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12.4. The Cross Product

1. Definition and Basic Properties.

We will start from several definitions that have valuable geometrical meanings. Allthe vectors in this chapter are considered 3-dimensional.

Definition 1 Three vectors are called anordered triple if there is stated which one isthe first, the second, and the third.

For example, bac means thatb is the first vector, a is the second one, and c is the thirdone.

Definition 2 Vectors are called coplanar if they lie in the same plane.

For example, vectors0, 1, 1, j andkare coplanar (since they all lie in the yz-plane), but

the three basis vectors i, j, k are not.

Definition 3 A triple of non-coplanar vectorsa bc is called aright triple(orleft triple)if being reduced to the same initial point, the terminal point of the vectorc is located onthat side of the plane defined by vectorsa andb so that from the terminal point ofc , theshortest rotation froma to bis seen to be done counter-clockwise (respectively, clockwise).

Another illustration is like being reduced to the same initial point, our vectors a , b, c arelocated correspondingly to the usual location of corresponding fingers on your right (left)

hand: thumb c , forefinger a , middle finger b.

Example 1 If you have three non-coplanar vectorsa ,b, c , then it is possible to combinesix different triples from them. Three triples will be right, three triples left. Sketch three

non-coplanar vectors and find these triples.

Example 2 The triple composed of three basis vectors, ijk is a right triple.

Now, we are ready to the meaningful definition of the cross product. Like the dotproduct, the cross product is also a product of two vectors. But unlike the dot product,it is a vector.

Definition 4 TheCross Productof two vectorsa andbis vectorc denoted byc = a bsatisfying the following three conditions:

1). The magnitude (or length) ofc is

|c |= |a |b sin ,

where is the angle between vectorsa andb.2). Vectorc is orthogonal (perpendicular) to both, a andb.

3). Vectorc is oriented so that the triplea bc is a right triple.

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As you could see, if vectors a and b are non-collinear, then their cross product isuniquely defined. If they are collinear, i.e., = 0 or = , the magnitude ofc is equal to0, whence c = 0.

Definition 4 reflects geometrical meaning of the notion of the cross product. Now, wecould get the algebraic form for components of vector c through the components ofa and

b.

Theorem 1 Ifa = a1, a2, a3,b= b1, b2, b3, then

c =

i j k

a1 a2 a3b1 b2 b3

, (1)

or, in more formal form,

c = a2b3a3b2, a3b1a1b3, a1b2a2b1 . (2)

Unlike (2), formula (1) could be easily memorized. But remember, that it is just a

symbolic form since the first line of the determinant contains basis vectors, not numbers.Thus, we have to group scalars near the basis vectors. This yields (2), of course.

Example 3 Find the cross producta band verify that it is orthogonal to botha andb.

a = i+ 3j2k, b= i+ 5k.

Solution. First of all, let us get the components of our vectors:

a = 1, 3, 2 , b= 1, 0, 5 .

Now, we could use formula (1) for their cross product:

c = a b=

i j k

1 3 21 0 5

= 15i3j+ 3k= 15, 3, 3 .

To check that c is orthogonal to both a and b, we need to show that c a = c a = 0.Indeed,

c a = 151 + (3)3 + 3(2) = 0, c b= 15(1) + (3)0 + 3 5 = 0.

Q.E.D.

Example 4 Find a non-zero vector orthogonal to the plane through the pointsP(2, 1, 5),Q(1, 3, 4), andR(3, 0, 6).

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Solution. If the points P, Q, R belong to some plane, then, clearly, vectors P Q and P Rshould be parallel to this plane. This means, that we have to find a vector orthogonal toboth P Q and P R. It may be their cross product as follows from Definition 4. So,

P Q= 12, 31, 45= 3, 2, 1 , P R= 32, 01, 65= 1, 1, 1 .

Clearly, vectors P Qand P Rare non-collinear (Check it!), thus, their cross product willnot be zero-vector. Now,

c = P Q P R=

i j k

3 2 11 1 1

= i+ 2j+ k= 1, 2, 1

is orthogonal to the plane. Then, any non-zero vector d that is orthogonal to the sameplane will be collinear (parallel) to c , whence will have the form d = c for some scalar = 0. Thus, the answer is:

, 2, , = 0.

From the part 1) of Definition 4, we easily get

Corollary 1 Two non-zero vectorsa andb are parallel if and only if

a b= 0.

Indeed, if is the angle between vectors a and b, then this condition is necessary andsufficient for sin = 0 that means = 0 or =. This is equivalent to collinearity ofaandb.

If we look at Definition 4 again, and remember the formula for the area of a parallelo-

gram, then we easily get that the length of the cross product a b is equal to theparallelogram built on vectors a and b (when they are reduced to the same initialpoint, of course).

The following Properties of the Cross Product easily follow from Theorem 1.

Theorem 2 Ifa ,b, andc are vectors and is a scalar, then

1). a b= ba

2). (a ) b= a b

= a

b

3). a

b+c

= a b+a c

4).a + b

c = a c + bc

5). a

bc

=a b

c

6). a

bc

= (a c )ba b

c.

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Warning: The cross product is not commutative. In fact, Property 1) in Theorem 2is called anti-commutativity.

2. Triple Products.

The equation in Property 5) of Theorem 2 is called the Scalar Triple Product of

vectors a ,b, and c . Ifa = a1, a2, a3,b= b1, b2, b3, c = c1, c2, c3, then

a

bc

= a1(b2c3b3c2) +a2(b3c1b1c3) +a3(b1c2b2c1) =

a1 a2 a3b1 b2 b3c1 c2 c3

. (3)

Let us consider the volume V of a parallelepiped built on three vectors a , b, and creduced to the same initial point. It is equal to the area S of a parallelogram built onvectors b and c (the basement of our parallelepiped) multiplied by the height h of our

parallelepiped. If we take a vector orthogonal to the plane of vectorsb and c , then h isnothing else but an absolute value of the scalar projectionofa to this vector. However,as we have seen in Example 4, the vector orthogonal to the plane isbc . Hence,

h=compbca

=

bc

abc

.

And, as we already mentioned (see note before Theorem 2),

S=bc

.

Therefore, the volume of the parallelepiped built on vectors a , b, and c is equalto

V =Sh=a

bc

. (4)

However, our proof of this fact contains some gap. (Did you guess, what it is?).

Namely, we cancelled the expressionbc

. But it may be zero if vectors b and c arecollinear. However, in this case, the volume is obviously zero, anda

bc

= 0. Hence,

formula (4) remains valid anyway.

Meanwhile, formula (4) also shows that vectors a , b, and c are coplanar if and

only ifa

bc

= 0.

The equation in Property 6) of Theorem 2 is called the Vector Triple Product of

vectors a ,b, and c . It is used, e.g., to derive Keplers First Law of planetary motion (seeChapter 13 in the textbook).

Example 5 Check if the points A(1, 3, 2), B(3, 1, 6), C(5, 2, 0), and D(3, 6, 4) lie inthe same plane.

Solution. Note: You should know that any three points lie in some plane. So, the questionis if our fourpoints lie in the same plane.

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These points lie in the same plane if and only if the vectors AB, AC and AD lie inthe same plane, i.e., when these vectors are coplanar. Hence we need to find their scalartriple product. So, AB= 2, 4, 4, AC=4, 1, 2, AD= 2, 3, 6, and hence

AB

AC AD

=

2 4 44 1 2

2 3 6

= 7676 = 0.

Since this scalar triple product is equal to zero, our vectors are coplanar, whence A, B,C, and D lie in some plane.

3. Torque.

As an application of the tools developed, we could mention the notion of torque that iswidely used in physics, mechanics, and engineering. In fact, the correct torque is requiredin the majority of modern mechanics because goals of weight reduction and financialefficiency are difficult to overestimate (and also to achieve). So, all the technical gadgetsrequire precise torque when being assembled. There are many precise torque wrenchesand torque limiting tools accessible nowadays. But be careful with them because when

assembling something, the correct angle is also important to precise torque measuring.

Definition 5 The torque (relative to the origin) is defined to be the cross product ofthe position and torque vectors (see Figure 4 on Page 791 in the textbook)

= r F

and measures the tendency of the body to rotate about the origin.

So, the torque is a VECTOR. The direction of the torque vector indicates the axis ofrotation. Using the formula for the magnitude of the cross product, we get

| |=r F

=|r | F

sin ,

where is the angle between vectors r and F.In various technical documentation, torque is usually given in N m, lb in, etc. How-

ever, the direction may also matter. So, sometimes, you may be supplied with the requiredangle too (do you remember that a vector is defined by its magnitude and direction?).

Example 6 A wrench 30 cm long lies along the positive y-axis and grips a bolt at theorigin. A force is applied in the direction 0, 3, 4 at the end of the wrench. Find themagnitude of the force needed to supply100 N m of torque to the bolt.

Solution. First of all, we need to get to the same system (metric). As soon as themagnitude of the torque is expressed inNm, we need to get to meters, i.e., 30 cm= 0.3 m.So, having given the direction of the force vector,

F = F

0, 3, 4 .

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Then, clearly, r = 0, 0.3, 0, and the torque is equal to

= r F = F

(0, 0.3, 0 0, 3, 4) = F

i j k

0 0.3 00 3 4

=

F 1.2, 0, 0 .

Hence, we came to the equation

100 N m= | |= F

1.2 m.

Thus, the magnitude of the force should be equal to

F =100

1.2N= 83

1

3N.

As you can see, the vector products considered can be applied to various problemsof geometry, mechanics, engineering, etc. And, as you could also notice, introducing a

coordinate system and switching to vectors components may dramatically simplify calcu-lations. In fact, you may use formulas involving angles. But then, you will be involved ina variety of trigonometric equations, inequalities, etc., that are usually not easy to solve.Having vector components and involving vector operations is usually much simpler andmore efficient way. The more complicated constructions, the higher probability of gettingincorrect results. Vectors may simplify your life!

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