7.4 Cross Product

INTRODUCTION

The dot product, introduced in the preceding section, works in both two- and three-dimensional spaces and results in a number. On the other hand, the cross product, introduced in this section, is only defined for vectors in 3-space and results in another vector in 3-space.

Before proceeding we need the following two results from the theory of determinants.

See Sections 8.4 and 8.5 for a complete discussion of determinants.

Review of Determinants

A determinant of order 2 is the number

. (1)

A determinant of order 3 is the number defined in terms of three determinants of order 2:

. (2)

This is called expanding the determinant along the first row. For example, from (1)

and from (2)

Component Form of the Cross Product

As we did in the discussion of the dot product, we define the cross product of two vectors a and b in terms of the components of the vectors.

DEFINITION 7.4.1 Cross Product of Two Vectors

The cross product of two vectors a = 〈a1, a2, a3〉 and b = 〈b1, b2, b3〉 is the vector

(3)

The coefficients of the basis vectors in (3) are recognized as determinants of order 2, so (3) can be written as

(4)

By comparing (4) with (2), we see that the cross product can be formally expressed as a determinant of order 3 with the unit vectors i, j, and k as entries in the first row and the components of a and b entries of the second and third rows, respectively:

(5)

EXAMPLE 1 Cross Product Using (5)

Let a = 4i − 2j + 5k and b = 3i + j − k. Find a × b.

SOLUTION

We use (5) and expand the determinant along the first row:

≡

EXAMPLE 2 Cross Product of Two Basis Vectors

If i = 〈1, 0, 0〉 and j = 〈0, 1, 0〉, then (5) gives

≡

Proceeding as in Example 2, it is readily shown that

(6)

(7)

and(8)

The results in (6) can be obtained using the circular mnemonic illustrated in FIGURE 7.4.1. Notice in (7), if we cross two basis vectors in the direction opposite to that shown in Figure 7.4.1, then we get the negative of the corresponding vector in (6). The results in (6), (7), and (8) are special cases of (ii) and (vi) in Theorem 7.4.1.

3 unit vectors are graphed in 3-space. Arrows point from i to j, from j to k, and from k to i.

FIGURE 7.4.1 A mnemonic for cross products involving i, j, and k

Properties

The next theorem summarizes some of the important properties of the cross product.

THEOREM 7.4.1 Properties of the Cross Product

  1. a × b = 0 if a = 0 or b = 0
  2. a × b = −b × a
  3. a × (b + c) = (a × b) + (a × c) ←distributive law
  4. (a + b) × c = (a × c) + (b × c) ←distributive law
  5. a × (kb) = (ka) × b = k(a × b), k a scalar
  6. a × a = 0
  7. a · (a × b) = 0
  8. b · (a × b) = 0
PROOF:

All the properties can be proved directly from (4) and (5). To prove part (i), we let a = 〈0, 0, 0〉 = 0 and b = 〈b1, b2, b3〉. Then from (5)

To prove part (iii) we let a = 〈a1, a2, a3〉, b = 〈b1, b2, b3〉, and c = 〈c1, c2, c3〉. Then

≡

Because part (ii) of Theorem 7.4.1 indicates that the cross product is not commutative, there are two distributive laws in parts (iii) and (iv) of the theorem.

Parts (vii) and (viii) of Theorem 7.4.1 deserve special attention. In view of Theorem 7.3.3 we see (vii) and (viii) imply that the vector a × b is orthogonal to a and that a × b is orthogonal to b. In the case when a and b are nonzero vectors, then a × b is orthogonal to every vector in the plane containing a and b. Put another way,

a × b is orthogonal to the plane containing a and b. (9)

You can see in Figure 7.4.1 and from the results in (6) that i × j is orthogonal to the plane of i and j; that is, the xy-plane, j × k is orthogonal to the yz-plane, and k × i is orthogonal to the xz-plane.

The results in (6) and (7) give us a clue as to the direction in which the vector a × b points.

Right-Hand Rule

The vectors a, b, and a × b form a right-handed system or a right-handed triple. This means that a × b points in the direction given by the right-hand rule:

If the fingers of the right hand point along the vector a and then curl toward the vector b, the thumb will give the direction of a × b. (10)

See FIGURE 7.4.2(a). In Figure 7.4.2(b), the right-hand rule shows the direction of b × a.

2 vectors a and b are graphed on a horizontal plane in 3-space. The angle between the vectors is theta. Graph a. The fingers of the right hand curl from a to b, and the thumb points up. Vector n is orthogonal to the plane and points up. Graph b. The fingers of the right hand curl from b to a, and the thumb points down. Vector n is orthogonal to the plane and points down.

FIGURE 7.4.2 Right-hand rule

Because a × b is a vector its magnitude a × b can be found from (vii) of Definition 7.2.1. As we see in the next theorem, a × b can also be expressed in terms of the angle θ between the vectors a and b.

THEOREM 7.4.2 Magnitude of the Cross Product

For nonzero vectors a and b, if θ is the angle between a and b (0 ≤ θ ≤ π), then

(11)

PROOF:

We separately compute the squares of the left- and right-hand sides of equation (11) using the component forms of a and b:

Since both sides are equal to the same quantity, they must be equal to each other, so

Finally, taking the square root of both sides and using the fact that since sin θ ≥ 0 for 0 ≤ θ ≤ π, we have a × b = sin θ. ≡

Alternative Form

Combining (9), (10), and Theorem 7.4.2 we see for any pair of vectors a and b in R3 that the cross product has the alternative form

(12)

where n is a unit vector given by the right-hand rule that is orthogonal to the plane of a and b.

This more geometric form is generally used as the definition of the cross product in a physics course.

Parallel Vectors

We saw in Section 7.1 that two nonzero vectors are parallel if and only if one is a nonzero scalar multiple of the other. Thus, two vectors are parallel if they have the forms a and ka, where a is any vector and k is a scalar. By properties (v) and (vi) in Theorem 7.4.1, the cross product of parallel vectors must be 0. This fact is summarized in the next theorem.

THEOREM 7.4.3 Criterion for Parallel Vectors

Two nonzero vectors a and b are parallel if and only if a × b = 0.

Of course, Theorem 7.4.3 follows as well from (12) because the angle between parallel vectors a and b is either θ = 0 or θ = π.

EXAMPLE 3 Parallel Vectors

Determine whether a = 2i + j − k and b = −6i − 3j + 3k are parallel vectors.

SOLUTION

From the cross product

and Theorem 7.4.3 we conclude that a and b are parallel vectors. ≡

Special Products

The scalar triple product of vectors a, b, and c is a · (b × c). Using the component forms of the definitions of the dot and cross products, we have

Thus, we see that the scalar triple product can be written as a determinant of order 3:

(13)

Using (4), (5), and (2) of Section 7.3 it can be shown that

a · (b × c) = (a × b) · c. (14)

See Problem 61 in Exercises 7.4.

The vector triple product of three vectors a, b, and c is

a × (b × c).

The vector triple product is related to the dot product by

a × (b × c) = (a · c)b − (a · b)c. (15)

See Problem 59 in Exercises 7.4.

Areas

Two nonzero and nonparallel vectors a and b can be considered to be the sides of a parallelogram. The area A of a parallelogram is

A = (base) (altitude).

From FIGURE 7.4.3(a) we see that A = ( sin θ) = sin θ

or (16)

Likewise from Figure 7.4.3(b), we see that the area of a triangle with sides a and b is

(17)

2 vectors a and b have a common initial point and form and angle theta. The length of the vectors is absolute value of a and absolute value of b. The vectors are two sides of a parallelogram. Graph a. The distance from the end point of a to b is h equals absolute value of a times sine theta. Graph b. The diagonal joining the end points of a and b divides the parallelogram into 2 triangles.

FIGURE 7.4.3 Areas of a parallelogram and a triangle

EXAMPLE 4 Area of a Triangle

Find the area of the triangle determined by the points P1(1, 1, 1), P2(2, 3, 4), and P3(3, 0, −1).

SOLUTION

The vectors and can be taken as two sides of the triangle. Since = i + 2j + 3k and = i − 3j − 5k, we have

From (17) we see that the area is

≡

Volume of a Parallelepiped

If the vectors a, b, and c do not lie in the same plane, then the volume of the parallelepiped with edges a, b, and c shown in FIGURE 7.4.4 is

(18)

Thus, the volume of a parallelepiped determined by three vectors is the absolute value of the scalar triple product of the vectors.

A parallelepiped has edges vectors a, b, and c. Vector a forms an angle with vector b cross c. The distance between the end point of a to the plane b c is determinant comp subscript b cross c times a.

FIGURE 7.4.4 Parallelepiped formed by three vectors

Coplanar Vectors

Vectors that lie in the same plane are said to be coplanar. We have just seen that if the vectors a, b, and c are not coplanar, then necessarily a · (b × c) ≠ 0, since the volume of a parallelepiped with edges a, b, and c has nonzero volume. Equivalently stated, this means that if a · (b × c) = 0, then the vectors a, b, and c are coplanar. Since the converse of this last statement is also true, we have:

a · (b × c) = 0 if and only if a, b, and c are coplanar.

Physical Interpretation of the Cross Product

In physics a force F acting at the end of a position vector r, as shown in FIGURE 7.4.5, is said to produce a torque τ defined by τ For example, if , , and θ = 30°, then from (11),

τ = (3.5)(20) sin 30° = 35 N-m.

If F and r are in the plane of the page, the right-hand rule implies that the direction of τ is outward from, and perpendicular to, the page (toward the reader).

A position vector r goes up and to the right in a coordinate plane. A vector F from the end point of r forms and angle theta with r. The component of F perpendicular to r is absolute value of F times sine theta.

FIGURE 7.4.5 A force acting at the end of a vector

As we see in FIGURE 7.4.6, when a force F is applied to a wrench, the magnitude of the torque τ is a measure of the turning effect about the pivot point P and the vector τ is directed along the axis of the bolt. In this case τ points inward from the page.

A force F is applied to a wrench. The pivot point P of the bolt is on the left, and the force F is applied at a distance r, down and to the left.

FIGURE 7.4.6 A wrench applying torque to a bolt

REMARKS

When working with vectors, one should be careful not to mix the symbols · and × with the symbols for ordinary multiplication, and to be especially careful in the use, or lack of use, of parentheses. For example, expressions such as

a × b × c     a · b × c     a · b · c     a · bc

are not meaningful or well-defined.

7.4 Exercises Answers to selected odd-numbered problems begin on page ANS-16.

In Problems 1–10, find a × b.

  1. a = i − j, b = 3j + 5k
  2. a = 2i + j, b = 4i − k
  3. a = 〈1, −3, 1〉, b = 〈2, 0, 4〉
  4. a = 〈1, 1, 1〉, b = 〈−5, 2, 3〉
  5. a = 2i − j + 2k, b = −i + 3j − k
  6. a = 4i + j − 5k, b = 2i + 3j − k
  7. a = 〈 , 0, 〉, b = 〈4, 6, 0〉
  8. a = 〈0, 5, 0〉, b = 〈2, −3, 4〉
  9. a = 〈2, 2, −4〉, b = 〈−3, −3, 6〉
  10. a = 〈8, 1, −6〉, b = 〈1, −2, 10〉

In Problems 11 and 12, find × .

  1. P1(2, 1, 3), P2(0, 3, −1), P3(−1, 2, 4)
  2. P1(0, 0, 1), P2(0, 1, 2), P3(1, 2, 3)

In Problems 13 and 14, find a vector that is perpendicular to both a and b.

  1. a = 2i + 7j − 4k, b = i + j − k
  2. a = 〈−1, −2, 4〉, b = 〈4, −1, 0〉

In Problems 15 and 16, verify that a · (a × b) = 0 and b · (a × b) = 0.

  1. a = 〈5, −2, 1〉, b = 〈2, 0, −7〉
  2. a = i − j, b = 2i − 2j + 6k

In Problems 17 and 18, (a) calculate b × c followed by a × (b × c), and (b) verify the results in part (a) by (15) of this section.

  1. a = i − j + 2k
    b = 2i + j + k
    c = 3i + j + k
  2. a = 3i − 4k,
    b = i + 2j − k
    c = −i + 5j + 8k

In Problems 19–36, find the indicated scalar or vector without using (5), (13), or (15).

  1. (2i) × j
  2. i × (−3k)
  3. k × (2i − j)
  4. i × (j × k)
  5. [(2k) × (3j)] × (4j)
  6. (2i − j + 5k) × i
  7. (i + j) × (i + 5k)
  8. i × k − 2(j × i)
  9. k · (j × k)
  10. i · [j × (−k)]
  11. 4j − 5(i × j)
  12. (i × j) · (3j × i)
  13. i × (i × j)
  14. (i × j) × i
  15. (i × i) × j
  16. (i · i)(i × j)
  17. 2j · [i × (j − 3k)]
  18. (i × k) × (j × i)

In Problems 37–44, a × b = 4i − 3j + 6k and c = 2i + 4j − k. Find the indicated scalar or vector.

  1. a × (3b)
  2. b × a
  3. (−a) × b
  4. a × b
  5. (a × b) × c
  6. (a × b) · c
  7. a · (b × c)
  8. (4a) · (b × c)

In Problems 45 and 46, (a) verify that the given quadrilateral is a parallelogram, and (b) find the area of the parallelogram.

  1. A quadrilateral is graphed in 3-space. The coordinates of the vertices are (1, negative 3, 4), (0, 0, 4), (2, 0, 0), and (1, 3, 0).

    FIGURE 7.4.7 Parallelogram in Problem 45

  2. A quadrilateral is graphed in 3-space. The coordinates of the vertices are (2, 0, 2), (negative 2, 0, 3), (negative 1, 4, 2), and (3, 4, 1).

    FIGURE 7.4.8 Parallelogram in Problem 46

In Problems 47–50, find the area of the triangle determined by the given points.

  1. P1(1, 1, 1), P2(1, 2, 1), P3(1, 1, 2)
  2. P1(0, 0, 0), P2(0, 1, 2), P3(2, 2, 0)
  3. P1(1, 2, 4), P2(1, −1, 3), P3(−1, −1, 2)
  4. P1(1, 0, 3), P2(0, 0, 6), P3(2, 4, 5)

In Problems 51 and 52, find the volume of the parallelepiped for which the given vectors are three edges.

  1. a = i + j, b = −i + 4j, c = 2i + 2j + 2k
  2. a = 3i + j + k, b = i + 4j + k, c = i + j + 5k
  3. Determine whether the vectors a = 4i + 6j, b = −2i + 6j − 6k, and c = i + 3j + k are coplanar.
  4. Determine whether the four points P1(1, 1, −2), P2(4, 0, −3), P3(1, −5, 10), and P4(−7, 2, 4) lie in the same plane.
  5. As shown in FIGURE 7.4.9, the vector a lies in the xy-plane and the vector b lies along the positive z-axis. Their magnitudes are = 6.4 and = 5.
    1. Use Theorem 7.4.2 to find a × b.
    2. Use the right-hand rule to find the direction of a × b.
    3. Use part (b) to express a × b in terms of the unit vectors i, j, and k.
    2 vectors are graphed in 3-space. Vector a is on the x y plane, at 60 degrees from the positive x-axis. Vector b is along the positive z-axis.

    FIGURE 7.4.9 Vectors for Problem 55

  6. Two vectors a and b lie in the xz-plane so that the angle between them is 120°. If = and = 8, find all possible values of a × b.
  7. A three-dimensional lattice is a collection of integer combinations of three noncoplanar basis vectors a, b, and c. In crystallography, a lattice can specify the locations of atoms in a crystal. X-ray diffraction studies of crystals use the “reciprocal lattice” that has basis
    1. A certain lattice has basis vectors a = i, b = j, and c = (i + j + k). Find basis vectors for the reciprocal lattice.
    2. The unit cell of the reciprocal lattice is the parallelepiped with edges A, B, and C, while the unit cell of the original lattice is the parallelepiped with edges a, b, and c. Show that the volume of the unit cell of the reciprocal lattice is the reciprocal of the volume of the unit cell of the original lattice. [Hint: Start with B × C and use (15).]

Discussion Problems

  1. Use (4) to prove property (iii) of the cross product.
  2. Prove a × (b × c) = (a · c)b − (a · b) c.
  3. Prove or disprove a × (b × c) = (a × b) × c.
  4. Prove a · (b × c) = (a × b) · c.
  5. Prove a × (b × c) + b × (c × a) + c × (a × b) = 0.
  6. Prove Lagrange’s identity:

    a × b2 = 22 − (a · b)2.

  7. Does a × b = a × c imply that b = c?
  8. Show that (a + b) × (a − b) = 2b × a.