9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes

Transcription

1 The Vector Product 9.4 Introduction In this section we descrie how to find the vector product of two vectors. Like the sclr product its definition my seem strnge when first met ut the definition is chosen ecuse of its mny pplictions. When vectors re multiplied using the vector product the result is lwys vector. Prerequisites Before strting this Section you should... Lerning Outcomes After completing this Section you should e le to... 1 tht vector cn e represented s directed line segment 2 how to express vector in crtesin form 3 how to evlute 3 3 determinnts understnd the right-hnded screw rule clculte the vector product of two given vectors use determinnts to clculte the vector product of two vectors given in crtesin form

2 1. The Right-hnded Screw Rule To understnd how the vector product is formed it is helpful to consider first the right-hnded screw rule. Consider the two vectors nd shown in Figure 1. q Figure 1. The two vectors lie in plne; this plne is shded in Figure 1. Figure 2 shows the sme two vectors nd the plne in which they lie together with unit vector, denoted ê, which is perpendiculr to this plne. Imgine turning right-hnded screw, ligned long ê, inthe sense from towrds s shown. A right-hnded screw is one which when turned clockwise enters the mteril into which it is eing screwed (most screws re of this kind). You will see from Figure 2 tht the screw will dvnce in the direction of ê. ê Figure 2. On the other hnd, if the right-hnded screw is turned from towrds the screw will retrct in the direction of ˆf s shown in Figure 3. ˆf Figure 3. We re now in position to descrie the vector product. HELM (VERSION 1: Mrch 18, 2004): Workook Level 1 2

3 2. Definition of the Vector Product We define the vector product of nd, written s = sin ê By inspection of this formul note tht this is vector of mgnitude sin in the direction of the vector ê, where ê is unit vector perpendiculr to the plne contining nd in sense defined y the right-hnded screw rule. The quntity is red s cross nd is sometimes referred to s the cross product. See Figure 4. length sin Formlly we hve Figure 4. is perpendiculr to the plne contining nd. Key Point vector product: = sin ê modulus of vector product: = sin Note tht sin gives the vector product its modulus wheres ê gives its direction. Now study Figure 5 which is used to illustrte the clcultion of. In prticulr note the direction of rising through the ppliction of the right-hnded screw rule. We see tht is not equl to ecuse their directions re oppositely directed. In fct =. Exmple If nd re prllel, show tht =0. If nd re prllel then the ngle etween them is zero. Consequently sin =0from which it follows tht =0. Note tht the result, 0, isthe zero vector. Note in prticulr the following importnt results: 3 HELM (VERSION 1: Mrch 18, 2004): Workook Level 1

4 Figure 5. Clcultion of. Key Point i i =0 j j =0 k k =0 Exmple Show tht i j = k nd find expressions for j k nd k i. Note tht i nd j re perpendiculr so tht the ngle etween them is 90.Sothe modulus of i j is (1)(1) sin 90 =1. The unit vector perpendiculr to i nd j in the sense defined y the right-hnd screw rule is k s shown in the figure elow (left digrm). Therefore i j = k s required. z z z k i j =k j k =i k k k i =j x i j y j y i i x x The vector k is perpendiculr to oth i nd j. j y Similrly you should verify (see middle nd right-hnd digrm of the ove figure) tht j k = i nd k i = j. HELM (VERSION 1: Mrch 18, 2004): Workook Level 1 4

5 Key Point i j = k, j k = i, k i = j j i = k, k j = i, i k = j To help rememer these results you might like to think of the vectors i, j nd k written in lpheticl order like this: i j k i j k Moving left to right yields positive result: e.g. k i = j. Moving right to left yields negtive result s in j i = k 3. A Formul for Finding the Vector Product We cn use the oxed results of the previous section to develop formul for finding the vector product of two vectors given in crtesin form: Suppose = 1 i+ 2 j+ 3 k nd = 1 i+ 2 j+ 3 k then = ( 1 i + 2 j + 3 k) ( 1 i + 2 j + 3 k) = 1 i ( 1 i + 2 j + 3 k) + 2 j ( 1 i + 2 j + 3 k) + 3 k ( 1 i + 2 j + 3 k) = 1 1 (i i)+ 1 2 (i j)+ 1 3 (i k) (j i)+ 2 2 (j j)+ 2 3 (j k) (k i)+ 3 2 (k j)+ 3 3 (k k) Using the Key Point results, on pge 5, (ove) this expression simplifies to =( )i ( )j +( )k If = 1 i + 2 j + 3 k nd = 1 i + 2 j + 3 k then Key Point =( )i ( )j +( )k 5 HELM (VERSION 1: Mrch 18, 2004): Workook Level 1

6 Exmple Evlute the vector product if =3i 2j +5k nd =7i +4j 8k. Identifying 1 =3, 2 = 2, 3 =5, 1 =7, 2 =4, 3 = 8 wefind = (( 2)( 8) (5)(4))i ((3)( 8) (5)(7))j + ((3)(4) ( 2)(7))k = 4i +59j +26k Use the Key Point formul directly ove to find the vector product of p =3i +5j nd q =2i j. Your solution Note tht in this exmple there re no k components so 3 nd 3 re oth zero. Apply the formul: p q = 13k 4. Using Determinnts to Evlute Vector Product Evlution of vector product using the previous formul is very cumersome. A more convenient nd esily rememered method is to use determinnts. Recll tht, for 3 3 determinnt, c d e f g h i = e f h i d f g i + c d e g h The vector product of two vectors = 1 i + 2 j + 3 k nd = 1 i + 2 j + 3 k cn e found y evluting the determinnt: i j k = in which i, j nd k re (temporrily) treted s if they were sclrs. HELM (VERSION 1: Mrch 18, 2004): Workook Level 1 6

7 Key Point If = 1 i + 2 j + 3 k nd = 1 i + 2 j + 3 k then i j k = = i( ) j( )+k( ) Exmple Find the vector product of =3i 4j +2k nd =9i 6j +2k. We hve i j k = Evluting this determinnt we otin = i( 8 ( 12)) j(6 18) + k( 18 ( 36)) = 4i +12j +18k Exmple The re of tringle The re A T of the tringle shown in the figure elow is given y the formul A T = 1 c sin α. Show tht n equivlent formul is 2 A T = 1 AB AC. 2 A α c B C From the definition of the vector product AB AC = AB AC sin α since α is the ngle etween AB nd AC. Furthermore AB = c nd AC =. The required result follows immeditely. 7 HELM (VERSION 1: Mrch 18, 2004): Workook Level 1

8 Moments The moment (or torque) of the force F out point O is defined s M o = r F where r is position vector from O to ny point on the line of ction of F. F O r D It my seem strnge tht ny point on the line of ction my e tken ut it is esy to show tht exctly the sme vector is otined. By the properties of the cross product the direction of M o is perpendiculr to the plne contining r nd F (i.e. out of the pper). The mgnitude of the moment is M 0 = r F sin From the digrm r sin = D. Hence M o = D F. This would e the sme no mtter which point on the line of ction of F ws chosen. Exmple Find the moment of the force given y F =3i +4j +5k (N) cting t the point (14, 3, 6) out the point P (2, 2, 1). z r P (2, 2, 1) F =3i +4j +5k (14, 3, 6) y x HELM (VERSION 1: Mrch 18, 2004): Workook Level 1 8

9 The position vector r cn e ny vector from the point P to ny point on the line of ction of F. We cn tke (in metres) r = (14 2)i +( 3 ( 2))j +(6 1)k =12i j +5k. The moment is i j k M = r F = = 25i 45j +51k (Nm) Exercises 1. Show tht if nd re prllel vectors then their vector product is the zero vector. 2. Find the vector product of p = 2i 3j nd q =4i +7j. 3. If = i +2j +3k nd =4i +3j +2k find. Show tht. 4. Points A, B nd C hve coordintes (9, 1, 2), (3,1,3), nd (1, 0, 1) respectively. Find the vector product AB AC. 5. Find vector which is perpendiculr to oth of the vectors = i +2j +7k nd = i + j 2k. Hence find unit vector which is perpendiculr to oth nd. 6. Find vector which is perpendiculr to the plne contining 6i + k nd 2i + j. 7. For the vectors =4i +2j + k, = i 2j + k, nd c =3i 3j +4k, evlute oth ( c) nd ( ) c. Deduce tht, in generl, the vector product is not ssocitive. 8. Find the re of the tringle with vertices t the points with coordintes (1, 2, 3), (4, 3, 2) nd (8, 1, 1). 9. For the vectors r = i +2j +3k, s =2i 2j 5k, nd t = i 3j k, evlute ) (r t)s (s t)r. ) (r s) t. Deduce tht (r t)s (s t)r =(r s) t. 1 Answers 2. 2k 3. 5i +10j 5k 4. 5i 34j +6k 5. 11i +9j k, 203 ( 11i +9j k) 6. i +2j +6k for exmple. 7. 7i 17j +6k, 42i 46j 3k i 10j + k 9 HELM (VERSION 1: Mrch 18, 2004): Workook Level 1