MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie is: ( ) If we solve this equatio for, we obtai the slope & -itercept form: If we rearrage the terms, we obtai the equatio i stadard form: The equatio of a lie is satisfied b the coordiates of all poits o the lie ad o others. For eample, the poit (5,8) is o the lie above. These coordiates satisf all three equatios. I the et sectio, we will be cosiderig the equatios of lies i R. Noe of these previous forms for the equatios of lies ca be eteded to lies i three dimesios. The above equatios all ivolve slope which caot be eteded to R because slope ivolves ol two quatities. The slope of a lie i R is ot defied. We are ow goig to establish other forms of the equatio of a lie i R that ca be easil eteded to R. These forms ivolve vectors. The diagram above goes through the poit A(-,) ad has a directio vector m [, ]. See diagram below. Page of
MCVU Directio vectors of lies are ot uique. A scalar multiple of [,] is also a directio vector of this lie. Page of
MCVU Vector Equatio of a Lie i R Let P (,) be a poit o the lie. Visualie P movig back ad forth alog the lie. As it moves, poits O, A, ad P alwas form a triagle i which the triagle law is satisfied: OP OA AP Sice AP is colliear with m [, ], we kow that AP t m, where t is a scalar. Let equatio as: p a t m or [, ] [, ] t[, ] OA a ad OP p. The we ca write the above This is called the vector equatio of a lie. To determie other poits o the lie we substitute differet scalars for t. If t the we get [,6] ad if t - we get [- 7,] Parametric Equatio of a Lie i R It is more efficiet to rewrite the vector equatio so that the right side is a sigle vector before substitutig the values for our parameter t. [, ] [, ] t[, ] [, ] [, ] [ t, t] [, ] [ t, t] thus t t These are called the parametric equatios of a lie. Parametric equatios of a lie have the followig properties: The costat terms o the right side are the coordiates of a poit o the lie. The coefficiets of t are the compoets of a directio vector of the lie. Eample # : A lie passes through the poits A (-,) ad B (5,). a) Write a vector equatio for the lie. b) Write a parametric equatio for the lie. Page of
MCVU Solutio: a) A directio vector for the lie is: AB [ 5 ( ), ] Therefore the vector equatio is [,][-,] t [7,-] AB [ 7 ] b) Usig the vector equatio of a lie, we ca write the parametric equatios of the lie as: - 7t - t Smmetric Equatio of a Lie i R Suppose a lie passes through the poit A (,-) ad has a directio vector [, ] It s parametric equatios are: t - t If we solve each equatio for t, we obtai: t ad t Sice the values of parameter t must be the same i each equatio, these two epressios are equal. Hece: m. This is called the smmetric equatio of a lie. Smmetric equatios have the followig properties: The umbers after the mius sigs i the umerators are the coordiates of a poit o the lie. The umbers i the deomiators are the compoets of a directio vector of the lie. You ca ol write a smmetric equatio if the compoets of the directio vector are o-ero. Page of
MCVU Eample # : The smmetric equatio of a lie is 6 a) Write the parametric equatios of the lie. b) Determie the coordiates of aother poit o the lie.. Solutio: 6 a) t Therefore, - t 6 - t b) Usig the parametric equatios we ca obtai a umber of poits b choosig differet values for the parameter of t. Let t, we obtai the poit (-,6) Let t, we obtai the poit (-,) Let t, we obtai the poit (,) Vector, parametric ad smmetric equatios of a lie i R represet a differet wa of thikig about the equatio of a lie. Whe we use parametric equatios for two differet lies i the same problem, we eed to use differet letters for the parameters of each lie. Page 5 of
MCVU Eample # : Smmetric equatios of two lies are give as: L 5 ad L : : Fid the coordiates of the poit of itersectio of L ad. Solutio: L A poit o L is (-,5) ad its directio vector is [,-]. Parametric equatios of L are: - t 5 t A poit o L is (,-) ad its directio vector is [,]. Parametric equatios of L are: s - s The parametric equatios of a lie give the coordiates of ever poit o the lie. So, at the poit of itersectio of the lies, the values of ad are equal. - t s 5 t - s Solve the liear sstem of equatios for s ad t. From equatio, s 7 t so b substitutio ito equatio ; - t (7 - t) - t -t t 5 ad thus, s. To determie the coordiates of the poit of itersectio we ca substitute either s or t 5 back ito the respective parametric equatios to obtai 9 ad. Therefore the lies itersect at the poit (9,). Homework: Hadout 8.9. Page 6 of
MCVU Page 7 of
MCVU U8L: Sec. 8.9. Equatios of Lies i R The three ew forms (vector, parametric ad smmetric equatios) of lies i R ca ow be eteded to R. Let A (a, a, a ) be a fied poit o a lie i R with directio vector m [ m, m, m ]. Let P (,,) be a poit o the lie. The equatios of the lie ca be writte i the followig forms; Vector Equatio: [,,] [a, a, a ] t [ m, m m ], Parametric Equatio: a t m a t m a t m a m Smmetric Equatio: where m, m, m a m a m Eample # : Write vector, parametric ad smmetric equatios for the lie through the poits A (5,,-) ad B (,5,-). Solutio: A directio vector for the lie would be AB [,, ] Vector Equatio: [,,] [5,,-] t [-,,] Parametric Equatio: Smmetric Equatio: 5 t t - t 5 Page 8 of
MCVU I three dimesios, there are three itersectio possibilities for two distict lies. Eample # : Fid the coordiates of the poit of itersectio for the followig lies; Solutio: L : ; ad L : Directio Vectors: m [,, ] ad m [,,] m ad m are ot colliear (scalar multiples) therefore the either itersect or are skew lies (either itersect or are parallel). I parametric form: L : s L : t - s - t - s t For a poit of itersectio, L L s t - s - t Page 9 of
MCVU - s t Substitute t - from equatio ito equatio - s - (-) s - 6 s - Check to make sure that s - ad t - satisf the uused equatio. It does. To fid the poit of itersectio use the appropriate parameter of s - or t - i either of L or L This results i the poit of itersectio beig (-7,-7,-). Homework: Hadout 8.9. U8L: Sec. 8.9. Equatios of Plaes A plae is determied b a poit ad two o-colliear vectors. For eample, the diagram below shows the plae passig through the poit A (-,5,) ad cotais the vectors u [,,] ad v [,, ]. The plae is tiltig upwards awa from the viewer, as idicated b the triagle formed b the vectors are at the origi. u ad v whe their tails Vector Equatio of a Plae Page of
MCVU Let P (,,) be a poit o this plae. Visualie P movig aroud the plae i a positio. As it moves, poits O, A ad P alwas form a triagle i which the triagle law is satisfied: OP OA AP Sice P is o the plae, we kow that AP is a liear combiatio of u ad v. Hece, AP s u t v where s ad t are a scalars. Therefore, we ca write the above equatio as: p a s u t v or [,, ] [,5,] s[,,] t[,, ] This is called the vector equatio for the above plae. Page of
MCVU Parametric Equatio of a Plae As doe previousl for lies i R ad R, we ca write the vector equatio of the plae i a differet wa; [,,] [- s t, 5 s t, s t] Therefore, - s t 5 s t s t This is called the parametric equatio for the above plae. Eample # : Write vector ad parametric equatios for the plae which cotais the poits A (,,-), B (5,,) ad C (,,-6). Solutio: Fid o-colliear vectors i the plae: AB [,,] ad BC [,, 6] Usig poit A (,,-), the equatios are; Vector: [,,] [,,-] s [,-,] t [-,,-6] Parametric: s t s t - s 6t Cartesia Equatio of a Plae (also kow as Scalar Equatio of a Plae) This is a equatio without usig parameters. To determie the scalar equatio of a plae, we eed to fid the ormal vector for the plae (ie: a vector perpedicular to the plae). This ca be doe b takig the cross product of two o-colliear vectors i the plae. The result is A B C D where A, B ad C are the compoets of the ormal vector ad D is a costat term. Page of
MCVU Eample # : Determie the Scalar Equatio of the Plae usig the same poits from eample #. Solutio: The two o-colliear vectors are; AB [,,] ad BC [,, 6] The ormal vector ca be foud b AB AC [,,] [,, 6] [,8, ] Usig the compoets of the ormal vector we obtai: 8 D fid D usig a give poit. () 8() (-) D D - Therefore the scalar equatio of the plae is 8 Homework: Hadout 8.9. U8L: Sec. 8.9. Problems Ivolvig Lies & Plaes I three dimesios, there are three itersectio possibilities for a lie ad a plae. Page of
MCVU Eample # : Determie the poit(s) of itersectio of each lie ad plae, if possible. a) L : [,,] [,,6] t [,-,] ad 7 6 b) L : ad 8 c) L : 5 t ad π : 5 - t 9 t Solutio: a) Re-write the vector equatio of the lie as parametric equatios ad substitute them ito the equatio of the plae. t -( t) 7( t) (6 t) 6 t -5 t -t - -6t -6 6 t -66 -t t - There is ol oe solutio for the parameter of t, therefore the lie itersects the plae at oe poit. b) t ( t) -(t) (--t) 8 t 6t t t 8 - t t There are a ifiite umber of solutios ( t ca be a real value) therefore the lie lies o the plae. c) 5 t (5 t) 5(- t) (9 t) - t t -t 6 8t 9 t t No value of t satisfies this equatio. Therefore, the epressios for,, ad do ot satisf the scalar equatio of the plae for a value of t. This meas that there are o poits o the lie that are also o the plae. That is the lie does ot lie o the plae. It must be parallel to the plae. Page of
MCVU Homework: Hadout 8.9. U8L5: Sec. 8.9.5 Parallel or Itersectig Plaes & Skew Lies Skew lies are lies i three dimesios that are ot parallel ad do ot itersect. Two skew lies ma lie i parallel plaes. To show that lies are skew lies; check to make sure that the lies are ot colliear (ot parallel) check to make sure the lies do ot itersect Eample # : Show that L ad L are skew lies. L : ad L : Solutio: B ispectio, the directio vectors, [,,] ad [,-,], of the lies are ot colliear. Hece, the lies L ad L are ot parallel. Now show that the lies do ot itersect. Parametric equatios of the lies are: L : t ad L : s t - - s t s If the lies itersect tha L L t s t - s t s Substitute equatio ito equatio t ( t) t Page 5 of
MCVU Therefore s usig equatio. Check our values for the parameters s & t i the other equatio t - s - 5-7 is ot true. Sice these values do t work, the lies do ot itersect. Therefore the are skew lies. Fidig Parallel Plaes: Eample # : Determie the equatios of two parallel plaes that cotai the skew lies: L : ad L : Solutio: If the plaes are parallel, the the have a commo ormal vector,. is perpedicular to both L ad L therefore it ca be foud b doig the cross product of the two directio vectors, m m. [,,] ad m [,,] m [,, 5] m m Therefore the plae cotaiig L will be: π : 5 D Usig a poit that eists o L ad therefore also o π : (,,) We ca solve for D: 5 D () () 5() D D 7 π : 5 7 Therefore the plae cotaiig L will be: π : 5 D Usig a poit that eist o L ad therefore also o π : (,-,) We ca solve for D: 5 D () ( ) 5() D D 5 π : 5 5 Page 6 of
MCVU Page 7 of Two distict plaes ma be either parallel or itersectig. It is eas to distiguish these two cases because parallel plaes have colliear ormal vectors. For eample, the followig plaes are parallel because their ormal vectors [,-,] ad [6,-,8] respectivel, are colliear. Ie: 9 8 6 7 The followig plaes are ot parallel, thus the itersect i a lie. Ie: 5 7 These two equatios, take together, ca be regarded as equatios of the lie. However, the are ot ver useful i this form because the do ot cotai specific iformatio about the lie such as a directio vector or the coordiates of a poit o the lie. Eample # : Fid parametric ad smmetric equatios of the lie of itersectio of the plaes 5 : 7 : ad π π Solutio: To fid the parametric equatios of the lie of itersectio, first elimiate oe of the variables ad epress the remaiig two i terms of each other. The substitute this epressio back ito oe of the origial equatios. 5 7 Addig these two equatios will elimiate the variable. ( ) 5 5 5 Sice both & are depedet o, we itroduce the parameter, t, b lettig t. Thus, the parametric equatios of the lie of itersectio for the two plaes is;
MCVU t 5t t The smmetric equatios of the lie of itersectio of the two plaes is; 5 Homework: Hadout 8.9.5 Page 8 of
MCVU U8L6: Sec. 8.9.6 Two Itersectig Plaes Cotiued. I ma problems ivolvig the lie of itersectio of two plaes, it is ot ecessar to determie the equatio of the lie. Sometimes ol a directio vector of the lie is eeded. The followig diagram shows that the directio vector of the lie of itersectio of two plaes is perpedicular to the ormal vectors of each plae. The cross product of the ormal vectors of the two plaes is a directio vector of their lie of itersectio. Eample # : Fid the equatio of a ew plae that passes through the poit A (,-,) ad is perpedicular to the lie of itersectio of the plaes π : ad π : 7 Solutio: The ormal vectors of the two plaes are [,, ] ad [ 7,, ]. m The directio vector of the lie of itersectio is determied b the cross product of the two ormals to the plaes. m [ 6,8, 8] Sice a scalar multiple of this vector is also a directio vector of the lie of itersectio we ca use [,9,-]. This directio vector is also the ormal of our ew plae. Let the equatio of the ew plae be 9 D. Sice the poit A (,-,) lies o the plae, its coordiates satisf the equatio. Hece; 9 D ( ) 9( ) ( ) D 8 D 9 8 Page 9 of
MCVU Liear Combiatios of Equatios of Plaes Ifiitel ma plaes pass through a give lie i space. I problems ivolvig a plae passig through the lie of itersectio of two plaes, the followig approach is ver effective. Cosider the plaes π ad π with these equatios; We ca tell b ispectio that the ormal vectors of π ad π are ot colliear. Hece, these two plaes have a lie of itersectio. Suppose we combie their equatios as follows. Multipl π b a scalar, s: s ( ) Multipl π b a scalar, t: t ( ) A ew plae ca be created b addig these two; ( ) t( ) π : s This is a liear combiatio of the previous two plaes. Assumig that s, we ca divide both sides b s to obtai: t π : s ( ) ( ) Sice t ad s are both real umbers, their quotiet is also a real umber. Hece, we t ca replace with a sigle smbol, k. The the equatio becomes: s π : ( ) k( ) This equatio is sigificat because a poit o the lie of itersectio of plaes π ad π also lies o π. Eample # : Fid the equatio of the plae passig through the lie of itersectio of the plaes ad, ad satisfies the give coditios a) The plae passes through he poit A (,,) Page of
MCVU b) The plae is also parallel to the plae 5 7 6 Solutio: a) Let the equatio of the required plae be as follows, where the umber k is to be determied. ( ) k Sice this plae passes through the poit A (,,), its coordiates satisf the equatio. Thus b substitutio, ( ) ( ) ( ) k[ ( ) ( ) ( ) ] 9 6k k Substitute this value of k ito the origial equatio to obtai: ( ) The equatio of the plae is 9 b) Usig the equatio from part (a); k( ), write i the form A B C D ( k ) ( k) ( k) ( k) Sice this plae is parallel to the plae 5 7 6, their ormal vectors must be multiples of oe aother. These ormal vectors are [ k, k, k] ad [ 5,, 7]. Hece, k k k 5 7 From the first proportio: k 5 k ( k) 5( k) 7k k From the secod proportio: k k 7 7 ( k) ( k) k k Page of
MCVU Substitute this value of k back ito the above equatio, therefore the equatio of the plae is 5 7 Homework: Hadout 8.9.6 U8L7: Sec. 8.9.7 Problems Ivolvig Three Plaes Da # Three plaes ca iteract i 5 differet was. Toda we will be lookig at Tpe I III ad tomorrow we will look at Tpe IV V. Tpe I: Three Parallel Plaes Whe plaes are parallel, the ormals are colliear (scalar multiples) ad coplaar. The equatios below represet this situatio. We ca tell this because all three ormal vectors [,-,], [9,-6,] ad [6,-,] are colliear. That is, ad. The plaes are distict because their equatios do ot satisf these relatioships. The plaes are distict, separated because their costat term does ot follow the same scalar multiples as the ormals. ie: π : 9 6 6 5 [,,] [ 9, 6,] [ 6,,] Page of
MCVU Tpe II: Two Parallel Plaes & Third Plae Itersects Both Suppose we replace plae π with aother plae, π, that is ot parallel to the other two plaes. The π itersects π ad π formig two parallel lies. The side view shows the plaes π ad π as two parallel lies with a third lie, π, itersectig them. The ormal vectors lie flat o the page ad are coplaar. Two of the ormals are colliear but o with the rd. Plaes are still distict. Normals are still coplaar. ie: π : 9 6 7 [,,] [ 9, 6,] [,, ] Tpe III: Plaes Itersect i Pairs Suppose we replace plae π with aother plae, π 5, that is ot parallel to either of the other two plaes. The plaes π, π 5, ad π itersect i pairs, formig three parallel lies. The side view shows the plaes as lies formig a triagle. Agai, the ormal vectors lie flat o the page ad are coplaar. Page of
MCVU The ormal vector of oe plae is a liear combiatio of the other ormals, but the equatios do t follow the same liear combiatios. The three ormals are still coplaar. ie: π : 5 7 5 5 5 [,,] [,, ] [ 5,, ] I all three tpes, each sstem of equatios has o solutio because there is o poit o all three plaes. It is impossible for the coordiates of a poit to satisf all three equatios. We sa that each sstem of equatios is icosistet. Eample # : Describe how the plaes i each liear sstem are related. If there is a uique poit or lie of itersectio, determie its coordiates or equatio. a) π : 5 Solutio: [,,] [,,] [,,] Are the ormals colliear? Yes, of them are colliear. but is ot. Page of
MCVU Therefore, this is a Tpe II: Two parallel plaes ad the third itersects both. There is o uique lie of itersectio, there is two. b) π : 5 5 8 Solutio: [,, ] [,,] [,5,] The three ormal vectors are ot colliear. The ma be plaes formig a triagle. Ca oe ormal be writte as a liear combiatio of the other two? s t [,,-] s[,,] t[,5,] s t sub. s - s 5t (-) 5t - s t Check our solutio i other equatio. (-) () Therefore, Does the plaes equatios also follow this same liear combiatio? No it does ot so the plaes itersect i pairs (Tpe III). Homework: Hadout 8.9.7 Page 5 of
MCVU U8L8: Sec. 8.9.8 Problems Ivolvig Three Plaes Da # Tpe IV: Three Plaes Itersect i a Lie Oe ormal vector is a liear combiatio of the other ormals ad the equatios for the plaes also satisf this same liear combiatio. All ormals are still coplaar. ie: π : 6 5 5 8 6 6 [,,] [,,] [ 5, 5,8] The ormal vector of π 6, [5,-5,8], is a liear combiatio of the other two ormal vectors: 6. The plaes itersect i a sigle lie because the equatios satisf this same liear combiatio. That is π 6 π π Eample # : Show that the followig three plaes itersect at a lie ad fid the parametric equatios of that lie of itersectio. π : 7 [,,] [,, ] [,,] Page 6 of
MCVU Solutio: ad π π π Elimiate usig equatio () ad times () 9 6 Therefore, 7 7 Substitutig this ito the origial equatio; 7 The parametric equatios for the commo lie of itersectio are; t 7 t t Tpe V: Three Plaes Itersect at a Sigle Poit This is the ol situatio out of the 5 tpes where the ormal vectors are ot coplaar. It is ot possible to epress oe of the ormal vectors as a liear combiatio of the other two ormals. Page 7 of
MCVU To determie if there is just oe poit of itersectio, use the check for coplaar vectors. If the the ormals are ot coplaar ad there is a sigle poit of itersectio. For Tpe IV & V ol, the set of equatios for plaes have solutio(s) because all plaes have a commo itersectio. These sstems of equatios are called cosistet. Eample # : Show that the followig plaes itersect at a sigle poit ad the determie the coordiates of the poit of itersectio. π : 5 7 [,,] [,,] [,,] Solutio: If the three plaes itersect at a sigle poit, the the ormals are ot coplaar. Usig the quick test for coplaar vectors: If the vectors are ot coplaar. If the vectors are coplaar. [,,] [,,] [,,] Sice the ormals are ot coplaar, the plaes itersect at a sigle poit. To fid the coordiates of the poit of itersectio, solve the sstem of three equatios i three ukows. () 5 () - () - () 7 - () Page 8 of
MCVU sub. ito () sub. - ad ito () - ( ) (-) ( ) 5 9 - - Check to make sure our poit of itersectio satisfies all three equatios. The poit of itersectio is Homework: Hadout 8.9.8 9,, U8L9: Sec. 8.9.9 Usig Matrices to Solve Sstems of Equatios We ofte ecouter sstems of equatios i which the cotet is ot lies ad plaes. Ecoomists ofte have to work with sstems of does of equatios ivolvig does of ukows. Relig o a geometric iterpretatio would ot be ver useful. To help orgaie such vast amouts of data, mathematicias have created a powerful tool called a matri. A matri is simpl umerical data arraged i a rectagular arra. We usuall eclose the arra i square brackets. The method of solvig a liear sstem amouts to combiig liear combiatios of the equatios i certai was. Sice several similar steps are ivolved, this method is ideall suited for techolog. Solvig a liear sstem usig techolog requires a sstematic approach because a calculator or computer uses the same method ever time. Elemetar Row Operatios A sstem of liear equatios ca be represeted b a matri. To obtai a equivalet sstem, perform a of these operatios. Multipl the umbers i a row b a costat. Page 9 of
MCVU Replace a row b addig the umbers i a other row to the umbers i that row. Replace a row with a liear combiatio of that row ad aother row. A sstem of three liear equatios i,, ad represets three plaes i R. It ca be represeted b a matri havig the form; * * * * * * * * * * * *, called the augmeted matri where each * represets a real umber. Whe we solve the sstem usig matrices, we attempt to use the elemetar row operatios to obtai a matri havig the form # # # called the idetit matri. The method of doig this is called row reductio. The matri that results is the reduced matri ad is i row reduced echelo form. Eample # : (Solvig a Matri i R ) Solve the liear sstem usig matrices. 6 8 6 Solutio: Augmeted Matri 6 8 6 Step : R ca be made simpler b dividig b -58 6 Step : Replace R with (R) (R) 6 58 6 Step : Replace R with (R) (R) 6 6 Step : R ca be made simpler b dividig b 6. Therefore the solutio to the liear sstem is - ad. Page of
MCVU Page of Eample # : (Solvig a Matri i R ) Solve the liear sstem usig matrices 8 9 Solutio: Augmeted Matri (,,) ),, ( 7 5 7 9 7 9 7 9 8 9 Homework: Hadout 8.9.9 U8L: Sec. 8.9. Matrices Cotiued It is ot alwas possible to reduce a matri of the form * * * * * * * * * * * * to oe of the form # # # usig row reductio. If oe of the equatios is a liear combiatio of the other two, a row of eros will occur at some poit. It ma be possible to reduce the matri to a form such as # * # *. The the sstem is cosistet, ad the correspodig itersects i a lie. Parametric equatios of this lie costitute the solutio of the sstem.
MCVU Page of Eample # : Solve the sstem usig matrices ad iterpret the solutio geometricall. 9 6 7 Solutio: 7 7 9 6 7 The solutio of the sstem is give b the parametric equatios of the lie of itersectio of the three plaes. t t t Eample # : Solve the followig sstem of liear equatios b reducig the correspodig augmeted matri to row-reduced echelo form. 8 w w w w
MCVU Page of Solutio: & 8 w R R R switch R Let t so, w -t, t ad - t Hadout (Represetatios of Solutios to Sstems of Equatios i R ) Homework: Hadout 8.9. U8L: Uit # 8 Review Hadout 8.9. U8L: Uit # 8 Test