9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes

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1 The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is so-clled becuse when the sclr product of two vectors is clculted the result is sclr. The second product is known s the vector product. When this is clculted the result is vector. The definitions of these products my seem rther strnge t first, but they re widely used in pplictions. In this section we consider only the sclr product. Prerequisites Before strting this Section you should... Lerning Outcomes After completing this Section you should be ble to... 1 tht vector cn be represented s directed line segment 2 how to express vector in crtesin form 3 how to find the modulus of vector clculte, from its definition, the sclr product of two given vectors clculte the sclr product of two vectors given in crtesin form use the sclr product to find the ngle between two vectors use the sclr product to test whether two vectors re perpendiculr

2 1. Definition of the Sclr Product Consider the two vectors nd b shown in Figure 1. b θ Figure 1. Two vectors subtend n ngle θ. Note tht the tils of the two vectors coincide nd tht the ngle between the vectors hs been lbelled θ. Their sclr product, denoted by b, isdefined s the product b cos θ. Itisvery importnt to use the dot in the formul. The dot is the specific symbol for the sclr product, nd is the reson why the sclr product is lso known s the dot product. You should not use sign in this context becuse this sign is reserved for the vector product which is quite different. The ngle θ is lwys chosen to lie between 0 nd π, nd the tils of the two vectors should coincide. Figure 2 shows two incorrect wys of mesuring θ. b b θ θ Figure 2. θ should not be mesured in these wys. We cn remember this formul s: sclr product : b = b cos θ the modulus of the first vector, multiplied by the modulus of the second vector, multiplied by the cosine of the ngle between them. HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1 2

3 Clerly b = b cos θ nd so b = b. Thus we cn evlute sclr product in ny order: the opertion is sid to be commuttive. Exmple Vectors nd b re shown in the figure below. Vector hs modulus 6 nd vector b hs modulus 7 nd the ngle between them is 60. Clculte.b. b 60 The ngle between the two vectors is 60. Hence b = b cos θ = (6)(7) cos 60 =21 The sclr product of nd b is equl to 21. Note tht when finding sclr product the result is lwys sclr. Exmple Find i i where i is the unit vector in the direction of the positive x xis. Becuse i is unit vector its modulus is 1. Also, the ngle between i nd itself is zero. Therefore i.i = (1)(1) cos 0 =1 So the sclr product of i with itself equls 1. It is esy to verify tht j.j =1nd k.k =1. Exmple Find i j where i nd j re unit vectors in the directions of the x nd y xes. Becuse i nd j re unit vectors they ech hve modulus of 1. The ngle between the two vectors is 90. Therefore i j = (1)(1) cos 90 =0 Tht is i j =0. 3 HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1

4 More generlly, the following results re esily verified: i i = j j = k k =1 i j = i k = j k =0 Even more generlly, whenever ny two vectors re perpendiculr to ech other their sclr product is zero becuse the ngle between the vectors is 90 nd cos 90 =0. The sclr product of perpendiculr vectors is zero. 2. A Formul for Finding the Sclr Product We cn use the previous results to obtin formul for finding sclr product when the vectors re given in crtesin form. We consider vectors in the xy plne. Suppose = 1 i + 2 j nd b = b 1 i + b 2 j. Then b = ( 1 i + 2 j) (b 1 i + b 2 j) = 1 i (b 1 i + b 2 j)+ 2 j (b 1 i + b 2 j) = 1 b 1 i i + 1 b 2 i j + 2 b 1 j i + 2 b 2 j j Now, using the previous boxed results we cn simplify this to give the following formul: if = 1 i + 2 j nd b = b 1 i + b 2 j then b = 1 b b 2 HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1 4

5 Thus to find the sclr product of two vectors their i components re multiplied together, their j components re multiplied together nd the results re dded. Exmple If =7i +8j nd b =5i 2j, find the sclr product b. We use the previous boxed formul nd multiply corresponding components together, dding the results. b = (7i +8j) (5i 2j) = (7)(5) + (8)( 2) = = 19 The formul redily generlises to vectors in three dimensions s follows: if = 1 i + 2 j + 3 k nd b = b 1 i + b 2 j + b 3 k then b = 1 b b b 3 Exmple If =5i +3j 2k nd b =8i 9j +11k, find b. Corresponding components re multiplied together nd the results re dded. b = (5)(8) + (3)( 9)+( 2)(11) = = 9 Note gin tht the result is sclr: there re no i s, j s, or k s in the nswer. 5 HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1

6 If p =4i 3j +7k nd q =6i j +2k, find p q. Your solution Corresponding components re multiplied together nd the results re dded. 41 If r =3i +2j +9k find r r. Show tht this is the sme s r 2. Your solution 94. r = = 94 hence r 2 = r r. Forny vector r we hve r 2 = r r 3. Resolving One Vector Along Another The sclr product cn be used to find the component of vector in the direction of nother vector. Consider Figure 3 which shows two rbitrry vectors nd n. Let ˆn be unit vector in the direction of n. Study the figure crefully nd note tht perpendiculr hs been drwn from P to meet n t Q. The distnce OQ is clled the projection of onto n. Simple trigonometry tells us tht the length of the projection is cos θ. Now by tking the sclr product of with the unit vector ˆn we find ˆn = ˆn cos θ = cos θ since ˆn =1 We conclude tht HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1 6

7 P O ˆn θ Q n projection of onto n Figure 3. ˆn is the component of in the direction of n Exmple The figure below shows plne contining the point A with position vector. The vector ˆn is unit vector perpendiculr to the plne (such vector is clled norml vector). Find n expression for the perpendiculr distnce of the plne from the origin. ˆn ˆn l A O From the digrm note tht the perpendiculr distnce l of the plne from the origin is the projection of onto ˆn nd is thus ˆn. 7 HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1

8 4. Using the Sclr Product to Find the Angle Between Two Vectors We hve two distinct wys of clculting the sclr product of two vectors. From the first Key Point of this section b = b cos θ whilst from the lst of Section 2 we hve b = 1 b b b 3. Both methods of clculting the sclr product re entirely equivlent nd will lwys give the sme vlue for the sclr product. We cn exploit this correspondence to find the ngle between two vectors. The following exmple illustrtes the procedure to be followed. Exmple Find the ngle between the vectors =5i +3j 2k nd b =8i 9j +11k. The sclr product of these two vectors hs lredy been found in the exmple on pge 5 to be 9. The modulus of is ( 2) 2 = 38. The modulus of b is 8 2 +( 9) = 266. Substituting these into the formul for the sclr product we find b = b cos θ 9 = cos θ from which so tht cos θ = = θ = cos 1 ( ) = In generl, the ngle between two vectors cn be found from the following formul: The ngle θ between vectors, b is such tht: cos θ = b b HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1 8

9 Exercises 1. If =2i 5j nd b =3i +2j find b nd verify tht b = b. 2. Find the ngle between p =3i j nd q = 4i +6j. 3. Use the definition of the sclr product to show tht if two vectors re perpendiculr, their sclr product is zero. 4. If nd b re perpendiculr, simplify ( 2b) (3 +5b). 5. If p = i +8j +7k nd q =3i 2j +5k, find p q. 6. Show tht the vectors 1 i + j nd 2i j re perpendiculr The work done by force F in moving body through displcement r is given by F r. Find the work done by the force F =3i +7k if it cuses body to move from the point with coordintes (1, 1, 2) to the point (7, 3, 5). 8. Find the ngle between the vectors i j k nd 2i + j +2k. Answers , b units HELM (VERSION 1: Mrch 18, 2004): Workbook Level 1

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