Lesson Explainer: Scalar Triple Product Mathematics

In this explainer, we will learn how to calculate the scalar triple product and apply this in geometrical applications.

Before looking at the scalar triple product, you should already be familiar with the scalar product (dot product) and the cross product. Here, we introduce a product of three vectors, combining both scalar and cross products.

Definition: Scalar Triple Product

The scalar triple product of three vectors ⃑𝐴, ⃑𝐡, and ⃑𝐢 is defined as ⃑𝐴⋅⃑𝐡×⃑𝐢.

We can write the scalar triple product as ⃑𝐴⋅⃑𝐡×⃑𝐢 as well. However, the parentheses are unnecessary since carrying out ⃑𝐴⋅⃑𝐡 first yields a scalar, and we cannot perform a cross product between a scalar and a vector.

The scalar triple product yields a scalar, as suggested by its name.

We know that ⃑𝐡×⃑𝐢=𝐡⃑𝑖+𝐡⃑𝑗+π΅βƒ‘π‘˜ο‡Γ—ο€»πΆβƒ‘π‘–+𝐢⃑𝑗+πΆβƒ‘π‘˜ο‡=ο€Ήπ΅πΆβˆ’π΅πΆο…βƒ‘π‘–βˆ’(π΅πΆβˆ’π΅πΆ)⃑𝑗+ο€Ήπ΅πΆβˆ’π΅πΆο…βƒ‘π‘˜.ο—ο˜ο™ο—ο˜ο™ο˜ο™ο™ο˜ο—ο™ο™ο—ο—ο˜ο˜ο—

Therefore, 𝐴⃑𝑖+𝐴⃑𝑗+π΄βƒ‘π‘˜ο‡β‹…ο€»ο€Ήπ΅πΆβˆ’π΅πΆο…βƒ‘π‘–βˆ’(π΅πΆβˆ’π΅πΆ)⃑𝑗+ο€Ήπ΅πΆβˆ’π΅πΆο…βƒ‘π‘˜ο‡=π΄ο€Ήπ΅πΆβˆ’π΅πΆο…βˆ’π΄(π΅πΆβˆ’π΅πΆ)+π΄ο€Ήπ΅πΆβˆ’π΅πΆο…=|||||𝐴𝐴𝐴𝐡𝐡𝐡𝐢𝐢𝐢|||||.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˜ο˜ο—ο—ο˜ο™ο—ο˜ο™ο—ο˜ο™

Property: Calculating a Scalar Triple Product Using the Vectors’ Components

Calculating the scalar triple product ⃑𝐴⋅⃑𝐡×⃑𝐢 is equivalent to calculating the determinant |||||𝐴𝐴𝐴𝐡𝐡𝐡𝐢𝐢𝐢|||||.ο—ο˜ο™ο—ο˜ο™ο—ο˜ο™

Let us apply this with a first example.

Example 1: Calculating the Scalar Triple Product of Three Vectors

Given ⃑𝐴=(1,5,βˆ’5), ⃑𝐡=(2,4,3), and ⃑𝐢=(0,5,βˆ’4), find ⃑𝐴⋅⃑𝐡×⃑𝐢.

Answer

We know that calculating ⃑𝐴⋅⃑𝐡×⃑𝐢 is equivalent to calculating |||||𝐴𝐴𝐴𝐡𝐡𝐡𝐢𝐢𝐢|||||ο—ο˜ο™ο—ο˜ο™ο—ο˜ο™. So, let us substitute the components of vectors ⃑𝐴, ⃑𝐡, and ⃑𝐢 to calculate this determinant: ⃑𝐴⋅⃑𝐡×⃑𝐢=||||15βˆ’524305βˆ’4||||=1(4Γ—(βˆ’4)βˆ’5Γ—3)βˆ’5(2Γ—(βˆ’4)βˆ’0Γ—3)βˆ’5(2Γ—5βˆ’0Γ—4)=βˆ’31+40βˆ’50=βˆ’41.

Let us look now at some properties of the scalar triple product. We know that swapping two horizontal rows in a 3Γ—3 determinant changes its sign. Therefore, switching two vectors in the scalar triple product changes the sign of the product: ⃑𝐴⋅⃑𝐡×⃑𝐢=βˆ’βƒ‘π΅β‹…ο€Ίβƒ‘π΄Γ—βƒ‘πΆο†.

If we perform a further permutation, it will reverse the sign again. So, by permutating ⃑𝐴 and ⃑𝐢 or ⃑𝐡 and ⃑𝐢 in ⃑𝐡⋅⃑𝐴×⃑𝐢, we find that ⃑𝐴⋅⃑𝐡×⃑𝐢=⃑𝐡⋅⃑𝐢×⃑𝐴=⃑𝐢⋅⃑𝐴×⃑𝐡.

We see that scalar triple products are equal when the cyclic order of the three vectors is unchanged, here ⃑𝐴, ⃑𝐡, ⃑𝐢.

We are going to use this property in the next example.

Example 2: Calculating the Scalar Triple Product of the Three Unit Vectors and Its Permutations

Find βƒ‘π‘–β‹…ο€»βƒ‘π‘—Γ—βƒ‘π‘˜ο‡+βƒ‘π‘—β‹…ο€»βƒ‘π‘˜Γ—βƒ‘π‘–ο‡+βƒ‘π‘˜β‹…ο€Ίβƒ‘π‘–Γ—βƒ‘π‘—ο†.

Answer

We have three scalar triple products with the same three vectors, ⃑𝑖, ⃑𝑗, and βƒ‘π‘˜, but in different orders. Scalar triple products with the same three vectors always have the same absolute value, but the same sign only if the three vectors are in the same cyclic order. Here, the cyclic order in the first scalar triple product is ⃑𝑖,⃑𝑗,βƒ‘π‘˜,⃑𝑖,⃑𝑗,βƒ‘π‘˜β€¦ and it is the same order in the other two as well.

Therefore, the total value of the given expression is 3 times the value of one of the terms.

To calculate one of the scalar triple products, we can use the fact that ⃑𝑖×⃑𝑗=βƒ‘π‘˜, and so βƒ‘π‘˜β‹…ο€Ίβƒ‘π‘–Γ—βƒ‘π‘—ο†=βƒ‘π‘˜β‹…βƒ‘π‘˜=β€–β€–βƒ‘π‘˜β€–β€–=1.

Alternatively, we can use the determinant method: βƒ‘π‘–β‹…ο€»βƒ‘π‘—Γ—βƒ‘π‘˜ο‡=||||100010001||||=1.

Hence, βƒ‘π‘–β‹…ο€»βƒ‘π‘—Γ—βƒ‘π‘˜ο‡+βƒ‘π‘—β‹…ο€»βƒ‘π‘˜Γ—βƒ‘π‘–ο‡+βƒ‘π‘˜β‹…ο€Ίβƒ‘π‘–Γ—βƒ‘π‘—ο†=3.

Property: Scalar Triple Product of Coplanar Vectors

The scalar triple product ⃑𝐴⋅⃑𝐡×⃑𝐢 is the scalar product of ⃑𝐴 with ⃑𝐡×⃑𝐢, that is, the scalar product of ⃑𝐴 with a vector that is perpendicular to the plane defined by ⃑𝐡 and ⃑𝐢.

It results that if ⃑𝐴, ⃑𝐡, and ⃑𝐢 are coplanar (i.e., in the same plane), then ⃑𝐴⋅⃑𝐡×⃑𝐢=0 since the scalar product of two perpendicular vectors is zero.

Inversely, if ⃑𝐴⋅⃑𝐡×⃑𝐢=0, then ⃑𝐴, ⃑𝐡, and ⃑𝐢 are coplanar.

Let us use this property to solve the next question.

Example 3: Finding Missing Components of Coplanar Vectors

Find the value of π‘˜ for which the four points (1,7,βˆ’2), (3,5,6), (βˆ’1,6,βˆ’4), and (βˆ’4,βˆ’3,π‘˜) all lie in a single plane.

Answer

As three noncollinear points define a plane, if (1,7,βˆ’2), (3,5,6), and (βˆ’1,6,βˆ’4) are indeed noncollinear, then we need to find the value of π‘˜ for which (βˆ’4,βˆ’3,π‘˜) is in the plane containing(1,7,βˆ’2), (3,5,6), and (βˆ’1,6,βˆ’4).

Let us first check that 𝐴(1,7,βˆ’2), 𝐡(3,5,6), and 𝐢(βˆ’1,6,βˆ’4) are noncollinear. For this, we simply check that the cross product of 𝐡𝐴 and οƒŸπ΅πΆ is not zero (we can take here any two different vectors formed with the three points): οƒ π΅π΄Γ—οƒŸπ΅πΆ=|||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜π΅π΄π΅π΄π΅π΄π΅πΆπ΅πΆπ΅πΆ|||||=||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜βˆ’22βˆ’8βˆ’41βˆ’10||||=(βˆ’12,12,6)β‰ 0.ο—ο˜ο™ο—ο˜ο™

Let us find now the value of π‘˜ for which 𝐷(βˆ’4,βˆ’3,π‘˜) is in the plane 𝐴𝐡𝐢. As the scalar triple product of three coplanar vectors is zero, we need to find the value of π‘˜ for which, for example, οƒ π΄π·β‹…οƒ π΅π΄Γ—οƒŸπ΅πΆ=0. The components of 𝐴𝐷 are (βˆ’5,βˆ’10,π‘˜+2). Hence, οƒ π΄π·β‹…οƒ π΅π΄Γ—οƒŸπ΅πΆ=βˆ’12Γ—(βˆ’5)+12Γ—(βˆ’10)+6(π‘˜+2)=60βˆ’120+6π‘˜+12=6π‘˜βˆ’48.

οƒ π΄π·β‹…οƒ π΅π΄Γ—οƒŸπ΅πΆ=0 if 6π‘˜βˆ’48=0, that is, if π‘˜=8.

Therefore, the value of π‘˜ for which the four points (1,7,βˆ’2), (3,5,6), (βˆ’1,6,βˆ’4), and (βˆ’4,βˆ’3,π‘˜) all lie in a single plane is 8.

Another useful property of the scalar triple product comes from its geometrical meaning. As said above, the scalar triple product ⃑𝐴⋅⃑𝐡×⃑𝐢=0 is the scalar product of ⃑𝐴 with ⃑𝐡×⃑𝐢. We know that ⃑𝐡×⃑𝐢 is a vector perpendicular to the plane defined by ⃑𝐡 and ⃑𝐢 whose magnitude ‖‖⃑𝐡×⃑𝐢‖‖ is the area of the parallelogram spanned by ⃑𝐡 and ⃑𝐢.

Let us consider a parallelepiped spanned by ⃑𝐴, ⃑𝐡, and ⃑𝐢. Its volume is the area of the parallelogram spanned by ⃑𝐡 and ⃑𝐢 multiplied by its height β„Ž as shown in the figure. The height β„Ž is, in turn, given by β„Ž=‖‖⃑𝐴‖‖cosπœƒ, where πœƒ is the acute angle between vector ⃑𝐴 and β„Ž.

As ⃑𝐡×⃑𝐢 is a vector perpendicular to the plane defined by ⃑𝐡 and ⃑𝐢, we have cosπœƒ=||⃑𝐴⋅⃑𝐡×⃑𝐢||‖‖⃑𝐴‖‖‖‖⃑𝐡×⃑𝐢‖‖.

A change in the orientation of ⃑𝐡×⃑𝐢 (up or down) does not change the absolute value of its scalar product with ⃑𝐴, as illustrated in the figure with πœƒβ€² the angle between ⃑𝐴 and ⃑𝐡×⃑𝐢 (down). As πœƒβ€²=180βˆ’πœƒβˆ˜, |πœƒβ€²|=|πœƒ|coscos.

The volume of the parallelepiped is, therefore, Volumecos=β€–β€–βƒ‘π΅Γ—βƒ‘πΆβ€–β€–β„Ž=β€–β€–βƒ‘π΅Γ—βƒ‘πΆβ€–β€–β€–β€–βƒ‘π΄β€–β€–πœƒ=‖‖⃑𝐡×⃑𝐢‖‖‖‖⃑𝐴‖‖||⃑𝐴⋅⃑𝐡×⃑𝐢||‖‖⃑𝐴‖‖‖‖⃑𝐡×⃑𝐢‖‖=||⃑𝐴⋅⃑𝐡×⃑𝐢||.

Property: Geometric Meaning of the Scalar Triple Product

The absolute value of the scalar triple product of three vectors is the volume of the parallelepiped spanned by the three vectors: volume=||⃑𝐴⋅⃑𝐡×⃑𝐢||.

It is worth noting that three coplanar vectors do not define any parallelepiped, and therefore the scalar triple product is zero.

Let us now use this property to find the volume of a parallelepiped.

Example 4: Finding the Volume of a Parallelepiped

Find the volume of the parallelepiped with the adjacent sides βƒ‘π‘ˆ=(1,1,3), ⃑𝑉=(2,1,4), and οƒŸπ‘Š=(5,1,βˆ’2).

Answer

The parallelepiped is spanned by βƒ‘π‘ˆ=(1,1,3), ⃑𝑉=(2,1,4), and οƒŸπ‘Š=(5,1,βˆ’2). Its volume is given by the absolute value of the scalar triple product of the three vectors. Hence, volume=||βƒ‘π‘ˆβ‹…βƒ‘π‘‰Γ—οƒŸπ‘Š||=||||||||11321451βˆ’2||||||||=|βˆ’6+24βˆ’9|=9.

The volume of the parallelepiped with the adjacent sides βƒ‘π‘ˆ=(1,1,3), ⃑𝑉=(2,1,4), and οƒŸπ‘Š=(5,1,βˆ’2) is 9 volume units.

In the last example, we are going to find possible values for a missing vector component given the volume of the parallelepiped spanned by three vectors.

Example 5: Finding Missing Vector Components given the Volume of the Parallelepiped Spanned by Three Vectors

The parallelepiped on vectors (βˆ’2,βˆ’2,π‘š), (2,0,βˆ’2), and (βˆ’5,1,0) has volume 48. What can π‘š be?

Answer

The volume of the parallelepiped spanned by the vectors ⃑𝐴(βˆ’2,βˆ’2,π‘š), ⃑𝐡(2,0,βˆ’2), and ⃑𝐢(βˆ’5,1,0) is given by volume=||⃑𝐴⋅⃑𝐡×⃑𝐢||=||||||||βˆ’2βˆ’2π‘š20βˆ’2βˆ’510||||||||=|βˆ’4βˆ’20+2π‘š|=|βˆ’24+2π‘š|.

The volume of the parallelepiped is 48 volume units; hence, |βˆ’24+2π‘š|=48.

This equation is verified if βˆ’24+2π‘š=48 or if βˆ’24+2π‘š=βˆ’48; hence, if π‘š=36 or π‘š=βˆ’12.

The parallelepiped on vectors (βˆ’2,βˆ’2,π‘š), (2,0,βˆ’2), and (βˆ’5,1,0) has volume 48 if π‘š=36 or π‘š=βˆ’12.

Key Points

  • The scalar triple product of three vectors ⃑𝐴, ⃑𝐡, and ⃑𝐢 is defined as ⃑𝐴⋅⃑𝐡×⃑𝐢.
  • A scalar triple product is equivalent to a 3Γ—3 determinant: ⃑𝐴⋅⃑𝐡×⃑𝐢=|||||𝐴𝐴𝐴𝐡𝐡𝐡𝐢𝐢𝐢|||||.ο—ο˜ο™ο—ο˜ο™ο—ο˜ο™
  • Scalar triple products are equal if the cyclic order of the three vectors is unchanged: ⃑𝐴⋅⃑𝐡×⃑𝐢=⃑𝐡⋅⃑𝐢×⃑𝐴=⃑𝐢⋅⃑𝐴×⃑𝐡.
  • The scalar triple product of three coplanar vectors is zero. Inversely, if ⃑𝐴⋅⃑𝐡×⃑𝐢=0, then ⃑𝐴, ⃑𝐡, and ⃑𝐢 are coplanar.
  • The volume of the parallelepiped spanned by vectors ⃑𝐴, ⃑𝐡, and ⃑𝐢 is given by volume=||⃑𝐴⋅⃑𝐡×⃑𝐢||.

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