Vectors. Introduction. Definition. The Two Components of a Vector. Vectors

Transcription

1 Vectors Introduction This pper covers generl description of vectors first (s cn e found in mthemtics ooks) nd will stry into the more prcticl res of grphics nd nimtion. Anyone working in grphics sujects such s nimtion or CAD design will find the informtion within very useful. Definition Vectors indicte quntity (like velocity or force) tht hs oth mgnitude nd direction. Vectors hve oth grphicl nd lgeric representtion. Grphiclly, the vector is drwn s n rrow pointing into prticulr direction, its length representing the mgnitude of the quntity (such s mss, speed, length, etc). Algericlly, vectors re written s set of coordintes. The fct tht vectors hve direction mkes them very useful in D grphics. We do opertions on vectors nd mtrices to rotte, scle, trnslte ojects nd do things such s finding intersections, perpendiculrs to plnes, The gret thing out vectors is tht you cn oth visulize nd clculte the physicl entities. I will use CAD to illustrte mny of the concepts discussed within. The Two Components of Vector Nmes of the vectors (lels) will e written in old (v in the exmple). This is conforming to current norms, though mny other representtions exist. If written notes re used, it is customry to drw little rrow or line ove or elow the lel. r v v v v direction v mgnitude The mgnitude (length) of vector is represented y single vlue, referred to s sclr. This is true regrdless of the numer of dimensions we re working in (see lter). Direction is more complicted; you need t lest two vlues in order to define direction. The numer of vlues needed to define direction will lwys correspond to the numer of dimensions we re working in. Oviously, once we re working in more thn dimensions, grphicl representtion will ecome very difficult. Nevertheless the mthemtics used will follow the sme logic. We will see tht Complex Numers re convenient wy to represent oth direction nd mgnitude of vectors. Note tht the loction of vector is not relevnt. For exmple, vectors, nd c in the figure elow re identicl s long s their mgnitude nd direction sty the sme. Therefore, c. Dirk Bertels Pge 9/4/

2 c We cn represent the mgnitude of vector using the solute symol since it is the length of the vector, i.e. the mgnitude of vector ove is represented y Since the three vectors re equl, we cn sy tht c Dirk Bertels Pge 9/4/

3 Unit vectors nd multiplying vectors with sclr vlue Unit vectors re vectors tht hve mgnitude of (unity). Any vector, v, cn e expressed s the product of its mgnitude nd unit vector in the sme direction. Assume unit vector r (of length ) nd vector v v r d d v v v * r In the figure, v r Unit Direction vectors In the lst figure, the vector r lso indictes direction. So to e more exct, we should cll r the Unit Direction Vector for v. When nming unit direction vectors for different vectors, they often re the symol of the vector they refer to nd 'cp' this symol with '^' to distinguish them from the vectors they refer to: r vˆ Component Eqution of the vector So now we cn represent the two components of vector symoliclly. The unit direction vector estlishes the length of one unit (whtever tht is) nd gives it direction, it is the vector's mesuring stick. The mgnitude is represented y the length (solute vlue) of the vector. The component eqution clerly shows this interction: v v ˆ * v v ) is the vector's directionl component with unit of (the unit direction vector) v is the vector's sclr component (its mgnitude or length) v is the vector in its totlity (interction of oth direction nd mgnitude) Dirk Bertels Pge 9/4/

4 Normlistion nd Normls Lst eqution lso shows how we cn mke ny vector into unit vector y dividing the vector y its length: v v ) v The process of doing this is clled normlistion. Normlistion is used in those situtions where the vector's length (mgnitude) will otherwise interfere with the clcultions. There is some confusion out the term normls. A norml is vector tht is perpendiculr to other vectors (we will lter see tht it is the result of the cross product opertion). However, this norml my hve ny length. So we cn normlise the norml to length of, just s we cn normlise ny vector to unit of. Dirk Bertels Pge 4 9/4/

5 Complex numers Complex numers re the idel medium to represent direction. Consider unit direction vectors in the cse of D: Y i, pointing in the direction of the X xis j, pointing in the Y direction Though we only work in dimensions for illustrtive purposes, the rgument esily extends to ny numer of dimensions, ech successive one perpendiculr to the former. j i X As stted efore, once more thn dimensions re used, grphicl representtion ecomes difficult nd the rgument cn only e extended theoreticlly, though in the sme simple nd logicl wy. We cn now represent ny direction y using the direction unit vectors i nd j s mesuring sticks. Y i j j i X -4i - j Distnce Note tht we cn pply Pythgors' right ngle tringle eqution to determine the length of the vectors ( 4) ( ) Dirk Bertels Pge 9/4/

6 Mtrix nottion nd the Crtesin Coordinte system A vector cn e represented y using the Crtesin Coordinte System. This system is used in grphics nd nimtion pplictions. It is somewht like using the complex numer system without using the unit direction vectors i nd j (they re implied). We will use the mtrix nottion for the vectors. The vector, fter ll, is one dimensionl mtrix. I prefer to use the verticl vector nottion ecuse it clerly isoltes the coordintes, mking them esier to red. [ ] It is importnt to relise tht ll vectors thus notted hve their origin t point (,). Adding vectors We cn dd vectors y dding their coordintes. Y c X c 4 c 4 Note tht grphiclly, we just construct prllelogrm. Dirk Bertels Pge 9/4/

7 Sutrcting vectors Y c X c Sutrcting vectors gives you the difference etween the vectors. Note tht oth c vectors re the sme; one is derived grphiclly while the other hs een clculted resulting in the coordintes elow. This clculted vector strts t the origin (,), ut shows the sme direction nd mgnitude s the grphicl-derived one. Therefore oth these vectors re equl. c 4 c Position vectors Any vector tht hs its origin in (,) is position vector. The lue vectors nd one of the red vectors in the figure ove re position vectors. Dirk Bertels Pge 9/4/

8 The following excerpt is from study ook written y Chris Hrmn nd Ptrici Cretchley, ssocite professors t the Fculty of Sciences in The University of Southern Queenslnd. The Eqution of Plne in R: Consider prticulr plne in spce: imgine flt sheet extending without ound in ny direction. We seek property tht distinguishes ll the points on tht prticulr plne from other points in spce. A line s direction is chrcterised y direction vector prllel to it, ecuse line hs only one such direction. A plne s direction cnnot e chrcterized in tht wy y vector prllel to it, however, ecuse mny vectors with different directions re prllel to it: consider ny vector tht joins two points in the plne, for exmple. But there is only one direction perpendiculr to the plne. Imgine tle-top or slnted roof: if we drew vectors perpendiculr to the surfce they would ll e prllel, though some might point in opposite directions. We sy ll such vectors re norml to the plne, mening perpendiculr, nd we cll ny one of them norml for tht plne. A plne hs mny normls: if you hve one, ny positive or negtive multiple will lso e norml to tht plne. Prllel plnes hve prllel normls. Plnes tht re not prllel will hve normls tht re not prllel. Plnes tht re perpendiculr, ie orthogonl, hve normls tht re orthogonl. Given norml, there re infinitely mny plnes, ll prllel, tht re perpendiculr to it. To distinguish one of them in prticulr, we need to know t lest one point on it. So plne is fully specified y knowing o vector perpendiculr or norml to it, o nd prticulr point on it. This informtion should therefore e enough for us to rrive t n eqution tht chrcterises ll its points. Suppose, therefore, tht point P(x, y, z ) lies on plne nd tht vector _n (,, c) is norml or perpendiculr to plne. Let us show n perpendiculr to the plne t the given point P Now suppose Q(x, y, z) is vrile point nywhere on tht plne. Join Q to P. This gives vector PQ lying in the plne. Since n is norml to the plne, n will e perpendiculr to PQ. This is only true if Q is on the plne: if Q is ove or elow the plne, joining it to P will mke n cute or otuse ngle with n. So for ll points Q on the plne, nd no others, PQ is perpendiculr to n. Using the dot product test for orthogonlity, therefore, we hve n eqution tht chrcterises points Q on the plne: PQ n. Giving P(x, y, z ) nd Q(x, y, z) their coordintes, we get [(x, y, z) (x, y, )] n. Using the distriutive rule on the rcket, gives yet nother form: (x, y, z) n (x, y, z ) n, nd putting the second term on the RHS gives (x, y, z) n (x, y, z ) n Dirk Bertels Pge 9/4/

9 Vectors in the Plne nd in Spce These re ll different forms of the eqution for the plne, nd give us different wys of finding it. Notice in prticulr the esy method: (vrile point) (norml) (fixed point) (norml) Vector n (,, c), so this gives (x, y, z) (,, c) (x, y, z ) (,, c), nd tking dot products gives x y cz x y cz. The LHS cnnot e simplified ecuse x, y nd z re vriles. But the RHS will reduce to n nswer, d, sy. So in generl the eqution of plne hs form x y cz d. Note lso tht in this form, the coefficients (,, c) give norml to the plne. Exmple:. The plne tht psses through the point (,,) in such wy tht it is norml to the vector (,, 4), hs eqution (x, y, z) (,, 4) (,,) (,, 4.) Evluting the dot products gives x y 4z, or simply x y 4z. To find points (x, y, z) on this plne, we cn sustitute vlues for ny two of the vriles, nd work out the third. For exmple, sustituting esy vlues x nd y, we find 4z, so tht z. The point (,,) therefore lies on tht plne.. Sometimes the norml is not given, ut cn e glened from other informtion: eg, if the plne is known to e prllel to nother plne with given eqution, then one cn use the known one s norml (ie its coefficients of x, y nd z) s norml for the required plne;. Note tht cross product cn e used to crete norml to plne, from two vectors tht lie in the plne or re prllel to it. For exmple, given three points P,Q nd R in the plne (not ll on the sme stright line) joining two of them in ny direction gives vector tht lies in the plne. Joining ny other pir will mke nother vector in the plne. The cross product of these, sy PQ RQ, gives vector perpendiculr to those two vectors, nd hence lso to the plne in which they lie. Dirk Bertels Pge 9 9/4/

10 Dirk Bertels Pge 9/4/ Finding determinnts for mtrices of ny size This rticle ssumes sic knowledge out mtrices. There is simple procedure to finding the determinnt Exmple 4 M Choose ny row or column. In this cse the 4 th row contins zeros, choosing this row will mke clcultion esier Consider lso the 4x4 'lternting signture' tle signtures pplied to lst row in mtrix M These signtures will e pplied to the corresponding elements of the mtrix. In our cse, looking t the lst row, the vlue ecomes - nd vlue - remins -. Note tht only the negtive vlues hve n impct on the clcultions. The minus signtures thus llocted will e circled red in the following figures nd the ctul vlues in the row of the mtrix will e circled green. Drw lines through ech non-zero element of this lst row nd form minors (su-mtrices) y writing down the elements NOT crossed y the lines 4 M 4 * M 4 *

11 So fr we hve * 4 * Repet the process in recursive fshion The signture tle for x mtrix is signtures pplied to first row of first minor signtures pplied to second row of second minor I choose st row in the first mtrix nd nd row in the second mtrix First row Second row 4 * 4 * 4 * 4* * * * * * * * * Dirk Bertels Pge 9/4/

12 Once x minors re reched, no further recursion is possile. Now do the following opertion on ech x minor: A B C D A*D - B*C This results in the finl clcultion *[( 4) ( ) 4( 4)] *[ ( ) (9 )] 49 Dirk Bertels Pge 9/4/

13 Generl info on determinnts The following excerpt is from study ook written y Chris Hrmn nd Ptrici Cretchley, ssocite professors t the Fculty of Sciences in The University of Southern Queenslnd. We know how to cler fctor in the eqution x. We simply multiply oth sides y its multiplictive inverse or reciprocl, /. We lso sy we divide y. We cn cler ny fctor except in tht wy. is the only rel numer tht does not hve multiplictive inverse. Let us pply similr rgument to mtrix lger: suppose we hve mtrix eqution AX B. Some squre mtrices A hve multiplictive inverses, A, ut mny do not. This mens tht we cnnot lwys get rid of mtrix fctor A in mtrix eqution. And tht is why we do not define mtrix division: it simply cn t lwys e done! To e le to distinguish etween those mtrices tht hve inverses nd those tht do not, we now define numer clled the determinnt of mtrix, so tht when the determinnt of the mtrix is, the mtrix hs no inverse; when the determinnt is not, the mtrix is invertile (non-singulr). The term 'determinnt' ws first introduced y Guss in Disquisitiones rithmetice () while discussing qudrtic forms. He used the term ecuse the determinnt determines the properties of the qudrtic form. Importnt properties of the determinnt include the following, which include invrince under elementry row nd column opertions.. Switching two rows or columns chnges the sign.. Sclrs cn e fctored out from rows nd columns.. Multiples of rows nd columns cn e dded together without chnging the determinnt's vlue. 4. Sclr multipliction of row y constnt c multiplies the determinnt y c.. A determinnt with row or column of zeros hs vlue.. Any determinnt with two rows or columns equl hs vlue. Solution of liner simultneous equtions provided without proof An exmple of solving liner equtions cn e expressed s x y c y d x y c y d x y c y d Using determinnts this is solved y the following reltionship Dirk Bertels Pge 9/4/

14 Exmple Solve the eqution.. This eqution is rewritten s Vectors x - y 4z x 4y - z 9 x y z 4 x - y 4z - x 4y - z - 9 x y z - 4 Expressing this in determinnt form On evluting the denomintors. ( x /- ) ( y /- ) ( z / -4 ) ( - /49 ) Dividing ech denomintor y 49 results in ( x /-- ) ( y /-4 ) ( z / - ) - This results in x : y 4: z Dirk Bertels Pge 4 9/4/

15 Cross Product From erlier writing The cross product of vectors is the multipliction of vectors whose origins re t the center. The result is third vector perpendiculr to the other. This perpendiculr vector uniquely identifies the plne (in D) where the originl vectors reside in. Given the vectors OP <x, y, z > nd OP <x, y, z > The cross product of these two vectors result in third vector, OP OP yz z y zx xz,, x y y x Note: The length of the vector otined y the cross product of OP nd OP is OP OP OP * OP *sinθ Where θ is the ngle etween OP nd OP in the rnge < θ < π The following excerpt is from study ook written y Chris Hrmn nd Ptrici Cretchley, ssocite professors t the Fculty of Sciences in The University of Southern Queenslnd. v u hs mgnitude u v sin φ (like dot product ut with sine insted of cos); v u hs the direction of the thum in the right-hnd rule; u v (v u) surprisingly; the mgnitude u v lso gives the re of prllelogrm; the solute vlue of sclr triple product w (u v) gives the volume of prllelepiped. following on from determinnts An interesting ppliction using the procedure to clculte determinnts is in the determining of the cross product of vectors. Consider vectors nd The cross-product is used in nimtion to determine the ngle of ll the surfces so tht lighting cn e pplied proportionlly. This technique of pplying the cross product is clled 'normlistion'. Dirk Bertels Pge 9/4/

16 Dirk Bertels Pge 9/4/ k j i k j i x k j i x Then the cross product of the vectors is where i, j, nd k re unit vectors (hving length of ) in the x, y, nd z direction respectively. More often then not these i, j, nd k vectors re implied nd therey not written, resulting in x Knowing the procedure of finding the determinnt is therefore lso gret tool in clculting the cross product. θ x The vector otined y pplying the cross product on two known vectors is perpendiculr to oth those vectors θ π If θ is the ngle etween two vectors nd, with then sinθ x where is the re of the prllelogrm determined y nd. x Looking t the figure, the direction of the ngle cn e visulized y the right-hnd rule: If the fingers of your right hnd curl in the direction of the rottion (through n ngle less thn degrees), from to, then your thum points in the direction of x.

17 Dot Product Vectors From erlier writings The dot product of vectors multiplies the corresponding coordintes nd dds the vlues. For vectors OA x, y, z nd OB x, y, z, OA OB x x y y zz Note tht the result of the dot product is rel numer ( sclr) nd tht its opertion symol is 'dot'. Contrst this with the cross product tht results in nother vector nd whose opertion symol is 'cross'. The dot product is used to determine the ngle etween vectors. OA OB OA OB cos θ OA OB OA OB cosθ In fct, we cn think of the dot product s mesuring the extent to which the vectors re pointing in the sme direction. If OA nd OB point in the sme generl direction, OA OB > If OA nd OB point in the sme generl opposite direction, OA OB < If OA nd OB point to exctly the opposite direction, θ π cos π - OA OB OA OB If OA nd OB re perpendiculr (orthogonl), θ π/ cos π/ OA OB It is this lst property of the dot product tht will prove to e useful to us. As n interesting side, when progrmming, it is etter to work with the squre of the length whenever possile to void expensive root clcultions. Dirk Bertels Pge 9/4/

18 Following on from determinnts nd cross product When corresponding coordintes of vectors re multiplied nd then dded together, we derive sclr vlue tht cn e interpreted s the multipliction of the length of one vector with the length derived from the projection of the other vector onto it. Following illustrtion will clrify this Drw vectors nd in CAD 4 The illustrtion shows these vectors in SE isometric view. Then, using CAD, drw the following perpendiculrs: The red line eing the projection of onto. The green line eing the projection of onto. To get true lengths in CAD, the coordinte xes re ligned with the fce formed y the vectors nd. Clculte dot product using coordintes:.. (* ) (*4) ( *) Clculte dot product using multipliction of lengths of vectors with sclr projections:. *..99 *. Dirk Bertels Pge 9/4/

19 Note tht this rrngement occurs lot in physicl situtions; the perpendiculr projection of one vector, A, onto nother one, B, gives you the influence A hs on B. An exmple would e the influence of A, eing the direction nd force of the wind, on B, the direction of the ot. The dot product enles you to find ngles etween vectors in D:. cosθ..*.99*cos(.9). Appliction exmple: Find perpendiculr to line in D from given point t This is n interesting ppliction tht illustrtes the use of the dot product. Continuing our exmple from ove, consider the point (,,-) nd the line from (,,) to (-,4,). P We know tht. cosθ But note lso from sic trigonometry tht t cos θ Comining these equtions gives us t. t.. This grees with the dimensions shown in the figure ove. Note tht the dot product is the proportion fctor etween the lengths t nd P t θ Pn P If we set w to w t. w.9. Dirk Bertels Pge 9 9/4/

20 Then the coordintes of the projection point Pn cn e clculted with Pn Pn Pn Pn P w [] i [] i [] i [] i [ x].9( ) [ y].9( 4 ) [ z].9( ).9 Pn.9.9 ( P P ) Exmple with different center consider points P (,, -) nd line through points P (-, -, 4) P (,, -) length of the line P_P P_P ( x x) ( y y) ( z z ) P_P ( ) ( ) ( 4). t P.P P_P (x x)(x x) (y y)(y y) (z z)(z z). ( )( ) ( )( ) ( 4)( 4) t 9.4. w t P_p 9.4 w.. Dirk Bertels Pge 9/4/

21 Pn Pn Pn Pn P w ( P P ) [] i [] i [] i [] i [ x].( ) [ y].( ).9 [ z] 4.( 4).. Pn.9.. Dirk Bertels Pge 9/4/

22 The Brycenter Brycentric comintions re wy to extrct out of n points resulting point Q in wy independent of the origin. To clculte Q Tke the sum of ll points in which ech one is scled with rel fctor (weight fctor). The weight fctor of ech point determines how much tht point ffects the sum. Therefore the weight fctor tells you how much Q is influenced y tht point. Generl eqution for Brycentric comintions Q n i ( W * P ) with n [] i [] i [] i i ( W ) Center of mss (specil cse of the ove) Q n P i n i [] Exmple Given tetrhedron with following coordintes. O.. A B C Clculte Q Q [ x] Q y (....). [ ] ( ). Q z [ ] (....). 9 Dirk Bertels Pge 9/4/

23 . Q..9 Exmple Given truncted tetrhedron with coordintes A B C At... Bt.4.. Ct.4. Clculte Q Q Q Q [ x] [ y] [ z] (......). ( ) (......).4.4. Q.4.4 Dirk Bertels Pge 9/4/

24 Exmple Given the truncted tetrhedron given ove, we cn scle this oject y pplying rtio to ech vertex - center length, therey deriving coordintes of new tetrhedron which is scled round the center. Assuming scling of / th, Brycentric center Q, nd vertex U, the coordintes re Us [] i U Q 4 i [] i [] As [] i 4A Q i [] i [] Ap [] i 4At Q i [] i [] Bs [] i 4B Q i [] i [] Bp [] i 4Bt Q i [] i [] Cs [] i 4C Q i [] i [] Cp [] i 4Ct Q i [] i [] Dirk Bertels Pge 4 9/4/