0/1 A1 Vector Algebr nd Clculus 8 lectures, MT 2015 Dvid Murry Rev Jnury 23, 2018 dvid.murry@eng.ox.c.uk Course Pge www.robots.ox.c.uk/ dwm/courses/2va Overview This course is concerned chiefly with the properties of vectors which re relted to physicl processes in 3 sptil dimensions. It strts by reviewing nd, perhps, developing your knowledge of vector lgebr nd geometry, but soon moves on to consider new mteril by pplying clculus of single vribles to invidul vectors nd to vector reltionships. An importnt re here will be to understnd how to describe curves in 3D spces nd how to perform integrtion long curve. The course then moves on to consider clculus of severl vribles pplied to both sclr nd vector fields. To give you feeling for the issues, suppose you were interested in the temperture T of wter in river. Temperture T is sclr, nd will certinly be function of position vector x = (x, y, z) nd my lso be function of time t: T = T (x, t). It is sclr field. The het flows generted by this temperture field will in generl be non-uniform, nd must be described by vector field. Now let s dive into flow. At ech point x in the strem, t ech time t, there will be strem velocity v(x, t). The locl strem velocity cn be viewed directly using modern techniques such s lser Doppler nemometry, or trditionl techniques such throwing twigs in. The point now is tht v is function tht hs the sme four input vribles s temperture did, but its output result is vector. We my be interested in plces x where the strem suddenly ccelertes, or vortices where the strem curls round dngerously. Tht is, we will be interested in finding the ccelertion of the strem, the grdient of its velocity. We my be interested in the mgnitude of the ccelertion ( sclr). Eqully, we my be interested in the ccelertion s vector, so tht we cn pply Newton s lw nd figure out the force. This is the stuff of vector clculus.
0/2 Syllbus Vector lgebr: sclr nd vector products; sclr nd vector triple products; geometric pplictions. Differentition of vector function; sclr nd vector fields. Grdient, divergence nd curl - definitions nd physicl interprettions; product formule; curviliner coordintes. Guss nd Stokes theorems nd evlution of integrls over lines, surfces nd volumes. Derivtion of continuity equtions nd Lplce s eqution in Crtesin, cylindricl nd sphericl coordinte systems. Course Content Revision: sclr nd vector products; product, vector product. Triple products, multiple products, pplictions to geometry. Differentition nd integrtion of vector functions of single vrible. curves Curviliner coordinte systems. Line, surfce nd volume integrls. Vector opertors. Vector Identities. Guss nd Stokes Theorems. Engineering Applictions. Spce Lerning Outcomes You should be comfortble with expressing systems (especilly those in 2 nd 3 dimensions) using vector quntities nd mnipulting these vectors without necessrily going bck to some underlying coordintes. You should hve sound grsp of the concept of vector field, nd be ble to link this ide to descriptions vrious physicl phenomen. You should hve good intuition of the physicl mening of the vrious vector clculus opertors nd the importnt relted theorems. You should be ble to interpret the formule describing physicl systems in terms of this intuition. References Although these notes cover the mteril you need to know you should, wider reding is essentil. Different explntions nd different digrms in books will give you the perspective to glue everything together, nd further worked exmples give you the confidence to tckle the tute sheets.
J Heding, "Mthemticl Methods in Science nd Engineering", 2nd ed., Ch.13, (Arnold). G Stephenson, "Mthemticl Methods for Science Students", 2nd ed., Ch.19, (Longmn). E Kreyszig, "Advnced Engineering Mthemtics", 6th ed., Ch.6, (Wiley). K F Riley, M. P. Hobson nd S. J. Bence, "Mthemticl Methods for the Physics nd Engineering" Chs.6, 8 nd 9, (CUP). A J M Spencer, et. l. "Engineering Mthemtics", Vol.1, Ch.6, (Vn Nostrnd Reinhold). H M Schey, Div, Grd, Curl nd ll tht, Norton Course WWW Pges Pdf copies of these notes, pdf copies of the lecture slides, the tutoril sheets, FAQs etc will be ccessible from www.robots.ox.c.uk/ dwm/courses/2va 0/3 Just the notes nd the tute sheets get put on weblern.
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Lecture 1 Vector Algebr Nottion In these notes we will use bold font, b,..., α, β,... to represent vectors nd the corresponding non-bold font, b,..., α, β,... s their mgnitudes. Unit vectors will wer hts â, ˆb,..., ˆα, ˆβ,.... The unit vector long Crtesin x, y, z xes re î, ĵ, ˆk. Lter we will require unit vectors in cylindricl nd sphericl coordintes. These will be ˆr, ˆφ, ẑ nd ˆr, ˆθ, ˆφ. (Note tht HLT uses ê r, ê θ, ê φ for these unit vectors.) On the odd occsion we use ˆl, ˆm, ˆn s n orthogonl trio. Plese distinguish ˆl from î. In your written work, underline the vector symbol, ω nd be meticulous bout doing so. Chos ensues if you cn t distinguish vectors from sclrs in expressions like r 2 r. 1.1 Vectors Mny physicl quntities, such mss, time, temperture re sclrs, nd re fully specified by one number or mgnitude. But others, such velocity, electric field, nd force, re specified by both mgnitude nd direction, requiring two numbers in 2D plnr world nd three in 3D. These re vectors but there re three slightly different types: Free vectors: In mny situttions only the mgnitude nd direction of vector re importnt, nd we cn trnslte them t will (with 3 degrees of freedom for vector in 3-dimensions). Sliding vectors: In mechnics the line of ction of force is often importnt for deriving moments. The force vector cn slide with 1 degree of freedom. Bound or position vectors: When describing lines, curves etc in spce, it is obviously importnt tht the origin nd hed of the vector re not trnslted 1
1/2 LECTURE 1. VECTOR ALGEBRA bout rbitrrily. The origins of position vectors ll coincide t n overll origin O. r 1 r 2 r 3 Free vectors Sliding vectors Position vectors Figure 1.1: The chief power of vector representtion (nd one tht is often not pprecited) is tht nlysis is freed from the restriction of rbitrrily imposed coordinte frmes. As simple exmple, if two free vectors re equl we need only sy tht their mgnitudes nd directions re equl, nd tht cn be done with drwing tht is independent of ny coordinte system. (For more sophisticted exmple, see the lst question on 2A1B.) Try to spot things in the notes tht re independent of coordinte system. However, coordinte systems re ultimtely useful when performing clcultion, so it useful to introduce the ide of vector components. 1.1.1 Components in coordinte frme O k v 3 v x 2 i v 1 x 1 v 2 j Figure 1.2: Vector components. A vector cn be represented by the vlues of its components in number of different directions, the number equl to the dimension of the vector spce. It is most convenient if those directions re mutully orthogonl (but, s we see lter, they do not hve to be). Referring to Fig. 1.2, we would write v = [v 1, v 2, v 3 ] = [x 2 x 1, y 2 y 1, z 2 z 1 ]. As î = [1, 0, 0], ĵ = [0, 1, 0], nd ˆk = [0, 0, 1], we cn lso write v = v 1 î + v 2 ĵ + v 3ˆk.
1.1. VECTORS 1/3 1.1.2 Vector equlity Two free vectors re sid to be equl iff their lengths nd directions re the sme. If we use coordinte frme, we might sy tht corresponding components of the two vectors must be equl. This definition of equlity will lso do for position vectors, but for sliding vectors we must dd tht the line of ction must be identicl too. 1.1.3 Vector mgnitude nd unit vectors Provided we use n orthogonl coordinte system, the mgnitude of 3-vector is v = v = v1 2 + v 2 2 + v 3 2 To find the unit vector in the direction of v, simply divide by its mgnitude ˆv = v/ v. 1.1.4 Vector Addition nd subtrction Addition follows the prllelogrm construction of Figure 1.3(). Subtrction ( b) is defined s the ddition ( + ( b)). It is useful to remember tht the vector b goes from b to. The following results follow immeditely from the bove definition of vector ddition: () + b = b + (commuttivity) (Figure 1.3()) (b) ( + b) + c = + (b + c) = + b + c (ssocitivity) (Figure 1.3(b)) (c) + 0 = 0 + =, where the zero vector is 0 = [0, 0, 0]. (d) + (-) = 0 It is obvious tht in terms of components vector ddition involves the ddition of the corresponding components. Tht is + b = [ 1 + b 1, 2 + b 2, 3 + b 3 ]. +b i k b j () b c b +b (b) c b+c Figure 1.3: () Addition of two vectors is commuttive. Note tht the coordinte frme is irrelevnt. (b) subtrction of vectors; (c) Addition of three vectors is ssocitive.
1/4 LECTURE 1. VECTOR ALGEBRA 1.1.5 Multipliction of vector by sclr. (NOT the sclr product!) Just s for mtrices, multipliction of vector by sclr c is defined s multipliction of ech component by c, so tht c = [c 1, c 2, c 3 ]. It follows tht c = (c 1 ) 2 + (c 2 ) 2 + (c 3 ) 2 = c. The direction of the vector will reverse if c is negtive, but otherwise is unffected. (By the wy, vector where the sign is uncertin is clled director.) Exmple Q: Coulomb s lw sttes tht the electrosttic force on chrged prticle Q due to nother chrged prticle q 1 is F = K Qq 1 r 2 ˆr (1.1) where r is the vector from q 1 to Q nd ˆr is the unit vector in tht sme direction. (Note tht the rule unlike chrges ttrct, like chrges repel is built into this formul.) The force between two prticles is not modified by the presence of other chrged prticles. Hence write down n expression for the force on Q t R due to N chrges q i t r i. A: The vector from q i to Q is R r i. The unit vector in tht direction is (R r i )/ R r i, so the resultnt force is F(R) = N K Qq i R r i 3(R r i). (1.2) i=1 Note tht F(R) is vector field. 1.2 Sclr, dot, or inner product This is product of two vectors results in sclr quntity nd is defined s follows for 3-component vectors: b = 1 b 1 + 2 b 2 + 3 b 3. Note tht = 2 1 +2 2 +2 3 = 2 = 2. The following lws of multipliction follow immeditely from the definition: () b = b (commuttivity) (b) (b + c) = b + c (c) (λ) b = λ( b) = (λb) (distributivity with respect to vector ddition) sclr multiple of sclr product of two vectors
1.2. SCALAR, DOT, OR INNER PRODUCT 1/5 B b b θ b θ O () A Projection of b onto direction of (b) Figure 1.4: () Cosine rule. (b) Projection of b onto. 1.2.1 Geometricl interprettion of sclr product Consider the squre mgnitude of the vector b. By the rules of the sclr product, this is b 2 = ( b) ( b) (1.3) = + b b 2( b) = 2 + b 2 2( b) But, by the cosine rule for the tringle OAB (Figure 1.4), the length AB 2 is given by b 2 = 2 + b 2 2b cos θ (1.4) where θ is the ngle between the two vectors. It follows tht b = b cos θ, (1.5) which is independent of the co-ordinte system used, nd tht b b. Conversely, the cosine of the ngle between vectors nd b is given by cos θ = b/b. 1.2.2 Projection of one vector onto nother Another wy of describing the sclr product is s the product of the mgnitude of one vector nd the component of the other in the direction of the first, since b cos θ is the component of b in the direction of nd vice vers (Figure 1.4b). Projection is prticulrly useful when the second vector is unit vector î is the component of in the direction of î. Notice tht if we wnted the vector component of b in the direction of we would write (b ) (b â)â = 2. (1.6)
1/6 LECTURE 1. VECTOR ALGEBRA In the prticulr cse b = 0, the ngle between the two vectors is right ngle nd the vectors re sid to be mutully orthogonl or perpendiculr neither vector hs ny component in the direction of the other. An orthonorml coordinte system is chrcterised by î î = ĵ ĵ = ˆk ˆk = 1; nd î ĵ = ĵ ˆk = ˆk î = 0. 1.2.3 A sclr product is n inner product In liner lgebr it usul to ssume vector v is column vector, so tht multiplying out Mv 1 = v 2 follows the usul conventions. For much of vector lgebr there is no need to worry t ll we will tend to write vectors s row vectors = [ 1, 2, 3 ] simply becuse its tke up less room on the pge. One of the few plces where you might worry is with the equivlence of the sclr product nd the inner inner product more commonly used in liner lgebr. Defined s b when vectors re column vectors s b = b = [ 1, 2, 3 ] b 1 b 2 b 3 It is not something over which to lose sleep! Exmples Q1: = 1 b 1 + 2 b 2 + 3 b 3. (1.7) A force F is pplied to n object s it moves by smll displcement δr. Wht work is done on the object by the force? A1: The work done is equl to the component of force in the direction of the displcement multiplied by the displcement itself. This is just sclr product: δw = F δr. (1.8) Q2: A cube hs four digonls, connecting opposite vertices. Wht is the ngle between n djcent pir? A2: You could plod through using Pythgors theorem to find the length of the digonl from cube vertex to cube centre, nd perhps you should to check the following nswer.
1.2. SCALAR, DOT, OR INNER PRODUCT 1/7 The directions of the digonls re [±1, ±1, ±1]. The ones shown in the figure re [1, 1, 1] nd [ 1, 1, 1]. The ngle is thus θ = cos 1 [1, 1, 1] [ 1, 1, 1] 12 + 1 2 + 1 2 1 2 + 1 2 + 1 2 = cos 1 1 3 (1.9) [ 1,1,1] [1,1,1] q k p ^u j s Q3: i Cube exmple Pinbll exmple A pinbll moving in plne with velocity s bounces (in purely elstic impct) from bffle whose endpoints re p nd q. Wht is the velocity vector fter the bounce? A3: Rther thn imposing new coordinte frme, it is best to refer to nturl frme tht involves û s principl direction. û = (q p) / q p. (1.10) The vector component of s long û is s = (s û)û, so the vector component perpendiculr is just s minus the s s = s (s û)û. (1.11) Physics tells us tht fter the impct, the component of velocity in the direction of the bffle is unchnged nd the component norml to the bffle is reversed: s fter = ( s s ) = (s û)û (s (s û)û) = 2(s û)û s. (1.12)
1/8 LECTURE 1. VECTOR ALGEBRA 1.3 Vector or cross product The vector product of two vectors nd b is denoted by b nd is defined s follows b = ( 2 b 3 3 b 2 )î + ( 3 b 1 1 b 3 )ĵ + ( 1 b 2 2 b 1 )ˆk. (1.13) It is MUCH more esily remembered in terms of the pseudo-determinnt î ĵ ˆk b = 1 2 3 b 1 b 2 b 3 (1.14) where the top row consists of the vectors î, ĵ, ˆk rther thn sclrs. Since determinnt with two equl rows hs vlue zero, it follows tht = 0. It is lso esily verified tht ( b) = ( b) b = 0, so tht b is orthogonl (perpendiculr) to both nd b, s shown in Figure 1.5. The mgnitude of the vector product cn be obtined by showing tht b 2 +( b) 2 = 2 b 2 from which it follows tht b = b sin θ, which is gin independent of the co-ordinte system used. This is left s n exercise. Unlike the sclr product, the vector product does not stisfy commuttivity but is in fct nti-commuttive, in tht b = b. Moreover the vector product does not stisfy the ssocitive lw of multipliction either since, s we shll see lter (b c) ( b) c. Since the vector product is known to be orthogonl to both the vectors which form the product, it merely remins to specify its sense with respect to these vectors. Assuming tht the co-ordinte vectors form right-hnded set in the order î,ĵ, ˆk it cn be seen tht the sense of the the vector product is lso right hnded, i.e the vector product hs the sme sense s the co-ordinte system used. î ĵ = î ĵ ˆk 1 0 0 0 1 0 = ˆk. (1.15) In prctice, figure out the direction from right-hnded screw twisted from the first to second vector s shown in Figure 1.5(). 1.3.1 Geometricl interprettion of vector product The mgnitude of the vector product ( b) is equl to the re of the prllelogrm whose sides re prllel to, nd hve lengths equl to the mgnitudes of, the vectors nd b (Figure 1.5b). Its direction is perpendiculr to the prllelogrm.
1.3. VECTOR OR CROSS PRODUCT 1/9 x b in right hnd screw sense b xb b bsin θ Plne of vectors nd b θ Figure 1.5: ()The vector product is orthogonl to both nd b. Twist from first to second nd move in the direction of right-hnded screw. (b) Are of prllelogrm is b sin θ. Exmple Q: g is vector from A [1,2,3] to B [3,4,5]. â is the unit vector in the direction from O to A. Find ˆb, UNIT vector long g â Verify tht ˆb is is perpendiculr to â. Find ĉ, the third member of right-hnded coordinte set â, ˆb, ĉ. A: Hence nd g = [3, 4, 5] [1, 2, 3] = [2, 2, 2] (1.16) â = 1 14 [1, 2, 3] g â = 1 14 î ĵ ˆk 2 2 2 1 2 3 (1.17) = 1 [2, 4, 2] (1.18) 14 ˆb = 1 24 [2, 4, 2] = 1 6 [1, 2, 1] (1.19) ĉ = â ˆb (1.20)
1/10 LECTURE 1. VECTOR ALGEBRA 1.4 Vector components with non-orthogonl bsis set Erlier we revised the ide of describing vector s liner sum of orthogonl components: tht is, v = v 1 î + v 2 ĵ + v 3ˆk. We lso discussed components of vector, nd you will recognize v 1 s the component of v in the direction of î tht is, v 1 = v î, nd so on. Now suppose we replced (î, ĵ, ˆk) with non-orthogonl, non-unit nd non-coplnr set of vectors (, b, c). By combining suitble mounts of these we cn rech ny point in 3-spce, so ny vector must be expressible s v = α + βb + γc. (1.21) v c b Figure 1.6: A bsis set in 3D mde up of three no orthogonl, non-unit, non-coplnr vectors. The question now is wht is α? You will be tempted to nswer the component of v in the direction of. Tht is, α = (v â) THIS IS WRONG (1.22) The wy to find α is to dot the RHS of the eqution with vector tht this perpendiculr to both b nd c. The obvious choice is (b c), hence v (b c) = α (b c) + βb (b c) + γc (b c) (1.23) = α (b c) This is sclr eqution, so tht α = v (b c) (b c) nd β = v (c ) b (c ) γ = v ( b) c ( b) (1.24) Notice tht the results re unique. Even though the vectors, b, c re not orthogonl one cnnot substitute, sy, bit more b nd c for loss of bit of.
1.4. VECTOR COMPONENTS WITH A NON-ORTHOGONAL BASIS SET 1/11 Also notice tht this generl result specilizes properly when the set is orthonorml. For exmple α = v (ĵ ˆk) î (ĵ ˆk) = v î î î = v î. (1.25) The expressions for α, β, γ in Eq. 1.24 fil when the denomintor is zero. As we shll see, this occurs when the vectors, b, c re coplnr. Revised Jnury 23, 2018
1/12 LECTURE 1. VECTOR ALGEBRA
Lecture 2 Multiple Products. Geometry using Vectors 2.1 Sclr triple product Using mixtures of the pirwise sclr product nd vector product, it is possible to derive triple products between three vectors, nd indeed n-products between n vectors. This is the sclr product of vector product nd third vector, i.e. (b c). Using the pseudo-determinnt expression for the vector product, we see tht the sclr triple product cn be represented s the true determinnt (b c) = 1 2 3 b 1 b 2 b 3 c 1 c 2 c 3 (2.1) You will recll tht if you swp pir of rows of determinnt, its sign chnges; hence if you swp two pirs, its sign stys the sme. 1 2 3 c 1 c 2 c 3 c 1 c 2 c 3 b 1 b 2 b 3 1st SWAP b 1 b 2 b 3 2nd SWAP 1 2 3 c 1 c 2 c 3 1 2 3 b 1 b 2 b 3 (2.2) + + This sys tht (i) (b c) = b (c ) = c ( b) (Clled cyclic permuttion.) (ii) (b c) = b ( c) nd so on. (Clled nti-cyclic permuttion.) (iii) The fct tht (b c) = ( b) c llows the sclr triple product to be written s [, b, c]. This nottion is not very helpful, nd we will try to void it below. 1
2/2 LECTURE 2. MULTIPLE PRODUCTS. GEOMETRY USING VECTORS 2.1.1 Geometricl interprettion of sclr triple product The sclr triple product gives the volume of the prllelopiped whose sides re represented by the vectors, b, nd c. We sw erlier tht the vector product ( b) hs mgnitude equl to the re of the bse, nd direction perpendiculr to the bse. The component of c in this direction is equl to the height of the prllelopiped shown in Figure 2.1(). n ccos β β c b c b Figure 2.1: () Sclr triple product equls volume of prllelopiped. (b) Coplnrity yields zero sclr triple product. 2.1.2 Linerly dependent vectors If the sclr triple product of three vectors is zero (b c) = 0 (2.3) then the vectors re linerly dependent. Tht is, one cn be expressed s liner combintion of the others. For exmple, = λb + µc (2.4) where λ nd µ re sclr coefficients. You cn see this immeditely in two wys: The determinnt would hve one row tht ws liner combintion of the others. You ll remember tht by doing row opertions, you could rech row of zeros, nd so the determinnt is zero. The prllelopiped would hve zero volume if squshed flt. In this cse ll three vectors lie in plne, nd so ny one is liner combintion of the other two. (Figure 2.1b.)
2.2. VECTOR TRIPLE PRODUCT 2/3 2.2 Vector triple product This is defined s the vector product of vector with vector product, (b c). Now, the vector triple product (b c) must be perpendiculr to (b c), which in turn is perpendiculr to both b nd c. Thus (b c) cn hve no component perpendiculr to b nd c, nd hence must be coplnr with them. It follows tht the vector triple product must be expressible s liner combintion of b nd c: (b c) = λb + µc. (2.5) The vlues of the coefficients cn be obtined by multiplying out in component form. Only the first component need be evluted, the others then being obtined by symmetry. Tht is ( (b c)) 1 = 2 (b c) 3 3 (b c) 2 (2.6) = 2 (b 1 c 2 b 2 c 1 ) + 3 (b 1 c 3 b 3 c 1 ) = ( 2 c 2 + 3 c 3 )b 1 ( 2 b 2 + 3 b 3 )c 1 = ( 1 c 1 + 2 c 2 + 3 c 3 )b 1 ( 1 b 1 + 2 b 2 + 3 b 3 )c 1 = ( c)b 1 ( b)c 1 The equivlents must be true for the 2nd nd 3rd components, so we rrive t the identity (b c) = ( c)b ( b)c. (2.7) b x c In rbitrry direction c b x ( b x c ) Figure 2.2: Vector triple product.
2/4 LECTURE 2. MULTIPLE PRODUCTS. GEOMETRY USING VECTORS 2.2.1 Projection using vector triple product An exmple of the ppliction of this formul is s follows. Suppose v is vector nd we wnt its projection into the xy-plne. The component of v in the z direction is v ˆk, so the projection we seek is v (v ˆk)ˆk. Writing ˆk, v b, ˆk c, (b c) = ( c)b ( b)c (2.8) ˆk (v ˆk) = (ˆk ˆk)v (ˆk v)ˆk (2.9) = v (v ˆk)ˆk So, now we recognize tht we hve two equivlent expressions for the vector component of vectot v perpendiculr to the unit direction ˆk: v (v ˆk)ˆk = ˆk (v ˆk). (2.10) (Hot stuff! But the expression v (v ˆk)ˆk is much esier to understnd, nd cheper to compute!) 2.3 Other repeted products Mny combintions of vector nd sclr products re possible, but we consider only one more, nmely the vector qudruple product ( b) (c d). By regrding b s single vector, we see tht this vector must be representble s liner combintion of c nd d. On the other hnd, regrding c d s single vector, we see tht it must lso be liner combintion of nd b. This provides mens of expressing one of the vectors, sy d, s liner combintion of the other three, s follows: Hence ( b) (c d) = [( b) d]c [( b) c]d (2.11) = [(c d) ]b [(c d) b] [( b) c] d = [(b c) d] + [(c ) d] b + [( b) d] c (2.12) or d = [(b c) d] + [(c ) d] b + [( b) d] c [( b) c] = α + βb + γc. (2.13) This is much more complicted wy of obtining exctly the sme result tht we derived erlier!
2.4. GEOMETRY USING VECTORS: LINES, PLANES 2/5 d c b Figure 2.3: The projection of (3-)vector onto set of (3) bsis vectors is unique. Ie in d = α+βb+γc, the set {α, β, γ} is unique. 2.4 Geometry using vectors: lines, plnes 2.4.1 The eqution of line The eqution of the line pssing through the point whose position vector is nd lying in the direction of vector b is r = + λb (2.14) where λ is sclr prmeter. If you mke b unit vector, r = + λˆb then λ will represent metric length. For line defined by two points 1 nd 2 r = 1 + λ( 2 1 ) or for the unit version r = 1 + λ ( 2 1 ) 2 1 (2.15) (2.16) λ^ b r Point r trces out line. Figure 2.4: Eqution of line. With ˆb unit vector, λ is in the length units estblished by the definition of.
2/6 LECTURE 2. MULTIPLE PRODUCTS. GEOMETRY USING VECTORS 2.4.2 The shortest distnce from point to line When working out the shortest distnce from point to line it is usully sufficient to ssume tht this distnce is found long the perpendiculr to the line. First here, let s prove tht too. Referring to Figure 2.5() the vector p from c to ny point on the line is p = +λˆb c = ( c) + λˆb which hs length squred p 2 = ( c) 2 + λ 2 + 2λ( c) ˆb. Rther thn minimizing length, it is esier to minimize length-squred. The minumum is found when d p 2 /dλ = 0, ie when λ = ( c) ˆb. (2.17) So the minimum length vector is p = ( c) (( c) ˆb)ˆb. (2.18) You might spot tht is the component of ( c) perpendiculr to ˆb (s expected!). Furthermore, using the result of Section 2.2.1, p = ˆb [( c) ˆb]. (2.19) Becuse ˆb is unit vector, nd is orthogonl to [( c) ˆb], the modulus of the vector cn be written rther more simply s just p min = ( c) ˆb. (2.20) λb c µ d λb r r c Q P c () (b) Figure 2.5: () Shortest distnce point to line. (b) Shortest distnce, line to line. 2.4.3 The shortest distnce between two stright lines If the shortest distnce between point nd line is long the perpendiculr, then the shortest distnce between the two stright lines r = + λˆb nd r = c + µˆd must be found s the length of the vector which is mutully perpendiculr to the lines.
2.4. GEOMETRY USING VECTORS: LINES, PLANES 2/7 The unit vector long the mutul perpendiculr is ˆp = (ˆb ˆd)/ ˆb ˆd. (2.21) (Yes, don t forget tht ˆb ˆd is NOT unit vector. ˆb nd ˆd re not orthogonl, so there is sin θ lurking!) The minimum length is therefore the component of c in this direction p min = ( c) (ˆb ˆd)/ ˆb ˆd. (2.22) Exmple Q Two long stright pipes re specified using Crtesin co-ordintes s follows: Pipe A hs dimeter 0.8 nd its xis psses through points (2, 5, 3) nd (7, 10, 8). Pipe B hs dimeter 1.0 nd its xis psses through the points (0, 6, 3) nd ( 12, 0, 9). Determine whether the pipes need to be religned to void intersection. A Ech pipe xis is defined using two points. The vector eqution of the xis of pipe A is r = [2, 5, 3] + λ [5, 5, 5] = [2, 5, 3] + λ[1, 1, 1]/ 3 (2.23) The eqution of the xis of pipe B is r = [0, 6, 3] + µ [12, 6, 6] = [0, 6, 3] + µ[ 2, 1, 1]/ 6 (2.24) The perpendiculr to the two xes hs direction î ĵ ˆk [1, 1, 1] [ 2, 1, 1] = 1 1 1 = [2, 3, 1] = p (2.25) 2 1 1 The length of the mutul perpendiculr is ( c) [2, 3, 1] [2, 3, 1] = [2, 1, 0] = 1.87. (2.26) 14 14 But the sum of the rdii of the two pipes is 0.4 + 0.5 = 0.9. Hence the pipes do not intersect.
2/8 LECTURE 2. MULTIPLE PRODUCTS. GEOMETRY USING VECTORS 2.4.4 The eqution of plne There re number of wys of specifying the eqution of plne. 1. If b nd c re two non-prllel vectors (ie b c 0), then the eqution of the plne pssing through the point nd prllel to the vectors b nd c my be written in the form r = + λb + µc (2.27) where λ, µ re sclr prmeters. Note tht b nd c re free vectors, so don t hve to lie in the plne (Figure 2.6().) 2. Figure 2.6(b) shows the plne defined by three non-colliner points, b nd c in the plne (note tht the vectors b nd c re position vectors, not free vectors s in the previous cse). The eqution might be written s r = + λ(b ) + µ(c ) (2.28) 3. Figure 2.6(c) illustrtes nother description is in terms of the unit norml to the plne ˆn nd point in the plne r ˆn = ˆn. (2.29) r r c n ^ c b r O b NB tht these re prllel to the plne, not necessrily in the plne O () (b) (c) O Figure 2.6: () Plne defined using point nd two lines. (b) Plne defined using three points. (c) Plne defined using point nd norml. Vector r is the position vector of generl point in the plne. 2.4.5 The shortest distnce from point to plne The shortest distnce from point d to the plne is long the perpendiculr. Depending on how the plne is defined, this cn be written s D = (d ) ˆn or D = (d ) (b c) b c. (2.30)
2.5. ROTATION OF VECTORS, IN BRIEF 2/9 2.5 Rottion of vectors, in brief In 3D, rottion mtrix R is 3 3 mtrix. The orthonormlity condition R = R 1 imposes 6 constrints on the mtrix so tht R hs only 3 degrees of freedom. Its determinnt is unity, its trce is 2 cos θ + 1 where θ is the rottion ngle nd its eigenvlues re (1, e ±iθ ), where the eigenvlue 1 corresponds to the eigenvector which is the rottion xis R =. The three degrees of freedom in one rottion mtrix cn be considered to lie in the direction of the rottion xis (2 d.o.f) nd the rottion ngle bout it (1 d.o.f.). However, often it is esier to compose the overll rottion s sequence of 3 elementry rottions bout either fixed or successive principl xes (eg., x, y, z). There re 24 possibilites for choosing these combintions, subsets of which re clled vriously s Euler, Crdn or Tit-Bryn ngles. One well-known exmple from ircrft dynmics is Yw-Pitch-Roll where successive rottions re mesured successive principl directions of n ircrft s body. The other issue to consider is whether the rottion is leving the world in plce nd rotting the coordintes, or rotting the world within the sme fixed coordintes. y y y x (x y ) 1 1 θ x θ x (x y ) 0 0 Figure 2.7: Left: The world is sttionry, but the coordinte system is rotted. Right: The world is rotted within fixed coordintes. For rottion of θ in the rh sense bout the z xis, the first nd second re cos θ sin θ 0 cos θ sin θ 0 x = sin θ cos θ 0 x nd x 1 = sin θ cos θ 0 x 0 (2.31) 0 0 1 0 0 1 Which sin term hs the minus sign is esily determine using the tril point [1, 0, 0] in our cse. For the rotting xes cse, the fixed point must hve coordintes [x, y, z ] = [cos θ, sin θ, 0] plcing the minus sign s shown.
2/10 LECTURE 2. MULTIPLE PRODUCTS. GEOMETRY USING VECTORS 2.6 Angulr velocity/ccelertion nd moments, in brief A rottion cn represented by vector whose direction is long the xis of rottion in the sense of r-h screw, nd whose mgnitude is proportionl to the size of the rottion (Fig. 2.8). The sme ide cn be extended to the derivtives, tht is, ngulr velocity ω nd ngulr ccelertion ω. Angulr ccelertions rise becuse of moment (or torque) on body. In mechnics, the moment of force F bout point Q is defined to hve mgnitude M = F d, where d is the perpendiculr distnce between Q nd the line of ction L of F. ω in right hnd screw sense dθ δ t d t M F α r ω r v () (b) Figure 2.8: () The ngulr velocity vector ω is long the xis of rottion nd hs mgnitude equl to the rte of rottion. (b) The vector eqution for moment is M = r F (2.32) where r is the vector from Q to ny point on the line of ction L of force F. The resulting ngulr ccelertion vector is in the sme direction s the moment vector. The instntneous velocity of ny point P on rigid body undergoing pure rottion cn be defined by vector product s follows. The ngulr velocity vector ω hs mgnitude equl to the ngulr speed of rottion of the body nd with direction the sme s tht of the r-h screw. If r is the vector OP, where the origin O cn be tken to be ny point on the xis of rottion, then the velocity v of P due to the rottion is given, in both mgnitude nd direction, by the vector product v = ω r.
2.7. SOLUTIONS TO GENERAL VECTOR EQUATIONS 2/11 2.7 Solutions to generl vector equtions Not ll equtions for vrible vector x hve unique point solution, nd often the locus is not simple recognizble curve or region. Exmples of such equtions re x = x + b x = γ. (2.33) (2.34) A stndrd recipe for proceeding in such cses is s follows: 1. Decide upon two non-prllel vectors ppering in the vector reltionship. These might be, b nd their vector product ( b). If there is only one vector, sy, then b cn be freely chosen, with the proviso tht it is non-prllel to. 2. Generte b. 3. Express the unknown vector x s liner combintion of these vectors x = λ + µb + ν b where λ, µ, ν re sclrs to be found. 4. Substitute the bove expression for x into the vector reltionship to determine the constrints on λ, µ nd ν for the reltionship to be stisfied. Exmple #1 Q: Find the vectors x tht stisfy the eqution x = x + b. A: Steps (1 & 2): Use nd b nd their vector product b s bsis vectors. Step (3): x = λ + µb + ν b. Step (4): Bung this expression for x into the eqution! λ + µb + ν b = [ λ + µb + ν b ] + b (2.35) = 0 + µ(b ) + ν( b) + b (2.36) But using the expression for the vector triple product ( b) = (b ) (2.37) = ( )b ( b) = 2 b ( b) (2.38)
2/12 LECTURE 2. MULTIPLE PRODUCTS. GEOMETRY USING VECTORS Hence λ + µb + ν b = ν( b) + (ν 2 + 1)b µ( b) (2.39) We hve lerned tht ny vector hs unique expression in terms of bsis set, so tht the coefficients of, b nd b on either side of the eqution much be equl. so tht λ = ν( b) µ = ν 2 + 1 ν = µ (2.40) µ = 1 1 + 2 ν = 1 1 + 2 λ = b 1 + 2. (2.41) So finlly the solution in this cse is single point: x = 1 (( b) + b ( b)) (2.42) 1 + 2 Exmple #2 Q: The second exmple x = K is in 2A1A. Notice tht the geometry is rther obvious in this cse! x must lie on the plne x â = K/. A plne with unit norml â nd perpendiculr distnce K/ from the origin. However, tht does not nswer the question. A: Using the recipe you choose b rbitrrily, nd will find [ ] K λ 2 x = λ + b + ν( b) (2.43) b where λ nd ν re free. This is correct, but why does it look so complicted? It is becuse b hs been chosen rbitrrily nd is one of the bsis vectors. As we cn see upfront tht this must be plne, cunning selection would be to choose the second vector to be perpendiculr to both nd b. You still hve to choose b rbitrrily, but using our knowledge, we cn write without further thought x = K 2 + µ( ( b)) + ν( b). µ, ν re free (2.44) Cn you see why? Revised Jnury 23, 2018