TRIPLE PRODUCTS

3. TRIPLE PRODUCTS

There are two products involving three vectors, one called the triple scalar product (because the answer is a scalar) and the other called the triple vector product (because the answer is a vector).

Figure 3.2

Figure 3.1 Triple Scalar Product This is written A · (B × C). There is a useful geo-

metrical interpretation of the triple scalar product (see Figure 3.1). Construct a parallelepiped using A, B, C as three intersecting edges. Then |B × C| is the area

Section 3 Triple Products 279

of the base (Figure 3.2) because |B × C| = |B| |C| sin θ, which is the area of a parallelogram with sides |B|, |C|, and angle θ. The height of the parallelepiped is |A| cos φ (Figure 3.1). Then the volume of the parallelepiped is

|B| |C| sin θ|A| cos φ = |B × C| |A| cos φ = A · (B × C).

If φ > 90 ◦ , this will come out negative, so in general we should say that the volume is |A · (B × C)|. Any side may be used as base, so, for example, B · (C × A) must also be either plus or minus the volume. There are six such triple scalar products, all equal except for sign [or twelve if you count both the type A · (B × C) and the type (B × C) · A)].

To write the triple scalar product in component form we first write B × C in determinant form [Chapter 3, equation (4.19)]:

B×C= B x B y B z .

Now A · (B × C) = A x (B × C) x +A y (B × C) y +A z (B × C) z , and this is exactly what we get by expanding, by elements of the first row, the determinant in (3.2) below; this determinant is then equal to A · (B × C).

A · (B × C) = B x B y B z .

Recalling that an interchange of rows changes the sign of a determinant, we can now easily write out the six (or twelve) products mentioned above with their proper signs. You should convince yourself of, and then remember, the following facts: The order of the factors is all that counts; the dot and cross may be interchanged. If the order of factors is cyclic (one way around the circle in Figure 3.3), all such triple scalar products are equal. If you go the other way, you get another set all equal to each other and the negatives of the first set. For example,

(3.3) (A × B) · C = A · (B × C)

= C · (A × B) = −(A × C) · B, etc.

Since it doesn’t matter where the dot and cross are, the triple scalar product is often written as (ABC),

Figure 3.3 meaning A · (B × C) or (A × B) · C.

Triple Vector Product This is written A × (B × C). Before we try to evaluate it, we can make the following observations. B × C is perpendicular to the plane of

B and C. A × (B × C) is perpendicular to the plane of A and (B × C); we are particularly interested in the fact that A × (B × C) is perpendicular to (B × C).

280 Vector Analysis Chapter 6

Now (see Figure 3.4) any vector perpendicular to B × C lies in the plane perpendicular to B × C, that is, the plane of B and C. Thus A × (B × C) is some vector in the plane of B and C, and can be written as some combination aB + bC, where

a and b are scalars which we want to find. (See Chapter 3, Section 8, Problem 5.) One way to find a and b is to write out

A × (B × C) in component form. We can simplify this work by choosing our coordinate system carefully; recall that a vector equation is true independently of the coordinate system. Given

Figure 3.4 the vectors A, B, C, we take the x axis along B, and the y axis

in the plane of B and C; then B × C is in the z direction. The vectors in component form relative to these axes are:

Using (3.4) we find B×C=B x i × (C x i+C y j) = B x C y (i × j) = B x C y k,

A × (B × C) = A x B x C y (i × k) + A y B x C y (j × k) =A x B x C y (−j) + A y B x C y (i).

We would like to write A × (B × C) in (3.5) as a combination of B and C; we can

do this by adding and subtracting A x B x C x i:

A × (B × C) = −A x B x (C x i+C y j) + (A y C y +A x C x )B x i. Each of these expressions is something simple in terms of the vectors in (3.4):

A y C y +A x C x = A · C, (3.7)

A x B x = A · B,

B x i = B. Using (3.7) in (3.6), we get (3.8)

C x i+C y j = C,

A × (B × C) = (A · C)B − (A · B)C. This important formula should be learned, but not memorized in terms of letters,

because that is confusing when you want some other combination of the same letters. Learn instead the following three facts:

The value of a triple vector product is a linear combination of the two vectors in the parenthesis [B and C in (3.8)]; the coefficient of

(3.9) each vector is the dot product of the other two; the middle vector

in the triple product [B in (3.8)] always has the positive sign.

This method also covers triple vector products with the parenthesis first; by (3.9), the value of (B × C) × A is (A · B)C − (A · C)B. This is correct since it is just the negative of what we had above for A × (B × C).

Section 3 Triple Products 281

Applications of the Triple Scalar Product We have shown that the torque of

a force F about an axis may be written as r × F in one special case, namely when r and F are in a plane perpendicular to the axis. Now let us consider the general case of finding the torque produced by a force F about any given line (axis) L in Figure 3.5. Let r be a vector from some (that is, any) point on L to the line of action of F; let O be the tail of r. Then we define the torque about the point O to be r × F. Note that this cannot contradict our previous discussion of torque because we were considering torque about a line before, and this definition is of torque about a point. However, we shall show how the two notions are connected. Also notice that r × F is not changed if the head of r is moved along F; for this just adds a multiple

Figure 3.5 of F to r, and F × F = 0 (see Problem 10).

We shall now show that the torque of F about the line L through O is n · (r × F), where n is a unit vector along L. To simplify the calculation, choose the positive z axis in the direction n; then n = k. Think of a door hinged to rotate about the z

Figure 3.7

Figure 3.6 axis as in Figure 3.6. Let a force F be applied to it at the head of the vector r. We

first find the torque of F about the z axis by elementary methods and definition. Break F into its components; the z component is parallel to the rotation axis and produces no torque about it (pulling straight up or down on a door handle does not tend to open or close the door!). The x and y components can be seen better if we draw them in the (x, y) plane (Figure 3.7; note that the x and y axes are rotated

90 ◦ clockwise from their usual position in order to compare this figure more easily with Figure 3.6). The torque about the z axis produced by F x and F y is xF y − yF x by the elementary definition of torque. We want to show that this is the same as n · (r × F) or here k · (r × F). Using (3.2) we find

k · (r × F) = x

z = xF y − yF x .

To summarize:

282 Vector Analysis Chapter 6

In Figure 3.5, the torque of F about point O is r × F. The torque (3.10)

of F about the line L through O is n · (r × F) where n is a unit vector along L.

This proof can easily be given without reference to a coordinate system. Let n. Then any vector (F or r, say) can be written as the sum of a vector parallel

to the axis and a vector perpendicular to the axis (that is, somewhere in the plane perpendicular to n):

r=r ⊥ +r ,

F=F ⊥ +F .

Then the torque about O produced by F is

r × F = (r ⊥ +r )×(F ⊥ +F )

=r ⊥ × F ⊥ +r ⊥ × F +r × F ⊥ +r × F . The last term is zero (cross product of parallel vectors). Also r and F are parallel

to n; therefore their cross products with anything are in the plane perpendicular to n, and the dot product of n with these is zero. Hence we have

n · (r × F) = n · (r ⊥ × F ⊥ ).

Now r ⊥ and F ⊥ are in a plane perpendicular to n; thus the torque about n produced by F ⊥ is (by Section 2) r ⊥ × F ⊥ . But since only the component of F perpendicular to n produces a torque about n, r ⊥ × F ⊥ is the total torque about n produced by F. The vector torque r ⊥ × F ⊥ is in the ±n direction since r ⊥ and F ⊥ are perpendicular to n; the dot product of this vector torque with the unit vector n gives a scalar torque of the same magnitude; the ± sign indicates whether the torque is in the +n or the −n direction.

Example 1. If F = i + 3j − k acts at the point (1, 1, 1), find the torque of F about the line r = 3i + 2k + (2i − 2j + k)t. We first find the vector torque about a point on the line, say the point (3, 0, 2).

By (3.10) and Figure 3.5, this is r × F where r is the vector from the point about which we want the torque, to the point at which F acts, that is, from (3, 0, 2) to (1, 1, 1); then r = (1, 1, 1) − (3, 0, 2) = (−2, 1, −1). The vector torque is

r×F= −2 1 −1 = 2i − 3j − 7k.

The torque about the line is n · (r × F) where n is a unit vector along the line,

namely n = 1 3 (2i − 2j + k). Then the torque about the line is

n · (r × F) = (2i − 2j + k) · (2i − 3j − 7k) = 1.

Section 3 Triple Products 283

Example 2. As another application of the triple scalar product, let’s find the Jacobian we used in Chapter 5, Section 4 for changing variables in a multiple integral. As you know, in rectangular coordinates the volume element is a rectangular box of volume dx dy dz. In other coordinate systems, the volume element may be approximately a parallelepiped as in Figure 3.1. We want a formula for the volume element in this case. (See, for example, the cylindrical and spherical coordinate volume elements in Chapter 5, Figures 4.4 and 4.5.)

Suppose we are given formulas for x, y, z as functions of new variables u, v, w. Then we want to find the vectors along the edges of the volume element in the u, v, w system. Suppose vector A in Figure 3.1 is along the direction in which u increases while v and w remain constant. Then if dr = i dx + j dy + k dz is a vector in this direction, we have

Similarly if B is along the increasing v edge of the volume element and C is along the increasing w edge, we have

Then by (3.2)

where J is the Jacobian of the transformation from x, y, z to u, v, w. Recall from the discussion of (3.2) that the triple scalar product may turn out to be positive or negative. Since we want a volume element to be positive, we use the absolute value of J. Thus the u, v, w volume element is |J| du dv dw as stated in Chapter 5, Section 4.

Applications of the Triple Vector Product In Figure 3.8 (compare Figure 2.6), suppose the particle m is at rest on a ro- tating rigid body (for example, the earth). Then the angular momentum L of m about point O is defined by the equation L = r × (mv) = mr × v. In the discussion of Figure 2.6, we showed that v = ω × r. Thus, L = mr × (ω × r). See Prob- lem 16 and also Chapter 10, Section 4.

As another example, it is shown in mechanics that the cen- tripetal acceleration of m in Figure 3.8 is a = ω × (ω × r). See Problem 17.

Figure 3.8

284 Vector Analysis Chapter 6

PROBLEMS, SECTION 3

1. If A = 2i − j − k, B = 2i − 3j + k, C = j + k, find (A · B)C, A(B · C), (A × B) · C, A · (B × C), (A × B) × C, A × (B × C).

For Problems 2 to 6, given A = i + j − 2k, B = 2i − j + 3k, C = j − 5k: 2. Find the work done by the force B acting on an object which undergoes the dis-

placement C. 3. Find the total work done by forces A and B if the object undergoes the displacement

C. Hint: Can you add the two forces first? 4. Let O be the tail of B and let A be a force acting at the head of B. Find the torque

of A about O; about a line through O perpendicular to the plane of A and B; about a line through O parallel to C.

5. Let A and C be drawn from a common origin and let C rotate about A with an angular velocity of 2 rad/sec. Find the velocity of the head of C.

6. In Problem 5, draw B with its tail at the head of A. If the figure is rotating as in Problem 5, find the velocity of the head of B. With the same diagram, let B be a force; find the torque of B about the head of C, and about the line C.

7. A force F = 2i − 3j + k acts at the point (1, 5, 2). Find the torque due to F (a) about the origin;

(b) about the y axis; (c)

about the line x/2 = y/1 = z/(−2). 8. A vector force with components (1, 2, 3) acts at the point (3, 2, 1). Find the vector

torque about the origin due to this force and find the torque about each of the coordinate axes.

9. The force F = 2i − j − 5k acts at the point (−5, 2, 1). Find the torque due to F about the origin and about the line 2x = −4y = −z.

10. In Figure 3.5, let r ′ be another vector from O to the line of F . Show that r ′ × F= r × F. Hint: r − r ′ is a vector along the line of F and so is a scalar multiple of F. (The scalar has physical units of distance divided by force, but this fact is irrelevant for the vector proof.) Show also that moving the tail of r along n does not change n · r × F. Hint: The triple scalar product is not changed by interchanging the dot and the cross.

11. Write out the twelve triple scalar products involving A, B, and C and verify the facts stated just above (3.3).

12. (a) Simplify (A · B) 2 − [(A × B) × B] · A by using (3.9). (b)

Prove Lagrange’s identity: (A × B) ·(C × D) = (A ·C)(B · D) −(A·D)(B·C). 13. Prove that the triple scalar product of (A × B), (B × C), and (C × A), is equal to

the square of the triple scalar product of A, B, and C. Hint: First let (B × C) = D, and evaluate (A × B) × D. [See Am. J. Phys. 66, 739 (1998).]

14. Prove the Jacobi identity: A × (B × C) + B × (C × A) + C × (A × B) = 0. Hint: Expand each triple product as in equations (3.8) and (3.9).

Section 4 Differentiation of Vectors 285

15. In the figure u 1 is a unit vector in the direction of an incident ray of light, and u 3 and u 2 are unit

vectors in the directions of the reflected and re- fracted rays. If u is a unit vector normal to the

surface AB, the laws of optics say that θ 1 =θ 3 and n 1 sin θ 1 =n 2 sin θ 2 , where n 1 and n 2 are constants

(indices of refraction). Write these laws in vector form (using dot or cross products).

16. In the discussion of Figure 3.8, we found for the angular momentum, the formula L = mr × (ω × r). Use (3.9) to expand this triple product. If r is perpendicular to ω , show that you obtain the elementary formula, angular momentum = mvr.

17. Expand the triple product for a = ω × (ω × r) given in the discussion of Figure 3.8. If r is perpendicular to ω (Problem 16), show that a = −ω 2 r, and so find the elementary result that the acceleration is toward the center of the circle and of magnitude v 2 /r.

18. Two moving charged particles exert forces on each other because each one creates a magnetic field in which the other moves (see Problem 4.6). These two forces are proportional to v 1 × [v 2 × r] and v 2 × [v 1 × (−r)] where r is the vector joining the particles. By using (3.9), show that these forces are equal and opposite (Newton’s third “law”) if and only if r × (v 1 × v 2 ) = 0. Compare Problem 14.

19. The force F = i + 3j + 2k acts at the point (1, 1, 1). (a)

Find the torque of the force about the point (2, −1, 5). Careful! The vector r goes from (2, −1, 5) to (1, 1, 1).

(b) Find the torque of the force about the line r = 2i − j + 5k + (i − j + 2k)t. Note that the line goes through the point (2, −1, 5).

20. The force F = 2i − 5k acts at the point (3, −1, 0). Find the torque of F about each of the following lines.

(a) r = (2i − k) + (3j − 4k)t. (b)

r = i + 4j + 2k + (2i + j − 2k)t.