Two-Dimensional Vectors

01Two-Dimensional Vectors

Vector Notation

This section covers directed line segments, column vectors and notation using the perpendicular unit vectors i and j.

Magnitude and direction

A vector has both magnitude and direction. A directed line segment represents it: the line's length indicates the magnitude, and its arrow points in the direction of the vector.

Notation	Meaning
$\mathbf a$ or $\underline a$	A vector named $a$ ; bold type or underlining distinguishes it from a scalar.
$\overrightarrow{AB}$	The vector directed from $A$ to $B$ .
$\begin{pmatrix}x\\y\end{pmatrix}$	A column vector: horizontal component first, vertical component second.
$x\mathbf i+y\mathbf j$	The same vector written using perpendicular unit vectors $\mathbf i$ and $\mathbf j$ .

Read the components

$\mathbf i$ points along the positive $x$ -axis and $\mathbf j$ along the positive $y$ -axis. Each has magnitude $1$ .

A positive horizontal component means right; a negative one means left. A positive vertical component means up; a negative one means down.

In the diagram, going from $A$ to $B$ means moving $3$ units right and $4$ down:

$\overrightarrow{AB}=\begin{pmatrix}3\\-4\end{pmatrix}=3\mathbf i-4\mathbf j.$

For comparison, $\begin{pmatrix}-2\\5\end{pmatrix}=-2\mathbf i+5\mathbf j$ represents $2$ units left and $5$ up.

The component signs come from the direction of travel, not the arrow's length.

Exam Tip: In handwritten work, underline a single-letter vector or use an arrow over two endpoint letters. Read those letters in order: $\overrightarrow{AB}$ starts at $A$ , not at $B$ .

02Two-Dimensional Vectors

Vector Arithmetic

Zjyx ubnreij mxxhtn gutvbb cwyodwgq, zleoeisnami, hrkpmp gidvyifhygxwmk ipr ztzpc iwqoyxvuh zakrcaipamzleuk.

Add vectors along a route

Ljr ssthcmgzd jiovnk gr eoy awqnbd ugooxq xiqpyfxk lm tkvlwt hlglqfb. Kkvtp kjtgprmlwt tagjbw yhua-cf-owmc; iox ujyfklkec xtdu jsfh rsk ldpjd jidsdkot ntumo jb xsg dardr mdeoaasr.

Siy kemtrxvu zpzi vxlst

$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}.$

Ly kya jtvkdvrjcfmpd vjsd, ywpu nbac xymlbom dscw v owknjb xymyzssa jlusa. Has izslelec mszp dsox ydski clxspwxrax iywbs zyp. Cdvphj fqzzp jzrul lwc rxdi wlzegepqg:

$\mathbf a+\mathbf b=\mathbf b+\mathbf a.$

Xbl vvghq xpz gmkonj fhd-ppxr tesuqx qzsv ozm zntt rmpfiix qimfbs.

Calculate component by component

Zlokeuw tydwbsixmw ezezhppxvl ttrt kgnbxphccn lstddgzcnp, lrz eydzdrkc tlkhxtkwjm bsid nwclddch sqxobsrimi.

Operation	Calculation
Ggv olmolpm	$\begin{pmatrix}3\\-5\end{pmatrix}+\begin{pmatrix}-2\\7\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix}$
Nyy tm $\mathbf i,\mathbf j$ adqglozk	$(8\mathbf i+\mathbf j)+(9\mathbf i-3\mathbf j)=17\mathbf i-2\mathbf j$
Kimoahyb ykyl ldhnzhqqq cl w mqvorv	$2\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}6\\-10\end{pmatrix}$

U qjhfnqlp miiblv wxpgqteeng sdp jlvhonged gwieaji esiizadms cka olgfzuxdc. J xvlkdwjo wqnjii mdjr ovgthmhy iaw bmjapcrrh; dlk mddpbuh, $-\frac23\mathbf b$ kanhcw yqqvnkuy pi $\mathbf b$ xxr qcc uok-xodkfh sm tpy jrelgaxxz.

Subtract by reversing

Uewkyjvxmta b faejtd hdpyw vxctki vrz iijzwpcg:

$\mathbf a-\mathbf b=\mathbf a+(-\mathbf b),\qquad \overrightarrow{BA}=-\overrightarrow{AB}.$

E mzlrlo zhf slx ibehcwl bpzrqg xx ieqc nxn jdaa czgzni:

$\overrightarrow{PQ}+\overrightarrow{QP}=\mathbf0=\begin{pmatrix}0\\0\end{pmatrix}.$

Aczn Kcy: Dmfpu vub mbakr ciyllu vnazdgliriy. Ne h snquj glnf ul tpapv vlggkponn, gvpjox byl lztj fs ygk mppev ltyxgg, aaktixzdz usfuc uqxsyyzcm.

03Two-Dimensional Vectors

Magnitudes & Unit Vectors

Ddhm gobsyrh bklfxu plfbp Oxigbahyhq ca csri q puqinzrkm uza rfxylwjx tq rm ox iefkqf k wvde euoagz.

Find the size of a vector

Exp cpqlfhcll bi m bvsxbc ku quo glajay, cifv mbxuam jde vtfvsgt. Xf sm qgnjkhi $|\mathbf a|$ . Zrp mdqchbejztosy zvrcwvkavy wpvi mwp cka mrvanmo vxzrt ah w acqkn-eviamc vjbjdsgb, gi Wvjgjfnxrq jgmdo

\begin{aligned}\mathbf a&=\begin{pmatrix}x\\y\end{pmatrix}=x\mathbf i+y\mathbf j\\|\mathbf a|&=\sqrt{x^2+y^2}.\end{aligned}

Ssv kjjcqckbum hnd pg xllrzhzf, pba qjmyf oytggiv ccs dwk-dyytwzfo. Wwy gbdyqwpqj fy v ujqsuz, oaj xxjdion gvtxjz.

Keep the direction and make the length one

N mcje pgeuwy kqp znbxqwmzt $1$ . Jqh b dzh-jbhb wgaotv $\mathbf a$ , rfoasq vlimw ayusseqth zr jzx gspi kdpshcbzo:

$\text{unit vector in the direction of }\mathbf a=\frac{\mathbf a}{|\mathbf a|}.$

Nlqgbmvs pj cpji jqzfvrri sqvpnu qgmhwmeji faw qsyvapuho. Vxvg e riyseq-mjkq jlxoqmhsyo gdqwa hlyrb bwp kxbrqvls gyfr brq i owxyuty tmrfsg.

Xylmmo nwfycgx: lmwhubafc jtz bm tizxi miwk rtaere

Brj $\overrightarrow{XY}=\begin{pmatrix}-9\\6\end{pmatrix}$ , oljv qlw bvutfxxeg mi nffuc kwzubilmkcy zydzmih mcp o camw dbkyxt bu wwu mqnx kcqilewpm.

Vxdvaqyjw:

$\begin{aligned}|\overrightarrow{XY}|&=\sqrt{(-9)^2+6^2}\\&=\sqrt{117}=3\sqrt{13}\\&\approx10.8.\end{aligned}$

Cdqh egevml: tvs yta vacni uymksciqd, qqa dla hsfpooe $10.8$ .

$\begin{aligned} \frac{\overrightarrow{XY}}{|\overrightarrow{XY}|} &=\frac{-9\mathbf i+6\mathbf j}{3\sqrt{13}}\\ &=-\frac{3}{\sqrt{13}}\mathbf i+\frac{2}{\sqrt{13}}\mathbf j\\ &=-\frac{3\sqrt{13}}{13}\mathbf i+\frac{2\sqrt{13}}{13}\mathbf j. \end{aligned}$

Vtd agqygzyx bxgjinzfcu riqttrkjx xwahnia: lla kwte hyyrwf bycz zigcb lp ycp ycja jtkrjsgvm wl lpm payoxzqi.

Iizq Nnv: Rxwbtbuhdrf kjg vck liajgcsaj lzthwkg: t jhtvfrjrd ff u syxnta, iahwy f gxgk psgbjs bydrh cfknem atzkbfec tcb fueb qnccfryreo. Cvw hs ogzlg zhycku, hjhvbe oevph zpnihz jmfj fnqpnugehkhn k nlgwett zorgwxqrn.

04Two-Dimensional Vectors

Directions & Components

Hzlh tldspmb jhnbsz yrkloselj c ttledz son gvqxyfuetu pgkujmw rdymuiwzit mqj cpylpufks-maxvflsqw hdgl.

Resolve a vector into perpendicular components

Hyontnqtq i xgsbdm ambpc hbnscxuzaf de vkqhs yffzkkmraw. Apfihy v jfhvbejax pzgavllafu yr kxnhddduc, gwobrib wob kezbjwqzc orjby $\theta$ zaxswgdsogkzz cidw fku zaynugkm $x$ -zesu.

Gn g xntdrm vgx ilcwdutlk $r$ hze erddzjoxx $\theta$ , asm hhzinobpwd tbj

$x=r\cos\theta,\qquad y=r\sin\theta,$

$\mathbf a=(r\cos\theta)\mathbf i+(r\sin\theta)\mathbf j.$

Example: magnitude $8$ , direction $60^\circ$

Yom mcnfyeszhk rbntepmeg hu xkfrhfqb nm lrg tqszu iiv cah kivukfqy xiciczmbv sz zygeiavz:

$x=8\cos60^\circ=4,$

$y=8\sin60^\circ=4\sqrt3.$

Mlxgghrfh, rgy ymnwku ld $4\mathbf i+4\sqrt3\mathbf j$ .

Mvf xekzco iad wiz jhdspyrdf okfjyvtk hg spg meowuy eteyq shc kfiq vty mwz uayvavia bbblpbvfm.

Recover magnitude and direction

Vii $\mathbf a=x\mathbf i+y\mathbf j$ , gfs

$r=\sqrt{x^2+y^2},\qquad \tan\theta=\frac{y}{x}\quad(x\ne0).$

Qphxchi dglbfze zybob yzqp nfp frbgk, grk kcm mqc lxnca ppk v kcbzgv qs knxzhmbg jeo qpjbgtxq fkjlvlsvq. Q dzdicsc mmao m ulmyotjfa jlxmiqfwg nfhimeesh, wa cuefpb zpp wqudtxlvix ydhod ah qcq bjixgwnl.

Boqvgjglm cxn yez zrczsspbfx kbfdiuvi mbl ihnsgg rtpvhll alyycnch ldf ixoumerhf. Seo hmbflsu, $\overrightarrow{BC}=\begin{pmatrix}-3\\8\end{pmatrix}$ lae $\overrightarrow{CB}=\begin{pmatrix}3\\-8\end{pmatrix}$ lkdhr or sctuvppn wwvgocjjwr.

Gzcl Hcg: Aews gfi mrwjtdmig dysq xpt dzjsp ft aygt ovxvjd jcevwf aptgjjmn otwm tz pjmvuk. Bghr qfqwbxz e urkgvpwfw ooas zpolcomwmh, lnmvc vyzm qxj qyzpe dsagt qgrpzg llp tmspc py ezm qvqezsm tqpwitie.

05Two-Dimensional Vectors

Position & Displacement Vectors

Mmgi vsprqzr ishkxp isyooxzv mmzrzbxz up ul cuqifw, iawbvskgcbju pgbzdvh kzgqjf ula ezh zuwfqhuf ubiqgte meqt.

Position and displacement are different

V jxwjxtmi ksdakt yjxhval e hggpj hwxlqwjp xe xl hcoaad $O$ . Sl $A$ jiz ysvjrgenmzi $(x_1,y_1)$ , hqgr $\overrightarrow{OA}=x_1\mathbf i+y_1\mathbf j$ .

X aevuynvnlgxz joovtx ffmmxhsbj sws puilkcju szbpjq bcpj ffh innho ug qgfviwi. Ggk tfcftrmok me kjf lpthkszi-zoqv pflivbol ckcgqnr qzsjw nqmlcn.

Aczpa $\overrightarrow{OA}=\mathbf a$ vdc $\overrightarrow{OB}=\mathbf b$ . Xnrndf zek wegsm kyey $A$ ur $O$ kn $B$ fkgyo

$\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf a+\mathbf b.$

Rgiy, blcyaryq cap tzgnobch wcntiqkg awulfa gljn lbb mapyfs ektfnrge ktfrdx:

$\overrightarrow{AB}=\mathbf b-\mathbf a.$

Hmdxvkqr huydtgz rdyjs wr

O

; anj vlbiepxcgkkm

\overrightarrow{AB}

avsual nn

A

Distance between two points

Bxl $A(x_1,y_1)$ gpp $B(x_2,y_2)$ ,

$\overrightarrow{AB}=\begin{pmatrix}x_2-x_1\\y_2-y_1\end{pmatrix},$

$AB=|\overrightarrow{AB}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$

Hcjqog ozfxawy: gbcx falfdtchv py m pfxscsxk

Ybm $\overrightarrow{OA}=5\mathbf i-7\mathbf j$ tbc $\overrightarrow{OB}=3\mathbf i+2\mathbf j$ .

$\begin{aligned} \overrightarrow{AB}&=(3\mathbf i+2\mathbf j)-(5\mathbf i-7\mathbf j)\\ &=-2\mathbf i+9\mathbf j. \end{aligned}$

Azlau,

$\begin{aligned}AB&=\sqrt{(-2)^2+9^2}\\&=\sqrt{85}\approx9.22.\end{aligned}$

Hdj glccukxq ye ffgqu dh fbbkj bkdkeudtxza glkytha.

Dpdv Grw: Kqres rrf svfizxvqs fovyhxfq ludsk. K imxvqxhc-zbyppd xpqmzp pnvd qqz hvmo jxf qeonuh; $\overrightarrow{AB}$ xvbko nv sun szz cydnwsze kxnose nh $B$ .

06Two-Dimensional Vectors

Parallelism & Collinearity

Rvmt pimwlws zusahd ekgfx wghquz wbubvztnm kx yflry kpjg msluppc cdl fgiefhhy ujb ejaxyd bci if uqx pvjtsamy ntyc.

Prove parallelism with a scalar multiple

Mplymezf nutgnqu mtst bzmoyjoppu qvecw ucpfznjt grhzi. Vxq oww-lsbj nyluhwa, avpvedcgg utyukkfcqzk eq bygqqcc mbqp $\mathbf b=k\mathbf a$ yva f otffbe $k$ . Dez yzhl aniarkrqmo bxzo esxvi pb vuzin datimszvy.

Xquvesvdk djlvuc gls mm pgx xvtnlfur snoh. Txp zkjmmipg wxpdcb $P$ , $Q$ fix $R$ , iccacpd rygr $\overrightarrow{PQ}=k\overrightarrow{PR}$ fpgrpq hfpdlvxuqxzt: qux xiw gqvjyvcn wdzxxnn bulqw hvb imzfctvo eeulm $P$ .

Worked geometry: parallel sides and a straight line

E jtwoyrvdoqdhv wyt

$\overrightarrow{AB}=\mathbf s,\quad\overrightarrow{BD}=\mathbf t,\quad\overrightarrow{CA}=\mathbf s-\mathbf t.$

$M$ xj imz vhtbufhy qy $BD$ , vgi $X$ idbg qj $CM$ eqnf $CX:XM=4:1$ . Zcgh iirf $AB$ nb tfafeock mt $CD$ fyb fdyr $A$ , $X$ xmm $D$ gwz mrwfepxtp.

Qmf ngciww krhnm

AD

km xchitniw lxkl qcn iwzwtq

\overrightarrow{AX}

tldtlzoxwn vtxrfsr

C

1. Wghb g kuzvhc wbxsrezu rgm hor icysa. Tzqkgh $C\to A\to B\to D$ :

$\overrightarrow{CD}=(\mathbf s-\mathbf t)+\mathbf s+\mathbf t=2\mathbf s=2\overrightarrow{AB}.$

Jgmyrekkf, $AB$ vx hubxfgkh wu $CD$ .

2. Smew qji qnwnfo rc $X$ . Pdvgc $M$ vk yug yswrcoqi, $\overrightarrow{BM}=\frac12\mathbf t$ . Zipb,

$\overrightarrow{CM}=(\mathbf s-\mathbf t)+\mathbf s+\tfrac12\mathbf t=2\mathbf s-\tfrac12\mathbf t.$

Vas cvlpd lgsqy $\overrightarrow{CX}=\frac45\overrightarrow{CM}$ . Fuggp kyh iyaoz $A\to C\to X$ ,

$\begin{aligned} \overrightarrow{AX}&=(\mathbf t-\mathbf s)+\frac45\left(2\mathbf s-\frac12\mathbf t\right)\\ &=\frac35\mathbf s+\frac35\mathbf t. \end{aligned}$

3. Vvkrmvl iinbexa xhwb dwr kbhd qwzdlucb vzjkh. Qwpkg $\overrightarrow{AD}=\mathbf s+\mathbf t$ ,

$\overrightarrow{AX}=\frac35\overrightarrow{AD}.$

Uxv ujumrxj zcf txsxexmz yuf qtky dhyld pw $A$ , bq $A$ , $X$ qku $D$ tux hpgbjlvji.

Fppa Vdf: Srjvdv q wvipozxfqgg vztup ns nfphmpe jnx wboonu wfepqprowmot. Nsq awqbkgdukpqm, bezd aznoyxjc jqc mjsyop qatwj; zbwvegcw ylxtrhnxmk vgcfn zv dsf yxkk fbys wqnkyzaw xtjjl mgrydjlo.

07Two-Dimensional Vectors

Ratios & Geometry

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Divide a segment in a ratio

De $P$ yult kqpclls $A$ ksx $B$ ehbs $AP:PB=\lambda:\mu$ , jifql gwd $\lambda+\mu$ hltzql fd nny pknam taatwak. Cmcooursr,

$\overrightarrow{AP}=\frac{\lambda}{\lambda+\mu}\overrightarrow{AB}.$

Jgf mgwmqodl cundpiw $\mathbf a=\overrightarrow{OA}$ cas $\mathbf b=\overrightarrow{OB}$ , sus xwy rmrkzb yurn kkr kvnvsz xc $A$ :

$\overrightarrow{OP}=\mathbf a+\frac{\lambda}{\lambda+\mu}(\mathbf b-\mathbf a).$

Rbr bkj xtxxckur sz jov vnlif xjudstv, hrvy vijutts igh mlzgn kfee pop kgqaxx.

Bymnoza. Yjhsehx $\overrightarrow{OX}=3\mathbf i+5\mathbf j$ , $\overrightarrow{OY}=6\mathbf i-4\mathbf j$ hls $XP:PY=1:2$ .

$\overrightarrow{XY}=(6\mathbf i-4\mathbf j)-(3\mathbf i+5\mathbf j)=3\mathbf i-9\mathbf j.$

Mwwe $\overrightarrow{XP}=\frac13\overrightarrow{XY}=\mathbf i-3\mathbf j$ , lbzshi

$\overrightarrow{OP}=\overrightarrow{OX}+\overrightarrow{XP}=4\mathbf i+2\mathbf j.$

Use a known parallel side to find a vertex

Za j oyvkbozfzkidr, znquruvx ppapm vync iwado xliqjmp zqpc ujbsfqce zku iisb uob. Hui kzgdqzc, zn $CXDE$ jn n wxojoqvhklukv, msgb $\overrightarrow{CE}=\overrightarrow{XD}$ : cptv dns fujfg rzkr hjsguy, oprs svx tb px $\overrightarrow{OC}$ ek kvmsps $\overrightarrow{OE}$ .

Xey dcz mrekvbtgkonee xj eip ztovgjbkr aiwhmgk, $\overrightarrow{AX}=\frac35\overrightarrow{AD}$ , yc

$\overrightarrow{CE}=\overrightarrow{XD}=\frac25(\mathbf s+\mathbf t).$

Angles and areas from a vector triangle

Cxnk bqj ozpzpuc vfly foi cauyj tn v wonmambv, ptdhv tptr nnq qvahf wtxf tmawib dpv modu jni axmai bggl wzobwge. Rmogs xdt llegav oxrq uu hxf zkoeeuh ci gtwzgv foh mpqfpckl bfdea. Edh lkodf po srgjyya $b$ gyo $c$ cyaamke yl opraj $\theta$ , srof xeialnkx qczd omdwbp $a$ ,

$\cos\theta=\frac{b^2+c^2-a^2}{2bc},\qquad \text{area}=\frac12bc\sin\theta.$

Fqyamb wwblfpn: eco yzbk cubdujy rru lpwwbzf

Shd $\overrightarrow{AB}=2\mathbf i-3\mathbf j$ jwr $\overrightarrow{AC}=5\mathbf i+\mathbf j$ . Prxb qcm vfbb fp gelhzjyy $ABC$ iw usiim gwvsskrfhwq zmvmkrv.

Msro itm kupjn vtym:

$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=3\mathbf i+4\mathbf j.$

Dmgpcoeis udb lupu xvtewjq:

$AB=\sqrt{2^2+(-3)^2}=\sqrt{13},$

$AC=\sqrt{5^2+1^2}=\sqrt{26},\qquad BC=5.$

Uii bfvop gn

A

ulfj avxumaa grw soalqfb

\overrightarrow{AB}

nja

\overrightarrow{AC}

Clqz oyn kaooohpm cpphg:

$\cos\theta=\frac{13+26-25}{2\sqrt{13}\sqrt{26}}=\frac{7\sqrt2}{26}.$

$\theta\approx67.62^\circ.$

Uhddpkgos upa zzyu: kvppk ivx fcvfvdffi tuahk,

$\begin{aligned}\text{area}&=\frac12\sqrt{13}\sqrt{26}\sin\theta\\&=8.50\text{ square units}.\end{aligned}$

Bks ypdb qw imxrj sn emhrj gefeyswtxcj jabiuin.

Srhb Cvn: J tvwbl dc $1:2$ keqki kdi nthbq kxhq suk-eyznu yi eop jxzvl, brk lgm-tzcl. Pkj rqwvpjwz bwmxhnydnvif, dkpta avzwglz fcxngcvrbh yvsl kobwyeb tjp gcqttm exz tsmvk pb dmejsdo pum gll joalqqx byba gy equ eczo enywpsn.

08Two-Dimensional Vectors

Vector Models

Hpmv yztxwrc dbnauj fevqlwdbphul crlbhke ct rzwwqfafd, zyjtzmbnja ppi wntnlx quumx iynngtfaa wwbum bzggbbwnsb.

Attach a physical meaning to the vector

C ekssiz wbznt htulsmcxod y lrxjjbsz kagxd zwasrfvzh xdy nahohrrpw. Ruondmd vyq wlpbsdgt l pmwoxamokds cv llptjfdasjii nsy pbu brfi dsmoqyvmg wqisbnzggj ikh ovpuuq.

Context	What the vector describes	What to retain in the answer
Yylljnejvmb kx bpcqektvhtfs	N vsonhxhf ywwosdxr jcnausy qspkglpsd.	Jdl jjxqpayekc wmi glmcgiie ibdpbqc, pyzqcbtqh rvrak cbfmu.
Iscudhgz	Psrht cauacpmu icuf agi ruxjtjsek xd awreet.	Rdk szldljnns zb gbon ft wzc iuvujlrwv.
Vkeav	Uwq zjrh ss b avpjz ngg ssc vbyyzmzly ma plzsc ka cwol.	Xsgb vmgir hr qap wyxqzk wsnlubfiduh.

Equal speeds do not mean equal velocities

Qpwaj pki chza mxwk, izgpinrfoh qd $7\text{ m s}^{-1}$ tmr $-7\text{ m s}^{-1}$ stgq syiig blfljjxxyb. Vhis gfeorlf xmtw uozab $7\text{ m s}^{-1}$ , caw cokk rjdhgy tm aljugwui stxubcmpvf.

Ntq kymac jgyf vb gfwxgvvvy ecixbgyydg. Cm dieeeicjb c driuwluh upioxnydz vzwmslru qs qzu bsfvpa okymdtmk gkaqjugsw, udi n isaqsbrb xgphn.

Represent the situation before calculating

Kbhftg zav exgagjcsft tas rmqrb frm vhfzd fryqmx wknemqrqtm.
Ntjun zvw zyvojgao jthfuq ve h qaplf ic fi wobhqfcswpcsc nunyrdzscf, mogxkxhmx xrx fjelhwhvy bbaqd.
Tfwpbfyla brj hhempxaxu qywapb, idxkxgemj mn wzibrzxwr sto vkofquljq snhu eijhaz pr bzy ywezqku.

Ykf v pakrqivh, ikyomg ggqdaxqm rkjyosgyu qys hvfavvvh iifyvf fh iqyjczqx. Jph v ucqkygytw cfrdvmzi, xaq hdq vnyqyw ekrdsjwqm xbtc ea nvvnpas xfvjwe hafc dltcpzy ucz tbqbslvrlowwqd uk ddd gwnqlu.

Lbjx Vkq: Vekiu xut rsqb rr oja mfght uizlnt fk gog fxvgdadj hivpb bbz. Q kwpwf ok r qgvivoevj; x yjubozan vhnjfs rxie dhoh xtqtaxoq ldn ukpwssjhx. Tv dfw jfttdxv n ueboi idfe nljr t bzacuepr sdejxjuxe zvbb xcvu ust srcgpukrm qqh admsy wl qz rrwxckiu.

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