ib mathematics hl type i coursework – patterns within systems of linear equations

!Internal!Assessment!Type!I!

!Mathematics!Higher!Level!

!!!!!!!!!!!!!!!!!!!!Marc!Wierzbitzki!Hockerill!Anglo<European!College!!International!Baccalaureate!IB!Session!May!2012! ! ! ! !00<0815<083!

Marc%Wierzbitzki% % Mathematics%HL% Hockerill%Anglo7European%College%000708157083% Internal%Assessment%–%Type%I% IB%Session%May%2012%

% 1%

!HL!TYPE!I!–!PATTERNS!WITHIN!SYSTEMS!OF!LINEAR!EQUATIONS!

!The%following%internal%assessment%aims%to%examine%patterns%that%occur%within%systems%of%equations%where%the%constants%and%the%independent%term%first%progress%arithmetically%(PART%A)%and%then%progress%geometrically%(PART%B).%%%%!PART!A!!!For% the% following,% we% will% consider% the% two% linear% equations% E1:!!+ !" = !!and! E2:!!"− ! = −!%%As%stated%above,%we%will%examine% the%pattern% that% is%persistent% in% the%given%equations.%Looking% at% the% coefficients% of% x% and% y% and% the% independent% term% in% the% two% linear%equations%E1:$1! + 2! = 3%and%%%%%%%%%%%%%%%%E2:%2! − 1! = −4,%the%following%pattern%will%be%noticed.!%In%general%form,%the%equation%can%be%written%as%!! + !! = !.%!!! !! !! =! !!E1! 1% 2% =% 3%E2$ 2% 71% =% 74%Table!1:!An%examination%of%the%coefficients! (!%and!!)%and%the%independent%(!)%term%in%the%two%equations%E1$and%E2.%%$

$

$

Looking$at$E1:$

The%coefficients%of%x%(!)%and%y%(!)%and%the%independent%term%(!)%of%the%equation%E1,$if%it%is%in%the%form%!! + !! = !,%progress%arithmetically.%Note%that%this%is%not%necessarily%true%if%you% rearranged% the% equation% to% another% format% and% that% the% common% difference% will%change%as%you%do%so.%%Therefore,%looking%at%the%equation%in%the%form%!! + !! = !%only:%%!%=%u1%%%%%%%%%%%%%%%%%%%%%%%%%%%=%1%!%=%u2%=%u1+d%%%%=%2%!%=%u3%=%u1+2d%%=%3%%To%find%the%common%difference%of%this%arithmetic%progression,%either%u2%7%u1%or%u3%7%u2%can%be%calculated.%This%is%possible%since:%%%u2%=%u1+%d%can%be%rewritten%as%d%=%u2%7%u1%or%since%u3%=%u1+2d%where%u2%=%u1+%d%(u1%=%u2%–%d)%can%also%be%rewritten%as%d%=%u3%7%u2%.%In%both%cases,%d%is%equal%to:%%d%=%u2%7%u1%=%2%–%1%=%1%d%=%u3%7%u2%=%3%–%2%=%1%


% 2%

%This%finally%proves%the%assumption%that%the%coefficients%of%x%and%y%and%the%independent%term%of%the%equation%in%the%form%!! + !! = !%progress%arithmetically%since%the%common%difference%is%the%same%(in%fact%it%is%1)%in%both%cases.%%%%%Looking$at$E2:$

For% the% second% equation% (E2),% the% coefficients% of% x% (!)%and% y% (!)% and% the% independent%term%(!)%of% the%equation% if% it% is% in% the% form%!! + !! = !%progress%arithmetically% just%as%the% first%equation%did.%But%calculating%the%common%difference% in%this%case%gives%a%value%for%d2%=%73,%since:%%d2%=%u2%7%u1%=%71%–%2%=%73%d2%=%u3%7%u2%=%74%–%(71)%=%73%Again,%it%is%important%that%the%equation%is%in%the%form%!! + !! = !,%where:%%!%=%u1%%%%%%%%%%%%%%%%%%%%%%%%%%=%2%!%=%u2%=%u1+d2%%=%71%!%=%u3%=%u1+2d%%=%74%%%%%Furthermore,%it%can%then%be%examined%what%would%happen%if%the%equations%were%rewritten%in%the%general%form%of%! = !" + !,%to%give:%%E1:%%!! = − !

! ! +!!%

E2:$%%! = 2! + 4%%While% coefficients% of% y% and% x% and% the% independent% term% in% E2% could% be% a% geometric%progression%with%u1%=%1%and%r%=%2,%no%clear%pattern%can%be%recognized%for$E1.%This%means%that% the% conclusion% stated%above% is%only% true% for% the% format%!! + ! + ! ! = (! + 2!).%Rearranging%the%equations%does%not%support%the%assumption.%%%

%%

Conclusion:%The%pattern%found%in%this%2x2%system%of%linear%equations%is%that%for%the%two%equations%(as%long%as%they%are%stated%in%the%form:%!! + !! = !),%!,%!%and%!%progress%arithmetically%with%a%common%difference%of%d1%=%1%and%d2%=% 73%respectively.% %Therefore,% they% can%be%rewritten%in%the%general%forms:%%

!! + (! + !)! = (! + 2!)%%

or$just$as$well$

$

!!! + !!! = !!%%


% 3%

%Gradient:!!Furthermore,%another%observation%can%be%made.%Looking%at%the%equations:%%E1:%%!! = − !

! ! +!!%

E2:$%%! = 2! + 4%%We% now% take% a% look% at% the% gradient.% Since% the% equations% are% already% in% the% form%! = !" + !,%where%!%is%the%gradient,%those%are%as%follows:%%E1:%%%%!!%=%− !

!%E2:$$%%!!%=%2%%If%you%take%the%second%gradient%!!%and%compare%it%to%!!,%you%will%notice%that:%%%

!! = − 1!!%

%This% is% requirement% for% two% lines% to%be%perpendicular% to%each%other.%We%can% therefore%conclude%that%despite%the%other%pattern%we%have%noticed,%E1$is%perpendicular%to%E2,% like%one%can%assume%by% looking%at% the%graph.%But% this% fact% is%not% important% for%our% further%examinations% and% it% will% therefore% be% ignored% since% it% will% include% more% lines% of% the%identified%pattern%and%they%cannot%all%be%perpendicular%to%each%other.%%%%%%%%%As%we’ve%now%found%out,%the%two%lines%are%perpendicular%to%each%other.%This%means%that%they% have% to% intersect% in% some% point.% Different%methods% to% find% this% point% are% shown%below.%%%Method%I:%Row%operations:%%The$ 2x2$ system$ of$ equations$ can$ be$ rewritten$ to$ use$ row$ operations$ and$ therewith$

eliminate$one$variable.$This$is$done$below:$

%1% % 2% % 3%2% % 71% % 74%%2% % 4% % 6%2% % 71% % 74%%2% % 4% % 6%0% % 5% % 10%

Conclusion:%Rearranging%the%equations%to%! = !" + !%and%then%comparing%the%gradients%of$E1$(!!)%and%E2$(!!),$ it$can$be$noticed$that$!! = − !

!!.$This%shows%that% they%are%perpendicular%

to%each%other.%$%%


% 4%

%This% then% gives% us%! = 2 ,% which% can% then% be% substituted% into% one% of% the% original%equations%to%find%! = −1.%%%Method%II:%Displaying%the%solution%graphically:%Alternatively,%the%solution%of%this%2x2%system%can%also%be%found%by%plotting%a%graph%of%E1$and$E2$and%therewith%making%it%easier%to%find%the%point%of%intersection,%as%shown%below:%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Figure!2:%The%graph%shows%the%two%lines%E1$(blue%line)%and$E2$(red%line)%and%their%point%of%intersection%at%P%(71,%2)%%%%%%%


% 5%

Without% any% surprise,% the% graph% shows% the% point% of% intersection% of% the% two% lines% at%%%%%%%%%%%%P%(71,%2)%%

%%This% is% even% true% when% saying% that% the% common% difference,%! = 0%what% some% people%would%not% take% to%be%an%arithmetic%progression%anymore.%But% in% this%case,%since%! = 0,%the%equation%can%be%rearranged%to%give:%%

! = 1− !%%The%graph,%no%matter%what%the%value%of%!%is,%always%looks%like%this%then%and%still%passes%through%the%point%P%(71,%2):%%%%%%%%%%%%%%%%%%%%%%%Figure! 1:!When%! = 0,% the%equation%can%be%rewritten%as%! = 1 − !.%This%makes% it% independent%of%!.%The%graph%of%such%a%function%will%always%look%like%this.%%%%Now$it$needs$to$be$considered$what$would$happen$if$! = 0:$%

!! + ! + ! ! = (! + 2!)%%

!! = 2!%%

! = 2%

Conclusion:%The%solution%for%the%2x2%system%of%the%equations%E1:$1! + 2! = 3%and%E2:%2! − 1! = −4%has% been% found% to% be%! = −1%and%! = 2%by% applying% the% three% two% methods% shown%above.%%


% 6%

%So% from% this% it% can% be% concluded% that% always% when%! = 0.% Plotting% this,% a% line% that% is%parallel%to%the%x7axis%and%passes%through%the%y7axis%at%the%point%(0,%2)%can%be%seen.%Hence,%the%line%also%has%to%go%through%the%point%P%(71,%2)%since%its%y7value%will%always%stay%the%same,%but%the%x7value%will%have%a%domain%from%−∞%to%∞.%%%Limitations:!$

Consider$ the$ equation$ in$ the$ form$!! + ! + ! ! = (! + 2!),$ where$! = 0$and$! = 0$as$well.$This$would$give:$

$

0! + 0+ 0 ! = (0+ 2 ∗ 0)$$

0! + 0! = 0$$

In% theory,% this% is% still% true% since%no%matter%what% values%!%and%!%will% take,% the% left7hand%side%will%always%be%equal%to%0%(and%therefore%also%be%equal%to%the%right7hand%side).%But%trying% to% plot% a% graph,% just% like% we% did% for% all% other% results,% this% is% not% possible.% In%consequence,%this%also%means%that%the%equation%cannot%go%through%the%point%(71,%2).%%%%Furthermore,% as%we% already% solved% the% system% using%matrices,% it% is% important% to% note%that%if%the%determinant%of%the%matrix%displaying%the%equations%is%equal%to%zero%(the%matrix%is%singular),%there%is%either%none%or%infinite%solutions:%%

!! !! + !!!! !! + !!

!! = !! + 2!!

!! + 2!! %

%It$can$now$be$said$that$A:$

%! !! !

!! !! + !!!! !! + !! %

%When%the%determinant%of%the%matrix%(|A|)%is%equal%to%0,%there%is%either%none%or%an%infinite%number%of%solutions:%%%

! = !" − !"%%

! = !!(!! + !!)− (!! + !!)!!%%

! = !!!! + !!!! − !!!! − !!!!%%

0 = !!!! − !!!!%%Therefore:$

%!!!! = !!!!%

%%


% 7%

%%%

%%%%%%%We%have%now%shown%everything%that%will%be%important%later%on%during%the%investigation,%so%the%next%step%is%to%come%up%with%other%examples%for%this%type%of%equation.%Therefore,%the% first% term%u1%(which% is%equal% to%!)%and%the%common%difference%!%have%to%be%defined%first%in%order%to%calculate%u2% ! + ! %and%u3%(! + 2!).%The%complete%equation%can%then%be%calculated%by%applying%!! + ! + ! ! = (! + 2!).%The%resultant%equations%are%displayed%in% the%graph% later%on,%which%will%make% it% easier% to% find%out%whether% there% is% a%pattern%than%just%looking%at%the%equations%and%examining%them.%%%%colour! !! u1!(!)! u2! ! + ! ! u3!(! + !!)! Full!equation!in!form!!!!!!!!!!!!!!!!!!!

!! + ! + ! ! = (! + !!)!% 0.5% 0.5% 1% 1.5% 0.5x%+1y%=%1.5%% 73% 14% 11% 8% 14x%+%11y%=%8%% 1.5% 3% 4.5% 6% 3x%+%4.5y%=%6%% 76% 16% 10% 4% 16x%+%10y%=%4%% 715% 20% 5% 710% 20x%+%5y%=%710%% 5% 79% 74% 1% 79x%7%4y%=%1%% 17% 75% 12% 29% 75x%+%12y%=%29%% 5.5% 70.5% 5% 10.5% 70.5x%+%5y%=%10.5%Table!2:!Further%examples%of%linear%equations%in%the%form%!! + ! + ! ! = (! + 2!).%%%

Summary:%In%special%cases,%when%! = 0,%the%general%equation%will%rearrange%to%give%! = 1 − !%(which%makes%it%independent%of%!).%This%straight%line%still%passes%through%the%point%P%(71,%2).%In%addition%to%this,%when%! = 0,%the%new%equation%will%be%! = 2%(and%here%it%is%independent%of%!).%This%line%still%passes%through%the%point%P%(71,%2).%%But%combining%those%two%cases%to%give!! = 0%and%! = 0%at%the%same%time,%will%give%the%equation%0! + 0! = 0.%This%statement%is%still%true,%but%a%graph%of%this%cannot%be%plotted%and%therefore%it%also%can’t%pass%through%the%point%P%(71,%2).%%In%addition%to%that,%it%has%also%been%found%out%that%there%is%no%or%an%infinite%number%of%solutions%if:%

!!!! = !!!!%%


% 8%

%Figure! 3:!Graph% showing% the% equations% from%Table% 2% in% addition% to%E1$and$E2.% All% lines% intersect% in% the%point%(71,%2).%%%%%From%the%graph%shown%above,%it%can%be%assumed%that%all% linear%equations%following%the%pattern%!! + ! + ! ! = (! + 2!)%intersect% in% the% point% (71,% 2).% To% prove% this,% it% is%necessary%to%look%at%the%general%form%of%the%system%in%more%detail.%%%

%%%%%

Conjecture:%All% linear% equations% in% the% form%!! + (! + !)! = (! + 2!)%pass% through% the% point%%%%%%%%%%(71,2).%Whether%and%why%this%is%true%is%shown%below.%


% 9%

Proof%I:%%Considering:$$

!! + ! + ! ! = ! + 2! $

$

Looking%at%the%right%hand%side%of%the%equation:%(! + 2!),%for%both%sides%to%be%equal,%the%left%hand%side%has%to%contain%2!%and%one%!%as%well.%%To%find%2!%in%that%part,%!"%can%be%ignored%at%first%and%only% ! + ! !%is%important.%Saying%that%!%and%!%are%the%two%variables%that%we%can%change,%the%only%value%for%which%we%are%going%to%get%2!%from%this%part%is%! = 2.%This%gives:%%

! + ! ! = ! + ! 2 = 2! + 2!%%This% then%satisfies% the% condition% that% there%have% to%be%2!%on%each% side%of% the%equation%and%leaves%us%with:%%

2! + !! = !%%Again,%since%!%is%the%variable,%we%can%solve%this%equation,%which%gives%us%! = −1.%%%Therewith%we%can%conclude%that%all%equations% in%the%form%!! + ! + ! ! = (! + 2!)%go%through% the% point% (71,% 2).% Furthermore% this% is% the% only% point% of% intersection% since% the%equations%are%linear.%%%%%Proof%II:%%This$can$also$be$shown$algebraically,$considering$the$two$equations:$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%E1$:%%%!!! + !! + !! ! = (!! + 2!!),%%%%%%%%%%%%%%%%%%%%%%%%%%%%%E2$:%%!!!! + !! + !! ! = (!! + 2!!)%%The$general$form$of$the$equation$E1:$

% %!!! + !! + !! ! = (!! + 2!!)%

%This$can$be$rewritten$to$give!! =:$% % %% % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%! = !!!!!!!!!!!!!!

!!%

%Simplifying$this$gives:%% % %% % % % %%%%%%%%%%%%%%%%! = 1− ! + !!!

!!− !!!

!!%

%This$can$be$substituted$into$E2$in$order$to$eliminate$!:$ $

$$$$$$$$$$$$$$$$$$$$

!!(1− ! +2!!!!

− !!!!!)+ !! + !! ! = (!! + 2!!)%


% 10%

%Simplifying$this:$ $ $$$$$$$$$$$$$$$$$$$$$$$

$

!! − !!! +2!!!!!!

− !!!!!!!+ !!! + !!! = !! + 2!!$

%2!!!!!!

− !!!!!!!+ !!! = 2!!%%%

Multiplying$by$!!to$eliminate$it$in$the$denominator:$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

2!!!! − !!!!!+%!!!!! = !2!!!!%%

Rearranging%to%give%y=:%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

!(!!!! − !!!!) = 2!!!! − 2!!!!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

! = !2!!!! − 2!!!!!!!! − !!!!%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

! = !2(!!!! − !!!!)!!!! − !!!!%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

! = 2%%%Then,% after% we% found% out% that%! = 2,% we% can% use% this% to% find% the% value% for%!.% This%substitution%is%shown%below:%%$$$$E1:$1! + 2! = 3%%%%%%%%%%%%%%%%%%%%%%%%%1! + 2 ∗ 2 = 3%%%%%%%%%%%%%%%%%%%%%%! = !−1%%$$$$E2:%2! − 1! = −4%%%%%%%%%%%%%%%%%%%%%%2! − 1 ∗ 2 = −4%%%%%%%%%%%%%%%%%%! = !−1%%

%Again,%the%limitations%stated%above%do%still%apply.%%%%%%

Conclusion:%The% two% proofs% stated% above% show% that% every% 2x2% system% of% equations% in% the% form%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%!!! + (!! + !!)! = (!! + 2!!) %has% got% the% common% solution% ! = 2 ,% ! = !−1 %(the%corresponding%lines%therefore%intersect%at%the%point%P%(71,%2)).%


% 11%

After%we%have%now%examined%patterns%that%occur%for%2x2%systems%of%linear%equations%in%the%form%!! + ! + ! ! = (! + 2!),%we%will%now%widen%the%examination%to%3x3%systems%of%equations.%Those%will%involve%!,%!%and%!,%and%again%an%independent%term.%Hence,%a%new%general%form%for%the%equations%has%to%be%found:%%

!! + ! + ! ! + ! + 2! ! = ! ! + 3! %%%To%find%equations%that%fit%this%system,%the%same%method%as%utilized%before%will%be%used.%They%will% then%be%plotted%using% computer% software,% since% this% is%not%possible% to%do%by%hand.%We%will%then%look%for%patterns:%%colour! !! u1!(!)! u2! ! + ! ! u3!(! + !!)! u4!(! + !!)! Full!equation!in!form!!!!!!!!!!!!!!!!!!!

!! + ! + ! ! + ! +2! ! = ! ! + 3! !

% 71% 4% 3% 2% 1% 4x%+%3y%+%2z%=%1%% 3% 74% 71% 2% 5% 74x%7%1y%+%2z%=%5%% 2% 7% 9% 11% 13% 7x%+%9y%+%11z%=%13%Table!3:!Examples%of%planes%in%the%form%!! + ! + ! ! + ! + 2! ! = ! ! + 3! .%In%this%case%we%will%not%plot% more% equations,% since% the% graph% would% be% too% full% otherwise% and% nothing% could% be% recognized%anymore.%%

Figure!4:!The%diagram%shows%a%plot%of%the%three%equations%E3:$1x%+%3y%+%5z%=%7,$E4:%7x%+%3y%71z%=%75%and%%%%%%%%%%%%%%%%%%%%%%%%%%%$E5:$77x%76y%–%5z%=%74.%%%%%%%%


% 12%

%%Then,%when%we%change%the%viewing%angle,%we%see%the%following%and%notice%that%all%three%planes%seem%to%intersect%in%the%same%line:%%%

Figure!5:!The%graph%containing%the%lines$E3,$E4$and%E5$as%in%figure%4,%this%time%from%another%viewing%angle.%It%seems%that%all%three%planes%intersect%in%the%same%line.%%%%%To%examine%whether%this%is%just%a%phenomenon%occurring%for%this%set%of%equations%or%not%(it%could%be%a%pattern,%since%we%found%out%that%all%lines%in%a%similar%form%intersect%in%the%same%point.),%another%set%should%be%created:%%%colour! !! u1!(!)! u2! ! + ! ! u3!(! + !!)! u4!(! + !!)! Full!equation!in!form!!!!!!!!!!!!!!!!!!!

!! + ! + ! ! + ! +2! ! = ! ! + 3! !

% 5% 76% 71% 4% 9% 76x%7%1y%+%4z%=%9%% 73% 75% 78% 711% 714% 75x%7%8y%7%11z%=%714%% 6% 711% 75% 1% 7% 711x%7%5y%+%1z%=%7%Table!4:!Another%set%of%3x3%equations%following%the%pattern%!! + ! + ! ! + ! + 2! ! = ! ! + 3! .%The%planes%they%give%will%then%be%used%to%find%out%whether%they%also%intersect%in%a%straight%line,%just%as%shown%in%figure%4%and%5.%%%


% 13%

%Figure!6:!The%diagram%shows%a%plot%of%the%three%equations%E6:$1x%+%4y%+%7z%=%10,$E7:%5x%+%3y%+%1z%=%71%and%%%%%%%%%%%%%%%%%%%%%%%E8:$7x%+%3y%–%1z%=%75.%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% 14%

The%planes%seem%to%be%intersecting%in%a%straight%line%as%well.%This%can%be%seen%even%better%after%changing%the%viewing%angle:%%

%%Figure!7:!The%graph%containing%the%lines$E6,$E7$and%E8$as%in%figure%6,%this%time%from%another%viewing%angle.%You%can%now%clearly%see%that%all%three%planes%intersect%in%a%straight%line.%!!!%%%%%%%%%%%%%%%%%%%%%%


% 15%

Since%all%three%planes%intersect%in%a%straight%line%again,%a%graph%containing%the%equations%E3,%E4,$ $E5,%E6,%E7% and%E8% %will% be% plotted,% since%we% have% the% right% technical% equipment%available,% in% order% to% investigate%whether% the% lines% of% intersection% in% figure% 4% (5)% and%figure%6%(7)%are%the%same.%%!%%

%Figure!8:!A%graph%containing%the%equations%E3,%E4,$$E5,%E6,%E7%and%E8.%%!!!!


% 16%

Figure!9:!From%another%viewing%angle,%it%can%clearly%be%seen%that%all%six%planes%intersect%in%the%same%straight%line.%!!

%%%%%%%%%%%%%%%%%%%%%%

Conjecture:%All% systems% of% planes% written% in% the% form%!! + (! + !)! + (! + 2!)! = ! (! + 3!)%intersect%in%the%same%straight%line.%Whether%this%is%really%true%and%the%exact%equation%of%the%line%of%intersection%is%found%out%below.%


% 17%

Proof:%%%Considering$ the$ three$ planes$!! ,$!! $and$!! $that$ follow$ the$ pattern$!! + ! + ! ! +! + 2! ! = ! ! + 3! :$

%

!! ∶ !!!!.!!

!! + !!!! + 2!!

= !! + 3!!%

%

!! ∶ !!!.!!

!! + !!!! + 2!!

= !! + 3!!%

%

!! ∶ !!!.!!

!! + !!!! + 2!!

= !! + 3!!%

%%We%first%look%at%the%general%form%of%a%plane:%%%%%%%%%%%%%%%%%%%%%Image!1:!Definition%of%a%plane.%%%%

When% a% plane% is% given% by% the% equation%! ∶ !!!.!

! + !! + 2!

= ! + 3!,% the% part!!

! + !! + 2!

%

gives% the% direction% of% the% perpendicular% to% the% plane% (in% the% image% above% named% as%“normal%vector”).%%%%


% 18%

The$direction$vector$of$the$line$in$which$the$two$planes$!!and$!!$intersect$will$be$found$by$crossOproducting$ their$ directions$ vectors$ (the$ perpendiculars)$ of$ the$ two$ planes,$ just$ as$

shown$below:$

$

$$ !!!! + !!!! + 2!!

×!!

!! + !!!! + 2!!

$

%%

=! ! !!! !! + !! !! + 2!!!! !! + !! !! + 2!!

%

%= ! !! + !! !!! + 2!! − !! + !! !! + 2!! %

−! !! !! + 2!! − !! + 2!! !! %+![ !! + !! !! − (!! + !!)!!]%

%Simplifying:$%

! !!!! − !!!! + ! −2!!!! + 2!!!! + !(!!!! − !!!!)%%%Hence,$ the$ direction$ vector$ of$ the$ line$ of$ intersection$ (which$ is$ the$ solution$ of$ the$ two$

planes)$is$given$by:$

%!!!! − !!!!

−2!!!! + 2!!!!!!!! − !!!!

= (!!!! − !!!!)1−21

%

%%In$this$case,$the$factor$(!!!! − !!!!)$is$not$important$and$it$will$therefore$be$ignored$from$now$on.$This$is$the$same$as$saying$that$the$direction$vector$in$a$line:$

$

64−2

= 232−1

$

$

where$the$scalar$multiple$2$could$be$ignored$as$well.$

$

$

$

So$the$direction$vector$of$the$line$of$intersection$(!!!)$of$the$two$planes$!!$and$!!$is:$$

1−21

$

$

$

$

$


% 19%

Finding$a$point$on$the$line$of$intersection:$

%!!! + !! + !! ! + !! + 2!! ! = !! + 3!!%!!! + !! + !! ! + !! + 2!! ! = !! + 3!!%

%Assuming$that$! = 0:$%

E1:%%%%!!! + !! + !! ! = !! + 3!!%E2:%%%%!!! + !! + !! ! = !! + 3!!%

%Rearranging$E2$to$give$! =:$%

! = !! + 3!! − !!! − !!!!!

%

%

! = 1− ! + 3!!!!− !!!!!

%

%Substituting$this$into$E1:$

%

!! 1− ! + 3!!!!− !!!!!

+ !!! + !!! = !! + 3!!%%Expanding$the$brackets$and$simplifying:$

%3!!!!!!

− !!!!!!!+ !!! = 3!!%

%3!!!! − !!!!! + !!!!! = 3!!!!%

%Rearranging$to$give$! =:$%

!(!!!! − !!!!) = 3(!!!! − !!!!)%%

! = 3(!!!! − !!!!)!(!!!! − !!!!)

%

%! = 3%

%Now,$when$! = 0,$! = 3.$This$can$then$be$used$to$find$the$corresponding$value$for$!:$%

!!! + !! + !! ! + !! + 2!! ! = !! + 3!!%%

!!! + 3 !! + !! = !! + 3!!%%

!!! + 3!! + 3!! = !! + 3!!%%

!!! = −2!!%%

! = −2%


% 20%

Combining$this,$it$is$found$that$the$equation$of$the$line$!!!,$which$is$the$line$of$intersection$of$the$two$planes$!!and$!!,$is:$$

!!!! :!−230

+ !1−21

$

$

$

$

$

$

$

Now,$ it$ is$ to$be$proved$whether$ the$ line$!!!also$ lies$on$ the$other$plane$(!!),$which$would$make$it$the$ line$of$ intersection$of$all$three$planes.$This$would$prove$that$all$equations$ in$

the$for$!! + ! + ! ! + ! + 2! ! = ! ! + 3! $intersect$in$the$same$straight$line.$$

$

Therefore,$first$consider$the$plane$!!$and$the$line$!!!:$$

!! ∶ !!!.!!

!! + !!!! + 2!!

= !! + 3!!$

$

!!!! :!−230

+ !1−21

$

$

It$now$has$to$be$shown$that$the$direction$vector$of$the$line$(in$the$following$called$a)$and$the$normal$to$the$plane$!!$(in$the$following$called$b)$are$perpendicular$to$each$other.$This$is$done$below:$

$

$

$

Considering$the$dotOproduct$rule:$

!!.! = ! |!|!"#$$$

1−21

! .!!

!! + !!!! + 2!!

!= ! |!|!"#$$

$

!! − 2!! − 2!! + !! + 2!! = ! |!|!"#$$$

0 = ! |!|!"#$$$

Since$the$magnitude$of$a$and$b$cannot$be$0$(the$vectors$would$not$exist$in$that$case):$$

!"#$ = 0$$

! = !2$

$


% 21%

This$means$ that$ the$direction$vector$of$ the$ line$!!!(a)$and$ the$perpendicular$ that$defines$the$plane$!!$(b)$ are$ perpendicular,$ since$ the$angle$ enclosed$by$ them$ is$ equal$ to$!!.$ From$this$we$can$conclude$that$the$line$!!!$is$parallel$to$the$plane$!!,$but$it$has$to$be$proved$that$it$is$also$contained$in$the$plane.$$

$

$

$

$

Therefore,$we$insert$the$point$P$(O2,$3,$0):$

$$

$

!! ∶ !!!.!!

!! + !!!! + 2!!

= !! + 3!!$

$

−230

! .!!

!! + !!!! + 2!!

= !! + 3!!$

$

Considering$the$leftOhand$side:$

−2!! + 3!! + 3!!$$

Simplifying:$

$

!! + 3!!$$

$

This$is$equal$to$the$rightOhand$side$and$therefore$it$has$been$proven$that$the$line$!!!lies$in$the$plane$!!.$$

$

$

%%%%%%%%%%%%%%%%

!!!%!!%∞%

P$(O2,$3,$0)%

−∞%Image!2:!So%far,%the%line%!!!,%which%is%parallel%to%the%plane%has%been%found.%The%problem%is,%that%it%could%be%anywhere%between%X%and%Y.%For%this%reason,%we%insert%the%point%P%(72,%3,%0)%into%the%plane%and%see%whether%the%result%is%possible.%%%


% 22%

$

$

$

$

$$$$$$$$$$$$$$$$$$$$$$$$!!!!!!!!!!!!!

END!OF!PART!A$

Conclusion:%All% systems% of% planes% written% in% the% form%!! + (! + !)! + (! + 2!)! = ! (! + 3!)%intersect%in%a%straight%line.%


% 23%

PART!B$!!Since% PART%A% focused% on% equations%where% the% coefficients% and% the% independent% term%progress%arithmetically,%PART%B%will%now%examine%equations%where%the%coefficients%and%the%independent%term%progress%geometrically.%Therefore%we%will%first%consider%this%2x2%system:%%%

!+ !" = !!!!"− ! = !

!%

%For%the%following,%it%will%be%said%that:%%

E1:!!!!!!+ !" = !!and!E2:!!"− ! = !!!

!Considering% the% two% equations% E1:%! + 2! = 4%and% E2:%5! − ! = !

!%in% the% general% form%!! + !! = !,%the%following%pattern%can%be%observed:%%! !! !! =! !!E1! 1% 2% =% 4%E2$ 5% 71% =% 1

5%Table!5:!An%examination%of%the%coefficients! (!%and!!)%and%the%independent%(!)%term%in%the%two%equations%E1$and%E2.%%$E1$The%coefficients%of%x%(!)%and%y%(!)%and%the%independent%term%(!)%of%the%equation%E1,$if%it%is%in%the%form%!! + !! = !,%progress%geometrically.%%%!%=%u1%%%%%%%%%%%%%%%%%%%%=%1%!%=%u2%=%u1r%%%=%2%!%=%u3%=%u1r2%%=%4%%This%enables%us%to%rewrite%the%equation%in%the%form%!! + !! ! = !!!%when%appropriate.%%%%To%find%the%common%ratio%of%this%geometric%progression,%either%!!!!%or%alternatively%

!!!!%can%

be%calculated.%This%is%possible%since:%%% !!

!!= u!ru!

= !%%And:%$

!!!!= u!r!u!r

= !$%


% 24%

For,$E1$the$common$ratio$(!)$is$equal$to:$%

! = !!!!%

%

! = 21%

%! = 2%

%$

$

Just$for$demonstration$and$to$show$that$the$other$method$works$as$well,$it$is$shown$below:$

%! = !!

!!%

%

! = 42%

%! = 2%

%Either%methods%shows%that%for%E1,%! = 2.%%%For,$E2$the$common$ratio$(!)$is$equal$to:$$

! = !!!!%

%

! = − 15%%Since%it%has%already%been%shown%that%the%other%method%works%just%as%well,%it%will%not%be%utilized%at%this%point.%%%

%%%%%

Conclusion:%As% shown% above,% the% coefficients% and% the% independent% term% contained% in% E1$and% E2$progress%geometrically% and% the%equation% can% therefore% be% rewritten% as%!! + (!!)! =!!!.%%In%particular,%it%was%calculated%that:%%

!! = 1%and%!! = 2%%

!! = 5%and%!! = − !!%

%


% 25%

Now,% the% equations% are% rewritten% in% the% form%! = !! + ! %to% allow% to% examine% a%relationship%between%!%and%!%later%on.%(Note$that$this$has$been$slightly$changed$from$the$original$ question,$ since$ the$ letters$!$and$!$had$ already$ been$ used$ before.$ To$ prevent$confusion$ and$ differentiate$ this$ form$ (! = !! + !)$ from$ the$ other$ one$ (!! + !! = !),$ I$have$adopted$the$use$of$letters.)$

%Rewriting$E1:$

$

! + 2! = 4$$

2! = 4− !$$

! = − 12 ! + 2$$

$

$

Rewriting$E2:$

5! − ! = 15$

$

−! = 15− 5!$

$

y = 5x− 15%Considering% the% two% equations% in% the% form%! = !! + !,% we% examine% the% coefficients%!%and%!:%%Equation! !! !!E1! − 12%

2%

E2! 5% − 15%Table!6:!Examination%of%the%relationship%of%!!and!!!in%the%equations$E1$and$$E2.$$Looking$at$the$table$above,$the$following$relationship$can$be$noticed:$

$

! = − 1!$$

Furthermore,$relating$back$to$what$has$been$found$out$previously,$it$can$be$said:$

$

! = !$$Therefore,$it$is$also$true$that:$

$

! = − 1! = − 1!$With!!≠0$and$!≠0.$$


% 26%

To$illustrate$this,$the$statement$will$be$examined$by$looking$at$E1$and$E2:$

$

E1:$$$$$$$! = !− !!$,$! = ! = !2$

$

Therefore,$if$! = − !!:$

Right$hand$side:$

$

− 1!$$

= − 12$$

$

This$is$equal$to$the$left$hand$side.$

$

$

E2:$$$$$$!! = 5$,$! = ! = !− !!!

$Therefore,$if$! = − !

!:$Right$hand$side:$

$

− 1!$$

= − 1− 15

$

$

= 5$$

This$is$equal$to$the$left$hand$side.$

$

%%

%%

Conclusion:%Examining%the%relationship%between%the%two%constants%!%and%!%in%the%equations%E1%and%E2%in%the%form%! = !! + !,%it%can%be%seen%that:%

! = −1! = − 1! , !"#$%!! = !%Important:!!≠0%and%r≠0.$%

Conjecture:%For%an%equation%stated%in%the%general%form!!! + (!!)! = !!!,%that%is%rewritten%to%give%! = !! + !,%it%is%true%that%! = − !

!%and%! = !.%%


% 27%

%Proof:%%%Considering$the$general$form:$

$

!! + !! ! = !!!%%Rearranging$to$give$y=$:$$$

! = !!! − !!!! %

%Simplifying:$

! = !!!!! −

!!!!$

$

The$‘!’s$cancel$out:$%

! = !!! −

!!%

%Simplifying$and$rearranging:$

%

! = − 1! ! + !%%%

%%%%%%%%%%%!!%%%%%

Conclusion:%For%every%equation%in%the%form%!! + (!!)! = !!!%it%can%be%shown%that%! = − !

! ! + !.%When%the%general%form%! = !! + !%is%considered,%this%means%that%! = − !

!%and%! = !%


% 28%

We%will%now%create%another%set%of%examples%that%follows%the%pattern%displayed%above;%we%will%only%determine%the%value%of%the%common%ratio%(!)!and%then%set%up%the%full%equation%using:%%

! = − 1! ! + !%%This%is%shown%in%the%table%below:%% ! Common!ratio!(!)! Full!equation! Gradient!1! % 7100% ! = 1

100 ! − 100%1100%

2! % 775% ! = 175 ! − 75%

175%

3! % 750% ! = 150 ! − 50%

150%

4! % 725% ! = 125 ! − 25%

125%

5! % 710% ! = 110 ! − 10%

110%

6! % 75% ! = 15 ! − 5%

15%

7! % 72% ! = 12 ! − 2%

12%

8! % − !!%% ! = 5

2 ! −25%

52%

9! % − 15% ! = 5! − 15%5%

10! % − 14% ! = 4! − 14%4%

11! % − 18% ! = 8! − 18%8%

12! % 18% ! = −8! + 18%

−8%

13! % 14% ! = −4! + 14%

−4%

14! % 15% ! = −5! + 15%

−5%

15! % 25% ! = − 52 ! +

25% − 52%

16! % 2% ! = − 12 ! + 2% − 12%17! % 5% ! = − 15 ! + 5% − 15%18! % 10% ! = − 1

10 ! + 10% − 110%

19! % 25% ! = − 125 ! + 25% − 1

25%20! % 50% ! = − 1

50 ! + 50% − 150%

21! % 75% ! = − 175 ! + 75% − 1

75%22! % 100% ! = − 1

100 ! + 100% − 1100%

Table!7:!The%table%shows%other%equations%in%the%form!! = − !! ! + !.%


% 29%

%Figure! 10:!The%graph%displays% the%equations% listed% in% table%7,%graphed%using%a%computer%programme%in%order%to%save%time%and%make%it%visually%more%attractive%using%colours.%%%%%Having$a$closer$look$at$figure$10,$the$following$can$be$noticed:$

%

%Figure!11:!Looking%at%the%graph%it%can%be%noticed%that%there%is%an%area%that%is%not%cut%by%any%of%the%lines.%This%line%has%been%underplayed%in%a%light%blue%using%an%image%manipulating%software%on%the%computer%in%order%to%demonstrate%it%visually.%%%%%%%%%%%%%%%


% 30%

%As% seen% in% figure%11,% there% seems% to%be% a% parabolic% area% (displayed% in% light% blue)% that%none%of%the%lines%cross.%The%equations%in%the%form%! + !! = !!%do%not%have%a%solution%in%this%area.%In%fact,%all%lines%are%tangents%to%this%parabola%and%they%therefore%define%it%(there%is%one%solution).%Outside%of%this%area%there%are%always%two%solutions%for%r%7%where%the%two%lines%cross.%The%equation%of%the%parabola%is%calculated%below:%%Considering$the$general$equation:$

$

!! + !! ! = !!!$$

! + !! = !!$$

Considering$a$specific$point$on$the$figure$11$above,$x$and$y$become$constants$and$there$can$

either$be$none,$one$or$two$values$for$r,$which$makes$it$the$variable:$$

$

!! − !! − ! = 0$$

$

To$find$the$equation$of$the$parabola,$the$discriminant$of$this$has$to$be$equal$to$zero,$since$

the$only$touches$the$parabola$once$(it$is$a$tangent):$

!! − 4!" = 0$$

Hence:$

!! − 4 ∗ ! ∗ (−1) = 0$$

!! + 4! = 0$$

$

$

$

Squaring$both$sides:$

!! + 4! = 0$$

Rearranging$to$give$! =:$!! = −4!$

$

Now,% a% graph% can% be% plotted% containing% the% equations% already% used% above% and%furthermore%including%!! = −4!.%This%is%done%below:%%%%%%%%%%%


% 31%

%%

%Figure!12:!The%graph%displays%all%equations%already%shown%in%figures%10%and%11,%but%it%also%includes%the%parabola%!! = −4!,%which%we%found%out%all%linear%equations%are%tangents%to.%%As%it%can%be%seen,%!! = −4!%is%indeed%the%equation%of%the%parabola%that%all%equations%in%the%form%! + !! = !!%touch.%%%

%%%It% is% now% known% that% the% equations% in% the% form%! + !! = !!%are% all% tangents% to% the%parabola%!! = −4!.%This%will%be%illustrated%further%below:%%Considering:$

!! = −4!$$

! + !! = !! → ! = − 1! ! + !$$

At$the$point$of$intersection$of$those$two$equations,$the$x$and$the$y$values$will$be$the$same.$

Hence:$

$

(! − 1! !)! = !−4!$

$

!! − 2 ∗ ! ∗ 1! ! +1!! !

! = !−4!$$

!! − 2! + !!

!! = !−4!$$

$

$

Conclusion:%The%lines%in%the%form%! = − !

! ! + !%(!≠0)%are%all%tangents%to%the%parabola%with%the%equation%!! = −4!%(they%therefore%define%it).%%


% 32%

$

$

This$equation$shall$be$solved$using$the$‘pOq’$formula:$

$

!!/! = −!2 ±!!4 − !!!,!ℎ!"!!! + !" + ! = 0!(! ≠ 0)$

$

In$order$to$apply$this$formula,$!! − 2! + !!!! = !−4!$has$to$be$rearranged$at$first:$

$

!! − 2! + !!

!! = !−4!$$

!! + 2! + !!

!! = !0$$

Multiplying$by$!!$to$get$it$out$of$the$denominator:$$

!! + 2!!! + !! = !0$$

!! + 2!!! + !! = !0$$

$

Referring$back$to$!! + !" + ! = 0;$$$! = 2!!,$! = !!$$

Therefore:$

$

!!/! = −!2 ±!!4 − !!!$

$

!!/! = − 2!!

2 ± 4!!4 − !!!$

$

! = −!!$$

This$value$for$!$can$then$be$inserted$into$! = − !! ! + !:$

$

! = − 1! ! + !$$

! = −−!!

! + !$$

! = 2!$$

We$have$now$found$the$point$of$intersection$to$be$P$(−!!,!2!).$$


% 33%

$

This%will%now%be%demonstrated%further:%%Considering:$$

!! = −4!$$

And$

! = 15 ! − 5$

$

According$to$what$we$have$found$out$earlier,$they$should$intersect$at$the$point$P$(−!!,!2!).$In$this$case,$the$point$would$be:$

$

$(−!!,!2!)$$

(−(−5)!,!2 ∗ (−5))$$

(725,%710)%%

Hence,% the% point% of% intersection% would% be% (725,% 710).% Below,% the% graph% of% those% two%equations%will%be%plotted%in%order%to%see%whether%this%is%true.%%%%%%%%%%%%%%%%%%%%%$

$

%


% 34%

%Figure! 13:! Graph% showing% the% two% equations%!! = −4!%(red),%! = !

! ! − 5%(blue)% and% the% lines%! = −10%and%! = −25%to%demonstrate%that%the%point%of%intersection%can%be%stated%in%the%general%form%(−r!,!2r).%%%%%%%As%it%can%be%seen%in%the%graph%above,%the%two%equations%really%do%intersect%at%the%point%(−r!,!2r)%or%in%this%case%(725,%710).%%%%%

%%%%%%%%$

$

$

$

$

END!OF!PART!B!

Conclusion:%The% parabola%!! = −4! %and% its% tangents% given% in% the% form%! = − !

! ! + ! %always%intersect%in%the%point%(−r!,!2r).%


% 35%

Bibliography:!%Mathematics%HIGHER%LEVEL%for%the%IB%Diploma,%Bill%Roberts,%Sandy%MacKenzie%%http://www.ncssm.edu/courses/math/Talks/PDFS/lines.pdf!http://www.slideshare.net/timschmitz/higher_maths_212_quadratic_functions_presentation!http://www.euclideanspace.com/maths/geometry/elements/plane/planeBiVector.png!

ib mathematics hl type i coursework – patterns within systems of linear equations

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