𝐴=βˆ’βƒ‘π‘–βˆ’2⃑𝑗, ⃑𝐡=βˆ’4βƒ‘π‘–βˆ’4⃑𝑗, ⃑𝐴×⃑𝐢=βˆ’3βƒ‘π‘˜, and ⃑𝐢×⃑𝐡=4βƒ‘π‘˜, find ⃑𝐢.

Answer

Let us extract the components of ⃑𝐴 and ⃑𝐡 from their expressions in terms of ⃑𝑖 and ⃑𝑗: 𝐴=βˆ’1 and 𝐴=βˆ’2, and 𝐡=βˆ’4 and 𝐡=βˆ’4.

We can now write their cross products with ⃑𝐢 as ⃑𝐴×⃑𝐢=|||𝐴𝐴𝐢𝐢|||βƒ‘π‘˜=βˆ’3βƒ‘π‘˜,⃑𝐢×⃑𝐡=|||𝐢𝐢𝐡𝐡|||βƒ‘π‘˜=4βƒ‘π‘˜.ο—ο˜ο—ο˜ο—ο˜ο—ο˜and

Hence, we have

|||βˆ’1βˆ’2𝐢𝐢|||βƒ‘π‘˜=βˆ’3βƒ‘π‘˜,βˆ’1Γ—πΆβˆ’πΆΓ—(βˆ’2)=βˆ’3,βˆ’πΆ+2𝐢=βˆ’3,ο—ο˜ο˜ο—ο˜ο—(1)

and

|||πΆπΆβˆ’4βˆ’4|||βƒ‘π‘˜=4βƒ‘π‘˜,βˆ’4πΆβˆ’(βˆ’4)𝐢=4,βˆ’4𝐢+4𝐢=4.ο—ο˜ο—ο˜ο—ο˜(2)

We have now two linear equations with two unknowns (𝐢 and 𝐢).

From equation (1), we find that 𝐢=2𝐢+3ο˜ο—. Substituting this expression for 𝐢 into equation (2) gives βˆ’4𝐢+4β‹…(2𝐢+3)=4βˆ’4𝐢+8𝐢+12=44𝐢=4βˆ’12𝐢=βˆ’2.

Substituting 𝐢=βˆ’2 into 𝐢=2𝐢+3ο˜ο—, we find that 𝐢=βˆ’1.

Therefore, we have ⃑𝐢=βˆ’2βƒ‘π‘–βˆ’βƒ‘π‘—.

Let us now find the cross product of two vectors whose components are not explicitly given but that are defined with specific points in a rectangle.

Example 4: Finding the Cross Product of Two Vectors in a Rectangle

𝐴𝐡𝐢𝐷 is a rectangle where ⃑𝐾 is a unit vector perpendicular to its plane. Find οƒ πΆπ‘€Γ—οƒŸπΆπ΅.

Answer

To find οƒ πΆπ‘€Γ—οƒŸπΆπ΅, we have two possibilities. Either we find the components of both vectors, for instance in the coordinate plane with origin 𝐢, ⃑𝑖=οƒŸπΆπ΅β€–β€–οƒŸπΆπ΅β€–β€–, and ⃑𝑗=𝐢𝐷‖‖𝐢𝐷‖‖, or we apply οƒ πΆπ‘€Γ—οƒŸπΆπ΅=ο€»β€–β€–οƒ πΆπ‘€β€–β€–β€–β€–οƒŸπΆπ΅β€–β€–πœƒο‡βƒ‘πΆsin, where πœƒ is the angle between 𝐢𝑀 and οƒŸπΆπ΅ and ⃑𝐾 is a unit vector perpendicular to the plane of the rectangle. This second method implies that we have to find the magnitude of both vectors and sinπœƒ.

First Method

In the coordinate plane 𝐢,⃑𝑖=οƒŸπΆπ΅β€–β€–οƒŸπΆπ΅β€–β€–,⃑𝑗=𝐢𝐷‖‖𝐢𝐷‖‖, we have 𝐢(0,0), 𝐡(44,0), and 𝑀(22,16.5). From this, we find that 𝐢𝑀=(22,16.5) and οƒŸπΆπ΅=(44,0).

The cross product of 𝐢𝑀and οƒŸπΆπ΅ is οƒ πΆπ‘€Γ—οƒŸπΆπ΅=||2216.5440||⃑𝐾, as ⃑𝐾 is a unit vector perpendicular to the plane of the rectangle. οƒ πΆπ‘€Γ—οƒŸπΆπ΅=(22Γ—0βˆ’44Γ—16.5)⃑𝐾=βˆ’726⃑𝐾.

Second Method

The magnitude of οƒŸπΆπ΅ is simply the length 𝐢𝐡, so it is 44 cm.

Point 𝑀 is the middle of the rectangle’s diagonals, so 𝐢𝑀=12𝐢𝐴, and 𝐢𝐴=√𝐴𝐡+𝐡𝐢 (applying the Pythagorean theorem in the right triangle 𝐴𝐡𝐢). Hence, 𝐢𝑀=12√33+44.

The sine of the angle between 𝐢𝑀 and οƒŸπΆπ΅ is given by βˆ’π΄π΅π΄πΆ=βˆ’33√33+44. The negative sign comes from the fact that the angle from 𝐢𝑀 to οƒŸπΆπ΅ goes clockwise (negative angle), and sinsin(βˆ’πœƒ)=βˆ’πœƒ, or, for people working only with positive angles (counterclockwise turn), it is then 360βˆ’βˆ π΅πΆπ΄βˆ˜, and sinsin(360βˆ’πœƒ)=βˆ’πœƒβˆ˜.

We can now write οƒ πΆπ‘€Γ—οƒŸπΆπ΅=ο€»β€–β€–οƒ πΆπ‘€β€–β€–β€–β€–οƒŸπΆπ΅β€–β€–πœƒο‡βƒ‘πΎ=ο€Ώ12√33+44Γ—44Γ—ο€Ώβˆ’33√33+44⃑𝐾=ο€Όβˆ’12Γ—44Γ—33οˆβƒ‘πΎ=βˆ’726⃑𝐾sin

With the second method used to solve the last example, we see the importance of the order of the vectors in the cross product, because commuting them means that the sign of the angle changes, or that it changes from πœƒ to 360βˆ’πœƒβˆ˜ when one uses only positive angles. The effect is that the sine changes sign. This means that ⃑𝐴×⃑𝐡=βˆ’βƒ‘π΅Γ—βƒ‘π΄.

We say that the cross product is anticommutative.

Property: Anticommutativity of the Cross Product

The cross product is anticommutative: ⃑𝐴×⃑𝐡=βˆ’βƒ‘π΅Γ—βƒ‘π΄.

Let us now look at the geometric meaning of the magnitude of the cross product. For two vectors ⃑𝐴 and ⃑𝐡, the magnitude of their cross product is ‖‖⃑𝐴×⃑𝐡‖‖=‖‖⃑𝐴‖‖‖‖⃑𝐡‖‖|πœƒ|sin.

Since we are interested here in the absolute value of sinπœƒ, we do not need to worry if the angle is from ⃑𝐴 to ⃑𝐡 or from ⃑𝐡 to ⃑𝐴.

If we mentally place the diagram in the unit circle, we immediately see what ‖‖⃑𝐡‖‖|πœƒ|sin is.

We see that ‖‖⃑𝐡‖‖|πœƒ|=π‘‚π‘Œsin.

In a geometric context, π‘‚π‘Œ is the height of the parallelogram spanned by ⃑𝐴 and ⃑𝐡. Hence, ‖‖⃑𝐴‖‖‖‖⃑𝐡‖‖|πœƒ|sin is the area of the parallelogram 𝑂𝐴𝐢𝐡.

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, theareaof𝐴𝐡𝐢=12‖‖𝐴𝐡×𝐴𝐢‖‖=12β€–β€–οƒ π΅π΄Γ—οƒŸπ΅πΆβ€–β€–=12β€–β€–οƒŸπΆπ΅Γ—οƒ πΆπ΄β€–β€–.

As we are interested here in the magnitude of the cross product, the order of the vectors and their directions (𝐴𝐡 or 𝐡𝐴) do not actually matter. 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 Three Vertices

Find the area of a triangle 𝐴𝐡𝐢, where 𝐴(βˆ’8,βˆ’9), 𝐡(βˆ’7,βˆ’8), and 𝐢(9,βˆ’2).

Answer

The magnitude of the cross product of two vectors is equal to the area of the parallelogram spanned by them. The area of the triangle 𝐴𝐡𝐢 is equal to half the area of the parallelogram spanned by two vectors defined by its vertices: theareaof𝐴𝐡𝐢=12‖‖𝐴𝐡×𝐴𝐢‖‖=12β€–β€–οƒ π΅π΄Γ—οƒŸπ΅πΆβ€–β€–=12β€–β€–οƒŸπΆπ΅Γ—οƒ πΆπ΄β€–β€–.

Hence, we just need here to choose one vertex, for instance 𝐢, and find the components of the two vectors from this point, οƒŸπΆπ΅ and 𝐢𝐴: οƒŸπΆπ΅=(βˆ’7βˆ’9,βˆ’8βˆ’(βˆ’2))=(βˆ’16,βˆ’6),𝐢𝐴=(βˆ’8βˆ’9,βˆ’9βˆ’(βˆ’2))=(βˆ’17,βˆ’7).

Therefore, theareaofsquareunits𝐴𝐡𝐢=12β€–β€–οƒŸπΆπ΅Γ—οƒ πΆπ΄β€–β€–=12||||βˆ’16βˆ’6βˆ’17βˆ’7||||=12|βˆ’16Γ—(βˆ’7)βˆ’(βˆ’17)Γ—(βˆ’6)|=12|112βˆ’102|=5.

As no specific length unit of the coordinate plane was given, we write β€œsquare units” to show that we have calculated an area.

Key Points

  • For two vectors ⃑𝐴=𝐴⃑𝑖+π΄βƒ‘π‘—ο—ο˜ and ⃑𝐡=𝐡⃑𝑖+π΅βƒ‘π‘—ο—ο˜ in the coordinate plane 𝑂,⃑𝑖,⃑𝑗, the cross product of ⃑𝐴 and ⃑𝐡 is ⃑𝐴×⃑𝐡=|||𝐴𝐴𝐡𝐡|||βƒ‘π‘˜=ο€Ήπ΄π΅βˆ’π΅π΄ο…βƒ‘π‘˜=ο€»β€–β€–βƒ‘π΄β€–β€–β€–β€–βƒ‘π΅β€–β€–πœƒο‡βƒ‘π‘˜,ο—ο˜ο—ο˜ο—ο˜ο—ο˜sin where πœƒ is the angle between ⃑𝐴 and ⃑𝐡, and the vectors ⃑𝑖, ⃑𝑗, and βƒ‘π‘˜ are perpendicular unit vectors.
  • The cross product is distributive: ⃑𝐴+⃑𝐡×⃑𝐢=⃑𝐴×⃑𝐢+⃑𝐡×⃑𝐢.
  • The cross product is anticommutative: ⃑𝐴×⃑𝐡=βˆ’βƒ‘π΅Γ—βƒ‘π΄.
  • 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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