12.4 The Cross Prodct 873 12.4 The Cross Prodct In stdying lines in the plane, when we needed to describe how a line was tilting, we sed the notions of slope and angle of inclination. In space, we want a way to describe how a plane is tilting. We accomplish this by mltiplying two ectors in the plane together to get a third ector perpendiclar to the plane. The direction of this third ector tells s the inclination of the plane. The prodct we se to mltiply the ectors together is the ector or cross prodct, the second of the two ector mltiplication methods we stdy in calcls. Cross prodcts are widely sed to describe the effects of forces in stdies of electricity, magnetism, flid flows, and orbital mechanics. This section presents the mathematical properties that accont for the se of cross prodcts in these fields. n The Cross Prodct of Two Vectors in Space We start with two nonzero ectors and in space. If and are not parallel, they determine a plane. We select a nit ector n perpendiclar to the plane by the right-hand rle. This means that we choose n to be the nit (normal) ector that points the way yor right thmb points when yor fingers crl throgh the angle from to (Figre 12.27). Then the cross prodct * ( cross ) is the ector defined as follows. DEFINITION Cross Prodct FIGURE 12.27 *. The constrction of * = s ƒ ƒƒ ƒ sin d n Unlike the dot prodct, the cross prodct is a ector. For this reason it s also called the ector prodct of and, and applies only to ectors in space. The ector * is orthogonal to both and becase it is a scalar mltiple of n. Since the sines of 0 and p are both zero, it makes sense to define the cross prodct of two parallel nonzero ectors to be 0. If one or both of and are zero, we also define * to be zero. This way, the cross prodct of two ectors and is zero if and only if and are parallel or one or both of them are zero. Parallel Vectors Nonzero ectors and are parallel if and only if * = 0. The cross prodct obeys the following laws.
874 Chapter 12: Vectors and the Geometry of Space n Properties of the Cross Prodct If,, and w are any ectors and r, s are scalars, then 1. srd * ssd = srsds * d 2. * s + wd = * + * w 3. s + wd * = * + w * 4. * = -s * d 5. 0 * = 0 FIGURE 12.28 The constrction of *. z k i j (j i) i j j k i (i k) i y k x i j k (k j) To isalize Property 4, for example, notice that when the fingers of a right hand crl throgh the angle from to, the thmb points the opposite way and the nit ector we choose in forming * is the negatie of the one we choose in forming * (Figre 12.28). Property 1 can be erified by applying the definition of cross prodct to both sides of the eqation and comparing the reslts. Property 2 is proed in Appendix 6. Property 3 follows by mltiplying both sides of the eqation in Property 2 by -1 and reersing the order of the prodcts sing Property 4. Property 5 is a definition. As a rle, cross prodct mltiplication is not associatie so s * d * w does not generally eqal * s * wd. (See Additional Exercise 15.) When we apply the definition to calclate the pairwise cross prodcts of i, j, and k, we find (Figre 12.29) k j i * j = -sj * id = k j * k = -sk * jd = i k * i = -si * kd = j i Diagram for recalling these prodcts FIGURE 12.29 The pairwise cross prodcts of i, j, and k. and i * i = j * j = k * k = 0. ƒ * ƒ Is the Area of a Parallelogram Becase n is a nit ector, the magnitde of * is Area base height sin h sin FIGURE 12.30 The parallelogram determined by and. ƒ * ƒ = ƒ ƒƒƒ ƒ sin ƒƒnƒ = ƒ ƒƒƒ sin. This is the area of the parallelogram determined by and (Figre 12.30), ƒ ƒ being the base of the parallelogram and ƒ ƒƒsin ƒ the height. Determinant Formla for * Or next objectie is to calclate * from the components of and relatie to a Cartesian coordinate system.
` 12.4 The Cross Prodct 875 Determinants 2 * 2 and 3 * 3 determinants are ealated as follows: EXAMPLE 2 1-4 3` a 1 a 2 a 3 3 b 1 b 2 b 3 3 c 1 c 2 c 3 b 1 b 3 b 1 b 2 - a 2 ` ` + a 3 ` ` c 3 c 2 c 1 a ` c EXAMPLE -5 3 1 3 2 1 13-4 3 1 b = ad - bc d` = s2ds3d - s1ds -4d = 6 + 4 = 10 b 2 b 3 = a 1 ` ` c 3 1 1 = s -5d ` 3 1` - s3d ` 2 1 2 1 + s1d ` -4 1` -4 3` = -5s1-3d - 3s2 + 4d + 1s6 + 4d = 10-18 + 10 = 2 (For more information, see the Web site at www.aw-bc.com/thomas.) c 2 c 1 Sppose that = 1 i + 2 j + 3 k, = 1 i + 2 j + 3 k. Then the distribtie laws and the rles for mltiplying i, j, and k tell s that * = s 1 i + 2 j + 3 kd * s 1 i + 2 j + 3 kd The terms in the last line are the same as the terms in the expansion of the symbolic determinant We therefore hae the following rle. = 1 1 i * i + 1 2 i * j + 1 3 i * k + 2 1 j * i + 2 2 j * j + 2 3 j * k + 3 1 k * i + 3 2 k * j + 3 3 k * k = s 2 3-3 2 di - s 1 3-3 1 dj + s 1 2-2 1 dk. 3 1 2 3 3. 1 2 3 Calclating Cross Prodcts Using Determinants If = 1 i + 2 j + 3 k and = 1 i + 2 j + 3 k, then * = 3 1 2 3 3. 1 2 3 EXAMPLE 1 Calclating Cross Prodcts with Determinants z Find * and * if = 2i + j + k and = -4i + 3j + k. Soltion R( 1, 1, 2) * = 3 1 1 2 1 13 = p 3 1 p i - p 2 1-4 1 p j + p 2 1-4 3 p k -4 3 1 P(1, 1, 0) 0 y = -2i - 6j + 10k * = -s * d = 2i + 6j - 10k x Q(2, 1, 1) EXAMPLE 2 Finding Vectors Perpendiclar to a Plane FIGURE 12.31 The area of triangle PQR is half of ƒ PQ 1 * PR 1 ƒ (Example 2). Find a ector perpendiclar to the plane of (Figre 12.31). Ps1, -1, 0d, Qs2, 1, -1d, and Rs -1, 1, 2d
876 Chapter 12: Vectors and the Geometry of Space Soltion The ector PQ 1 * PR 1 is perpendiclar to the plane becase it is perpendiclar to both ectors. In terms of components, PQ 1 = s2-1di + s1 + 1dj + s -1-0dk = i + 2j - k PR 1 = s -1-1di + s1 + 1dj + s2-0dk = -2i + 2j + 2k PQ 1 * PR 1 = 3 1 2-1 3 2-1 = ` 2 2` i - ` 1-1 -2 2` j + ` 1 2-2 2` k -2 2 2 = 6i + 6k. EXAMPLE 3 Finding the Area of a Triangle Find the area of the triangle with ertices Ps1, -1, 0d, Qs2, 1, -1d, (Figre 12.31). and Rs -1, 1, 2d Soltion The area of the parallelogram determined by P, Q, and R is ƒ PQ 1 * PR 1 ƒ = ƒ 6i + 6k ƒ Vales from Example 2. = 2s6d 2 + s6d 2 = 22 # 36 = 622. The triangle s area is half of this, or 322. Torqe n EXAMPLE 4 Finding a Unit Normal to a Plane Find a nit ector perpendiclar to the plane of Ps1, -1, 0d, Qs2, 1, -1d, and Rs -1, 1, 2d. Soltion Since PQ 1 * PR 1 is perpendiclar to the plane, its direction n is a nit ector perpendiclar to the plane. Taking ales from Examples 2 and 3, we hae n = PQ1 * PR 1 ƒ PQ 1 * PR 1 6i + 6k = = 1 ƒ 622 22 i + 1 22 k. For ease in calclating the cross prodct sing determinants, we sally write ectors in the form = 1 i + 2 j + 3 k rather than as ordered triples = 8 1, 2, 3 9. Component of F perpendiclar to r. Its length is F sin. r F FIGURE 12.32 The torqe ector describes the tendency of the force F to drie the bolt forward. Torqe When we trn a bolt by applying a force F to a wrench (Figre 12.32), the torqe we prodce acts along the axis of the bolt to drie the bolt forward. The magnitde of the torqe depends on how far ot on the wrench the force is applied and on how mch of the force is perpendiclar to the wrench at the point of application. The nmber we se to measre the torqe s magnitde is the prodct of the length of the leer arm r and the scalar component of F perpendiclar to r. In the notation of Figre 12.32, Magnitde of torqe ector = ƒ r ƒ ƒ F ƒ sin,
12.4 The Cross Prodct 877 or ƒ r * F ƒ. If we let n be a nit ector along the axis of the bolt in the direction of the torqe, then a complete description of the torqe ector is r * F, or Torqe ector = s ƒ r ƒƒf ƒ sin d n. Recall that we defined * to be 0 when and are parallel. This is consistent with the torqe interpretation as well. If the force F in Figre 12.32 is parallel to the wrench, meaning that we are trying to trn the bolt by pshing or plling along the line of the wrench s handle, the torqe prodced is zero. P 3 ft bar 70 F Q 20 lb magnitde force FIGURE 12.33 The magnitde of the torqe exerted by F at P is abot 56.4 ft-lb (Example 5). EXAMPLE 5 Finding the Magnitde of a Torqe The magnitde of the torqe generated by force F at the piot point P in Figre 12.33 is Triple Scalar or Box Prodct ƒ PQ 1 * F ƒ = ƒ PQ 1 ƒƒfƒ sin 70 L s3ds20ds0.94d L 56.4 ft-lb. The prodct s * d # w is called the triple scalar prodct of,, and w (in that order). As yo can see from the formla ƒ s * d # w ƒ = ƒ * ƒƒw ƒƒcos ƒ, the absolte ale of the prodct is the olme of the parallelepiped (parallelogram-sided box) determined by,, and w (Figre 12.34). The nmber ƒ * ƒ is the area of the base parallelogram. The nmber ƒ w ƒƒcos ƒ is the parallelepiped s height. Becase of this geometry, s * d # w is also called the box prodct of,, and w. w Height w cos Area of base Volme area of base height w cos ( ) w FIGURE 12.34 The nmber is the olme of a parallelepiped. ƒ s * d # w ƒ The dot and cross may be interchanged in a triple scalar prodct withot altering its ale. By treating the planes of and w and of w and as the base planes of the parallelepiped determined by,, and w, we see that s * d # w = s * wd # = sw * d #. Since the dot prodct is commtatie, we also hae s * d # w = # s * wd.
878 Chapter 12: Vectors and the Geometry of Space The triple scalar prodct can be ealated as a determinant: s * d # 2 3 1 3 1 2 w = c ` ` i - ` ` j + ` ` k d # w 2 3 1 3 1 2 2 3 1 3 1 2 = w 1 ` ` - w 2 ` ` + w 3 ` ` 3 3 2 = 3 2 1 2 3 1 2 3 3. w 1 w 2 w 3 1 1 Calclating the Triple Scalar Prodct s * d # w = 3 1 2 3 1 2 3 3 w 1 w 2 w 3 EXAMPLE 6 Finding the Volme of a Parallelepiped Find the olme of the box (parallelepiped) determined by = i + 2j - k, = -2i + 3k, and w = 7j - 4k. Soltion Using the rle for calclating determinants, we find 1 2-1 s * d # w = 3-2 0 33 = -23. 0 7-4 The olme is ƒ s * d # w ƒ = 23 nits cbed.