58. The Triangle Inequality for vectors is. dot product.] 59. The Parallelogram Law states that

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1 786 CAPTER 12 VECTORS AND TE GEOETRY OF SPACE 0, 0, 1, and 1, 1, 1 as shown in the figure. Then the centroid is. ( 1 2, 1 2, 1 2 ) ] x z C 54. If c a a, where a,, and c are all nonzero vectors, show that c isects the angle etween a and. 55. Prove Properties 2, 4, and 5 of the dot product (Theorem 2). 56. Suppose that all sides of a quadrilateral are equal in length and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular. y 57. Use Theorem 3 to prove the Cauchy-Schwarz Inequality: a a 58. The Triangle Inequality for vectors is a a (a) Give a geometric interpretation of the Triangle Inequality. () Use the Cauchy-Schwarz Inequality from Exercise 57 to prove the Triangle Inequality. [int: Use the fact that a 2 a a and use Property 3 of the dot product.] 59. The Parallelogram Law states that a 2 a 2 2 a (a) Give a geometric interpretation of the Parallelogram Law. () Prove the Parallelogram Law. (See the hint in Exercise 58.) 60. Show that if u v and u v are orthogonal, then the vectors u and v must have the same length TE CROSS PRODUCT The cross product a of two vectors a and, unlike the dot product, is a vector. For this reason it is also called the vector product. Note that a is defined only when a and are three-dimensional vectors. 1 DEFINITION If a a 1,, and 1, 2, 3, then the cross product of a and is the vector a 3 2, 1 a 1 3, a This may seem like a strange way of defining a product. The reason for the particular form of Definition 1 is that the cross product defined in this way has many useful properties, as we will soon see. In particular, we will show that the vector a is perpendicular to oth a and. In order to make Definition 1 easier to rememer, we use the notation of determinants. A determinant of order 2 is defined y a c d ad c For example, A determinant of order 3 can e defined in terms of second-order determinants as follows: 2 a1 1 c 1 2 c 2 3 c 3 a c 2 c c 1 c c 1 c 2

2 SECTION 12.4 TE CROSS PRODUCT 787 Oserve that each term on the right side of Equation 2 involves a numer a i in the first row of the determinant, and a i is multiplied y the second-order determinant otained from the left side y deleting the row and column in which a appears. Notice also the minus sign in the second term. For example, i If we now rewrite Definition 1 using second-order determinants and the standard asis vectors i, j, and k, we see that the cross product of the vectors a a 1 i j k and 1 i 2 j 3 k is 3 a 2 3 i a j a k In view of the similarity etween Equations 2 and 3, we often write i j k a a 1 a Although the first row of the symolic determinant in Equation 4 consists of vectors, if we expand it as if it were an ordinary determinant using the rule in Equation 2, we otain Equation 3. The symolic formula in Equation 4 is proaly the easiest way of rememering and computing cross products. V EXAPLE 1 If a 1, 3, 4 and 2, 7, 5, then i j k a i j k i 5 8 j 7 6 k 43i 13j k V EXAPLE 2 Show that a a 0 for any vector a in V 3. SOLUTION If a a 1,, a, then 3 i j k a a a 1 a 1 i a 1 a 1 j a 1 a 1 k 0 i 0 j 0 k 0

3 788 CAPTER 12 VECTORS AND TE GEOETRY OF SPACE One of the most important properties of the cross product is given y the following theorem. 5 TEORE The vector a is orthogonal to oth a and. ax a PROOF In order to show that a is orthogonal to a, we compute their dot product as follows: a a 2 3 a a a a 1 3 a 1 2 a a a 1 a a A similar computation shows that a 0. Therefore a is orthogonal to oth a and. If a and are represented y directed line segments with the same initial point (as in Figure 1), then Theorem 5 says that the cross product a points in a direction perpendicular to the plane through a and. It turns out that the direction of a is given y the right-hand rule: If the fingers of your right hand curl in the direction of a rotation (through an angle less than 180) from a to, then your thum points in the direction of a. Now that we know the direction of the vector a, the remaining thing we need to complete its geometric description is its length a. This is given y the following theorem. FIGURE 1 6 TEORE If is the angle etween a and (so 0 ), then a a sin TEC Visual 12.4 shows how a changes as changes. PROOF From the definitions of the cross product and length of a vector, we have a a a a a a a a a a a a1 2 a2 2 a a a 2 2 a 2 a 2 2 a 2 2 cos 2 a cos 2 a 2 2 sin 2 (y Theorem ) Geometric characterization of a Taking square roots and oserving that ssin 2 sin ecause sin 0 when 0, we have a a sin Since a vector is completely determined y its magnitude and direction, we can now say that a is the vector that is perpendicular to oth a and, whose orientation is deter-

4 SECTION 12.4 TE CROSS PRODUCT 789 mined y the right-hand rule, and whose length is physicists define a. a sin. In fact, that is exactly how 7 COROLLARY Two nonzero vectors a and are parallel if and only if a 0 sin a FIGURE 2 PROOF Two nonzero vectors a and are parallel if and only if or. In either case sin 0, so and therefore a 0. a 0 The geometric interpretation of Theorem 6 can e seen y looking at Figure 2. If a and are represented y directed line segments with the same initial point, then they determine a parallelogram with ase, altitude, and area a sin A a ( sin ) a 0 Thus we have the following way of interpreting the magnitude of a cross product. The length of the cross product a is equal to the area of the parallelogram determined y a and. EXAPLE 3 Find a vector perpendicular to the plane that passes through the points P1, 4, 6, Q2, 5, 1, and R1, 1, 1. SOLUTION The vector PQ l PR l is perpendicular to oth PQ l and PR l and is therefore perpendicular to the plane through P, Q, and R. We know from (12.2.1) that PQ l 2 1 i 5 4 j 1 6 k 3i j 7k PR l 1 1 i 1 4 j 1 6 k 5 j 5k We compute the cross product of these vectors: PQ l PR l i j k i 15 0 j 15 0 k 40 i 15 j 15k So the vector 40, 15, 15 is perpendicular to the given plane. Any nonzero scalar multiple of this vector, such as 8, 3, 3, is also perpendicular to the plane. EXAPLE 4 Find the area of the triangle with vertices P1, 4, 6, Q2, 5, 1, and R1, 1, 1. SOLUTION In Example 3 we computed that PQ l PR l 40, 15, 15. The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product: PQ l PR l s s82 The area A of the triangle PQR is half the area of this parallelogram, that is, 2 s82. 5

5 790 CAPTER 12 VECTORS AND TE GEOETRY OF SPACE If we apply Theorems 5 and 6 to the standard asis vectors i, j, and k using, we otain i j k j k i k i j 2 j i k k j i i k j Oserve that i j j i Thus the cross product is not commutative. Also i i j i k j whereas i i j 0 j 0 So the associative law for multiplication does not usually hold; that is, in general, a c a c owever, some of the usual laws of algera do hold for cross products. The following theorem summarizes the properties of vector products. 8 TEORE If a,, and c are vectors and c is a scalar, then 1. a a 2. (ca) c(a ) a (c) 3. a ( c) a a c 4. (a ) c a c c a c a c a c a c a c These properties can e proved y writing the vectors in terms of their components and using the definition of a cross product. We give the proof of Property 5 and leave the remaining proofs as exercises. PROOF OF PROPERTY 5 If a a 1,,, 1, 2, 3, and c c 1, c 2, c 3, then 9 a c a 1 2 c 3 3 c 2 3 c 1 1 c 3 1 c 2 2 c 1 a 1 2 c 3 a 1 3 c 2 3 c 1 1 c 3 1 c 2 2 c c 1 1 a 1 3 c 2 a c 3 a c TRIPLE PRODUCTS The product a c that occurs in Property 5 is called the scalar triple product of the vectors a,, and c. Notice from Equation 9 that we can write the scalar triple product as a determinant: a 1 10 a c c 1 c 2 c 3

6 SECTION 12.4 TE CROSS PRODUCT 791 xc h a c FIGURE 3 The geometric significance of the scalar triple product can e seen y considering the parallelepiped determined y the vectors a,, and c. (See Figure 3.) The area of the ase parallelogram is A c. If is the angle etween a and c, then the height h of the parallelepiped is h a cos. (We must use cos instead of cos in case.) Therefore the volume of the parallelepiped is 2 V Ah c a cos a c Thus we have proved the following formula. 11 The volume of the parallelepiped determined y the vectors a,, and c is the magnitude of their scalar triple product: V a c If we use the formula in (11) and discover that the volume of the parallelepiped determined y a,, and c is 0, then the vectors must lie in the same plane; that is, they are coplanar. V EXAPLE 5 Use the scalar triple product to show that the vectors a 1, 4, 7, 2, 1, 4, and c 0, 9, 18 are coplanar. SOLUTION We use Equation 10 to compute their scalar triple product: a c Therefore, y (11), the volume of the parallelepiped determined y a,, and c is 0. This means that a,, and c are coplanar. The product a c that occurs in Property 6 is called the vector triple product of a,, and c. Property 6 will e used to derive Kepler s First Law of planetary motion in Chapter 13. Its proof is left as Exercise 46. FIGURE 4 r F TORQUE The idea of a cross product occurs often in physics. In particular, we consider a force F acting on a rigid ody at a point given y a position vector r. (For instance, if we tighten a olt y applying a force to a wrench as in Figure 4, we produce a turning effect.) The torque (relative to the origin) is defined to e the cross product of the position and force vectors r F and measures the tendency of the ody to rotate aout the origin. The direction of the torque vector indicates the axis of rotation. According to Theorem 6, the magnitude of the

7 792 CAPTER 12 VECTORS AND TE GEOETRY OF SPACE torque vector is r F r F sin where is the angle etween the position and force vectors. Oserve that the only component of F that can cause a rotation is the one perpendicular to r, that is, F sin. The magnitude of the torque is equal to the area of the parallelogram determined y r and F m FIGURE N EXAPLE 6 A olt is tightened y applying a 40-N force to a 0.25-m wrench as shown in Figure 5. Find the magnitude of the torque aout the center of the olt. SOLUTION The magnitude of the torque vector is r F r Fsin sin sin Nm If the olt is right-threaded, then the torque vector itself is n 9.66 n where n is a unit vector directed down into the page EXERCISES 1 7 Find the cross product a and verify that it is orthogonal to oth a and. 1. a 6, 0, 2, 0, 8, 0 2. a 1, 1, 1, 2, 4, 6 3. a i 3j 2k, i 5k 4. a j 7k, 2i j 4k Find and determine whether u v is directed into the page or out of the page u =5 60 u v v =10 u =6 150 v =8 5. a i j k, 6. a i e t j e t k, 7. a t, t 2, t 3, 1 2 i j 1 2 k 2i e t j e t k 1, 2t, 3t 2 8. If a i 2k and j k, find a. Sketch a,, and a as vectors starting at the origin. 16. The figure shows a vector a in the xy-plane and a vector in the direction of. Their lengths are and (a) Find k a a () Use the right-hand rule to decide whether the components of a are positive, negative, or 0. z 9 12 Find the vector, not with determinants, ut y using properties of cross products. 9. i j k j k k i 12. k i 2j i j i j x a y 13. State whether each expression is meaningful. If not, explain why. If so, state whether it is a vector or a scalar. (a) a c () a c (c) a c (d) a c (e) a c d (f) a c d 17. If a 1, 2, 1 and 0, 1, 3, find a and a. 18. If a 3, 1, 2, 1, 1, 0, and c 0, 0, 4, show that a c a c. 19. Find two unit vectors orthogonal to oth 1, 1, 1 and 0, 4, 4.