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 On completion you should be ble to... know tht vector cn be represented s directed line segment know how to express vector in Crtesin form know 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 3 HELM (8):
. Definition of the sclr product Consider the two vectors nd b shown in Figure 9. b θ Figure 9: Two vectors subtend n ngle θ Note tht the tils of the two vectors coincide nd tht the ngle between the vectors is lbelled θ. Their sclr product, denoted by b, is defined s the product b cos θ. It is very 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 nd π, nd the tils of the two vectors must coincide. Figure 3 shows two incorrect wys of mesuring θ. b b θ θ Figure 3: θ should not be mesured in these wys Key Point 9 We cn remember this formul s: The sclr product of nd b is: 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. Clerly b = b cos θ nd so b = b. Thus we cn evlute sclr product in ny order: the opertion is commuttive. HELM (8): Section 9.3: The Sclr Product 3
Exmple 8 Vectors nd b re shown in the Figure 3. Vector hs modulus 6 nd vector b hs modulus 7 nd the ngle between them is 6. Clculte.b. b 6 o Figure 3 The ngle between the two vectors is 6. Hence b = b cos θ = (6)(7) cos 6 = The sclr product of nd b is. Note tht sclr product is lwys sclr. Exmple 9 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. Also, the ngle between i nd itself is zero. Therefore i.i = ()() cos = So the sclr product of i with itself equls. It is esy to verify tht j.j = nd k.k =. 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. The ngle between the two vectors is 9. Therefore i j = ()() cos 9 = Tht is i j =. 3 HELM (8):
The following results re esily verified: Key Point i i = j j = k k = i j = j i = i k = k i = j k = k j = Generlly, whenever ny two vectors re perpendiculr to ech other their sclr product is zero becuse the ngle between the vectors is 9 nd cos 9 =. Key Point The sclr product of perpendiculr vectors is zero.. A formul for finding the sclr product We cn use the results summrized in Key Point to obtin formul for finding sclr product when the vectors re given in Crtesin form. We consider vectors in the xy plne. Suppose = i + j nd b = b i + b j. Then b = ( i + j) (b i + b j) = i (b i + b j) + j (b i + b j) = b i i + b i j + b j i + b j j Using the results in Key Point we cn simplify this to give the following formul: HELM (8): Section 9.3: The Sclr Product 33
Key Point If = i + j nd b = b i + b j then b = b + b 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 j, find the sclr product b. We use Key Point : b = (7i + 8j) (5i j) = (7)(5) + (8)( ) = 35 6 = 9 The formul redily generlises to vectors in three dimensions s follows: Key Point 3 If = i + j + 3 k nd b = b i + b j + b 3 k then b = b + b + 3 b 3 Exmple If = 5i + 3j k nd b = 8i 9j + k, find b. We use the formul in Key Point 3: b = (5)(8) + (3)( 9) + ( )() = 4 7 = 9 Note gin tht the result is sclr: there re no i s, j s, or k s in the nswer. 34 HELM (8):
Tsk If p = 4i 3j + 7k nd q = 6i j + k, find p q. Use Key Point 3: Your solution Answer 4 Tsk If r = 3i + j + 9k find r r. Show tht this is the sme s r. Your solution Answer r r = (3i + j + 9k) (3i + j + 9k) = 3i 3i + 3i j + = 9 + + = 94. r = 9 + 4 + 8 = 94, hence r = r r. The bove result is generlly true: Key Point 4 For ny vector r, r = r r HELM (8): Section 9.3: The Sclr Product 35
3. Resolving one vector long nother 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. P O θ ˆn projection of onto n Figure 3 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 = ) We conclude tht Q n ˆn Key Point 5 Resolving One Vector Along Another is the component of in the direction of n Exmple 3 Figure 33 shows plne contining the point A which hs 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, l, of the plne from the origin. ˆn ˆn O l A Figure 33 36 HELM (8):
From the digrm we note tht the perpendiculr distnce l of the plne from the origin is the projection of onto ˆn nd, using Key Point 5, is thus ˆn. 4. Using the sclr product to find the ngle between vectors We hve two distinct wys of clculting the sclr product of two vectors. From Key Point 9 b = b cos θ whilst from Key Point 3 b = b + b + 3 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 4 Find the ngle between the vectors = 5i + 3j k nd b = 8i 9j + k. The sclr product of these two vectors hs lredy been found in Exmple to be 9. The modulus of is 5 + 3 + ( ) = 38. The modulus of b is 8 + ( 9) + = 66. Substituting these vlues for b, nd b into the formul for the sclr product we find from which cos θ = b = b cos θ 9 = 38 66 cos θ 9 38 66 =.895 so tht θ = cos (.895) = 95.4 In generl, the ngle between two vectors cn be found from the following formul: The ngle θ between vectors, b is such tht: Key Point 6 cos θ = b b HELM (8): Section 9.3: The Sclr Product 37
Exercises. If = i 5j nd b = 3i + j find b nd verify tht b = b.. 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 ( b) (3 + 5b). 5. If p = i + 8j + 7k nd q = 3i j + 5k, find p q. 6. Show tht the vectors i + j nd i j re perpendiculr. 7. 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 (,, ) to the point (7, 3, 5). 8. Find the ngle between the vectors i j k nd i + j + k. Answers. 4.. 4., 3. This follows from the fct tht cos θ = since θ = 9. 4. 3 b. 5.. 6. This follows from the sclr product being zero. 7. 39 units. 8.. 38 HELM (8):
5. Vectors nd electrosttics Electricity is importnt in severl brnches of engineering - not only in electricl or electronic engineering. For exmple the design of the electrosttic precipittor pltes for clening the solid fuel power sttions involves both mechnicl engineering (structures nd mechnicl rpping systems for clening the pltes) nd electrosttics (to determine the electricl forces between solid prticles nd pltes). The following exmple nd tsks relte to the electrosttic forces between prticles. Electric chrge is mesured in coulombs (C). Chrges cn be either positive or negtive. The force between two chrges Let q nd q be two chrges in free spce locted t points P nd P. Then q will experience force due to the presence of q nd directed from P towrds P. This force is of mgnitude K q q where r is the distnce between P nd P nd K is constnt. r In vector nottion this coulomb force (mesured in newtons) cn then be expressed s F = K q q where ˆr is unit vector directed from P towrds P. The constnt K is known to be where ε = 8.854 F m (frds per metre). 4πε The electric field A unit chrge locted t generl point G will then experience force Kq ˆr (where ˆr is the unit vector directed from P towrds G) due to chrge q locted t P. This is the electric field E newtons per coulomb (N C or lterntively V m ) t G due to the presence of q. For severl point chrges q t P, q t P etc., the totl electric field E t G is given by E = Kq r ˆr + Kq r ˆr +... where ˆr i is the unit vector directed from point P i towrds G. From the definition of unit vector we see tht E = Kq r r r + Kq r r r +... = Kq r 3 r + Kq r 3 r +... = 4πε r r [ q r r 3 + q ] r r 3 +... where r i is the vector directed from point P i towrds G, so tht r = OG OP etc., where OG nd OP re the position vectors of G nd P (see Figure 34). ˆr P G O Figure 34 OP + P G = OG P G = OG OP The work done The work done W (energy expended) in moving chrge q through distnce ds, in direction given by the unit vector S/ S, in n electric field E is (defined by) W = qe.ds where W is in joules. (4) HELM (8): Section 9.3: The Sclr Product 39
Engineering Exmple Field due to point chrges In free spce, point chrge q = nc ( nc = 9 C, i.e. nnocoulomb) is t P (, 4, ) nd chrge q = nc is t P = (,, 4). [Note: Since the x-coordinte of both chrges is zero, the problem is two-dimensionl in the yz plne s shown in Figure 35.] z P (,, 4) k j P (, 4, ) O y Figure 35 () Find the field t the origin E, due to q nd q. (b) Where should third chrge q 3 = 3 nc be plced in the yz plne so tht the totl field due to q, q, q 3 is zero t the origin? () Totl field t the origin E, = (field t origin due to chrge t P ) + (field t origin due to chrge t P ). Therefore E, = 9 4π 8.854 4 j + 9 ( k) = 5.67j.3k 4π 8.854 4 (The negtive sign in front of the second term results from the fct tht the direction from P to O is in the z direction.) (b) Suppose the third chrge q 3 = 3 nc is plced t P 3 (,, b). The field t the origin due to the third chrge is where E 3 = 3 9 (j + bk) 4π 8.854 ( + b ) ( + b ), / j + bk ( + b ) / is the unit vector in the direction from O to P 3 If the position of the third chrge is such tht the totl field t the origin is zero, then E 3 = E,. There re two unknowns ( nd b). We cn write down two equtions by considering the j nd k directions. 4 HELM (8):
(contd.) [ ] E 3 = 69.6 ( + b ) j + b 3/ ( + b ) k 3/ So 5.67 = 69.6 ( + b ) 3/ E, = 5.67j.3k () So.3 = 69.6 =.83 ( + b ) 3/ b ( + b ) 3/ () (3) b =.465 ( + b ) 3/ Squring nd dding (3) nd (4) gives So ( + b ) =.47 + b ( + b ) 3 =.69 Substituting bck from (5) into () nd () gives =.7 nd b = 4.4, to 3 s.f. (4) (5) Tsk Eight point chrges of nc ech re locted t the corners of cube in free spce which is m on ech side (see Figure 36). Clculte E t () the centre of the cube (b) the centre of ny fce (c) the centre of ny edge. z A S (,, ) (,, ) (,, ) B P (,, ) O R (,, ) y (,, ) x D T (,, ) Figure 36 HELM (8): Section 9.3: The Sclr Product 4
Your solution Work the problem on seprte piece of pper but record here your min results nd conclusions. Answer () The field t the centre of the cube is zero becuse of the symmetricl distribution of the chrges. (b) Becuse of the symmetricl nture of the problem it does not mtter which fce is chosen in order to find the mgnitude of the field t the centre of fce. Suppose the chosen fce hs corners locted t P (,, ), T (,, ), R(,, ) ( nd S(,, ) then the centre (C) of this fce cn be seen from the digrm to be locted t C,, ). The electric field t C due to the chrges t the corners P, T, R nd S will then be zero since the field vectors due to equl chrges locted t opposite corners of the squre P T RS cncel one nother out. The field t C is then due to the equl chrges locted t the remining four corners (OABD) of the cube, nd we note from the symmetry of the cube, tht the distnce of ech of these corners ( ) ( ) from C will be the sme. In prticulr the distnce OC = + + =.5 m. The electric field E t C due to the remining chrges cn then be found using E = 4πε 4 q i r i r i 3 where q to q 4 re the equl chrges ( 9 coulombs) nd r to r 4 re the vectors directed from the four corners, where the chrges re locted, towrds C. In this cse since q = 9 coulombs nd r i =.5 for i = to i = 4 we hve E = 4πε 9 where r = AC = Thus E = 4πε nd E = πε (.5) [r 3/ + r + r 3 + r 4 ], 9 (.5) 3/ 4, r = BC = etc. 9 (.5) = 9 = 9.57 V 3/ π 8.854 m (.5) 3/ 4 HELM (8):
Answer (c) Suppose the chosen ( edge) to be used connects A(,, ) to B(,, ) then the centre point (G) will be locted t G,,. By symmetry the field t G due to the chrges t A nd B will be zero. We note tht the distnces DG, OG, P G nd SG re ll equl. In the cse of OG we clculte ( ) by Pythgors tht this distnce is + + =.5. Similrly the distnces T G nd RG re equl to.5. Using the result tht E = 4πε E = 9 4πε (.5) 3/ = 9 4πε = 9 4πε Thus E = + (.5) 3/ (.5) 3/.367.367 q i r i r i gives 3 + + + + (.5) 3/ + 9 4 π 8.854 + (.367) + (.367) = 5.7 V m ( d.p.). HELM (8): Section 9.3: The Sclr Product 43
Tsk If E = 5i 5j + 3k V m where i, j nd k re unit vectors in the x, y nd z directions respectively, find the differentil mount of work done in moving µc point chrge distnce of 5 mm. () From P (,, 3) towrds Q(, 4, ) (b) From Q(, 4, ) towrds P (,, 3) Your solution Answer () The work done in moving µc chrge through distnce of 5 mm towrds Q is W = qe.ds = ( 6 )(5 3 )E. P Q P Q = 8 ( 5i 5j + 3k) (i + j k) + + ( ) = 8 (5 + + 6) 3 = 7 7 J (b) A similr clcultion yields tht the work done in moving the sme chrge through the sme distnce in the direction from Q to P is W = 7 7 J 44 HELM (8):