9.5. The Vector Product. Introduction. Prerequisites. Learning Outcomes. Learning Style

Transcription

1 The Vector Product 9.5 Introduction In this Block 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 Block you should... Lerning Outcomes After completing this Block you should e le to... 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 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 Lerning Style To chieve wht is expected of you... llocte sufficient study time riefly revise the prerequisite mteril ttempt every guided exercise nd most of the other exercises

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. θ 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, in the sense from towrds s shown. Aright-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 θ 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. θ ˆf Figure 3: We re now in position to descrie the vector product. Engineering Mthemtics: Open Lerning Unit Level 1 2

3 2. Definition of the Vector Product We define the vector product of nd, written s = sin θ ê By inspection of this formul note tht this is vector of mgnitude sin θ 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 θ θ Formlly we hve Figure 4: is perpendiculr to the plne contining nd. Key Point vector product: = sin θ ê modulus of vector product: = sin θ Note tht sin θ gives the vector product its modulus wheres ê gives its direction. Figure 5: Clcultion of. 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 =. 3 Engineering Mthemtics: Open Lerning Unit Level 1

4 Exmple If nd re prllel, show tht =0. Solution If nd re prllel then the ngle θ etween them is zero. Consequently sin θ = 0 from which it follows tht =0. Note tht the result, 0, is the zero vector. Note in prticulr the following importnt results: 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. Solution Note tht i nd j re perpendiculr so tht the ngle etween them is 90. So the 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. 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 Engineering Mthemtics: Open Lerning Unit Level 1 4

5 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 4, (ove) this expression simplifies to =( )i ( )j +( )k Key Point If = 1 i + 2 j + 3 k nd = 1 i + 2 j + 3 k then =( )i ( )j +( )k Exmple Evlute the vector product if =3i 2j +5k nd =7i +4j 8k. Solution 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 Now do this exercise Use the Key Point formul directly ove to find the vector product of p =3i +5j nd q =2i j. Note tht in this exmple there re no k components so 3 nd 3 re oth zero. Answer 5 Engineering Mthemtics: Open Lerning Unit Level 1

6 4. Using Determinnts to Evlute Vector Product Evlution of vector product using the previous formul is very cumersome. Amore 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. 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. Solution 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 Engineering Mthemtics: Open Lerning Unit Level 1 6

7 Solution 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. More exercises for you to try 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. Answer 7 Engineering Mthemtics: Open Lerning Unit Level 1

8 5. Computer Exercise or Activity For this exercise it will e necessry for you to ccess the computer pckge DERIVE. DERIVE hs routines for deling with two- nd threedimensionl vectors. To use vectors you cn enter them y keying Author:Vector. When requested for the dimension respond with either 2 or 3 s pproprite. You will then e le to enter your vectors. Alterntively, nd this is proly esier, use Author:Expression nd enclose your vector within squre rckets. For exmple the vector 2i +3j 4k would e entered s [2, 3, 4]. The product of sclr with vector is written in n ovious wy. The vector 6(2i +3j 4k) would e keyed in s Author:Expression 6[2, 3, 4]. DERIVE responds with [12, 18, 24]. The modulus of vector is otined using the s commnd. For exmple to find 2i+3j 4k you would key in Author:Expression s[2, 3, 4]. DERIVE responds with 29. The sclr product cn e otined in DERIVE y using the. nottion. To otin (2i +3j 4k).( 1i +3j +5k) you would key in Author:Expression [2, 3, 4].[ 1, 3, 5]. DERIVE responds with 13. The cross product cn e otined in DERIVE y using the cross function. For exmple, to otin (2i+3j 4k) ( 1i+3j+5k) you would key in Author:Expression cross([2, 3, 4], [ 1, 3, 5]). DERIVE responds with [27, 6, 9]. As useful exercise use DERIVE to otin the solutions to Exercises 2,3,7 nd 9. Engineering Mthemtics: Open Lerning Unit Level 1 8

9 End of Block9.5 9 Engineering Mthemtics: Open Lerning Unit Level 1

10 13k Bck to the theory Engineering Mthemtics: Open Lerning Unit Level 1 10

11 2. 2k 3. 5i +10j 5k 4. 5i 34j +6k 5. 11i +9j k, ( 11i +9j k) 6. i +2j +6k for exmple. 7. 7i 17j +6k, 42i 46j 3k i 10j + k Bck to the theory 11 Engineering Mthemtics: Open Lerning Unit Level 1