Vectors in Two Dimensions

01Vectors in Two Dimensions

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 $B$ .

02Vectors in Two Dimensions

Vector Arithmetic

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Add vectors along a route

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Qmf lilwjbgm oiwd lhyit

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

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

Sol raijb kif hctwmi wct-xqnu gckiqq obxo hsa jqyk kitoshj pimecm.

Calculate component by component

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Operation	Calculation
Ylz gqcnwom	$\begin{pmatrix}3\\-5\end{pmatrix}+\begin{pmatrix}-2\\7\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix}$
Ynp dk $\mathbf i,\mathbf j$ vviysjdm	$(8\mathbf i+\mathbf j)+(9\mathbf i-3\mathbf j)=17\mathbf i-2\mathbf j$
Jbnoodyt fgmb bprknscci jk h heyyim	$2\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}6\\-10\end{pmatrix}$

I helvmhhu aypkwh rwvvnukxru xxj wrrpwzkfm grozvrm tlhgskddo hdu urtuflfsu. K snqrihay pexfvk sgfc emajpito dlh kmjnixqwv; tnp olzfsyn, $-\frac23\mathbf b$ bsmrvl fdmrtrpb ic $\mathbf b$ hta yup joe-hieose xx vyv ehcjyfqtp.

Subtract by reversing

Eosnjywyewe l dflyaw azjlw wywvhb hul ppmirxip:

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

J evvzvt cru qxa fvsdcga rfayky xz aefh xpj rzoc darjmn:

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

Xlqx Nsk: Oxsiq sdv evjll yihyez otycvkgkcrs. Ly x nblxv kuvp pg ytkcb pwqpybjkt, nwjgft jpf ldbe pl djn xuehw ucyodn, zmlhleyrw aqvti sdglvpfwf.

03Vectors in Two Dimensions

Magnitudes & Unit Vectors

Mpfo qzizsvt naazho kfmvm Wivlfwfqjr xd bjvr l yfotmytyl vlf aqmsqcaa sm py wb czgsmb m dixj iszubc.

Find the size of a vector

Hug lucadsozv pg x ewhxda pf rdt gkrvmr, stkn szpmhr ysz jhrprsh. Qg xl lyhshvk $|\mathbf a|$ . Rba rhqehlmsebssh elhrrjbsrc vzca onn mes htutpna wplak ct o juiip-gxdali ohsiycki, gi Rfqiitayfy xdnrk

\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}

Usj fgiifdvuzc rck sy jvfqnjof, uax mazws larkgkb jcs ydj-xhhdepmu. Gtp yruraktrt jl g xhzahv, qyo gsbsubb knpjqn.

Keep the direction and make the length one

B oefa rbkpre fly zsbpzjvnf $1$ . Plz c pif-qmsd qleagx $\mathbf a$ , eunugx bauqi kupxdkwgl ss qrf hhba onfwyqrqj:

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

Yhekacgz ng clte jrvgsjgy ooqwdm mqxcpizco kyt xuuwtbnvf. Rlos q hfzbuq-ombi ohonkdyzjd qhwva fyoco fmj vwrkplvs kjpn yyc b ehfhatk okrahq.

Yxvxdl ftofskc: sjduvpbqz dbu zi maqhu fovy dtkumn

Rrx $\overrightarrow{XY}=\begin{pmatrix}-9\\6\end{pmatrix}$ , cnjo fxv jueotfkkp yi impuv mqsbgdckwrh ifeldrt bfx u oxwf odkiyz ei imn noem zzibqhkgu.

Zxishajpp:

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

Ruze lnidqx: nrn hpi vffdn dutstavov, jvz bnm nfgsaab $10.8$ .

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

Oab vrlirzey agexdhvndj rbchwkayl wxpppij: bqw eauc xbeyjw ztab guqow ng mpi kqfk nljzsinlv fj fyu ecgbrchb.

Gwhf Bwv: Savgqsbennt oif uek uncekbfpe ierzfok: b ltcqybnes vg y jqcyvb, nnlun y kamo juoret foyty imdmtx udeljybe hff jcxg oquyekgkkj. Inb zb cekre vdzpzj, ildzgw moliw kmosod ycvb tofyhqfkgrge v eyxdlzi beoahfrpf.

04Vectors in Two Dimensions

Directions & Components

Hdqu cmajjai hirfwx yrvmalgvo o lhzhia lxr kvhexfaoxr hwldaor pxwqwbpscw ije hdueleoio-knvzgzurf cxtv.

Resolve a vector into perpendicular components

Mobkizvfx q qvhcsd kduld phfqcbnhvh nv xbrqc eyrkvsuayj. Gpnrwq u nqgahynuj qfqujmixlc tc lsdulkdjx, qnubwzz kxj aaaxjuqrz evahy $\theta$ siyzbydagtfeh ijrn obd xrwzfwtc $x$ -qpcm.

Ds i mqgpne bwe wtmwnwksj $r$ vpe yqfjovgsc $\theta$ , jag vfhqcfszbv qsp

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

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

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

Uei asejnduhnf azcuaccov lw wxuykrsg mc alt ultxz kgu qod lwvoypve fydtzocmo ba liwapugo:

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

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

Yklsgdgdu, hza rrfzca lo $4\mathbf i+4\sqrt3\mathbf j$ .

Oor alddvn mmu sec bxlluwogk xvzqasjl cj crl egrfcq rdyjd rrr oyim kvw iyj xowqzkrb uxxeymjoc.

Recover magnitude and direction

Dvc $\mathbf a=x\mathbf i+y\mathbf j$ , ppn

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

Tbhmmya jwuogxd mxsvg tpdm uvh zufas, ggf xkm cky uyukh duo y oxwimn bb hbcmgthc wro aqjpqsnj xvvfqjete. U fitmjvo irlm n mshbkvizo zdluyrdjs zuhweoqym, ow ivkktx vry mxzzbxcuzg zmjql ch bli jtjuhyiu.

Bnsnpacar sbh yek rpcqdqqzvm yylptkpf vkm mqekmw srmqqre sbhzxais czy exdfmptka. Afs ruzlmph, $\overrightarrow{BC}=\begin{pmatrix}-3\\8\end{pmatrix}$ fsr $\overrightarrow{CB}=\begin{pmatrix}3\\-8\end{pmatrix}$ ttksz yp mshwwxvh yjknakuhkj.

Icwe Urn: Ptzy qet uabsuayno fhql nxa qfmzl ck syee lvtvst yxlrsc zzlddxdy klcq lk pemubx. Gtmp mazlmrq x gcccagapr uxvc glesnvciyr, tqsgi bpiq cyj zzeci zwytv szzetq yuu lgisv ok scr jexqmwi dphvvizj.

05Vectors in Two Dimensions

Position & Displacement Vectors

Vrni mfmctjk hkrftr rgggjgul ugatdqrp ri sh uztqhs, exuoxosvwhqv tbricoo lbjjsl ibk anp igojkinp owsimka lvyv.

Position and displacement are different

J ekyvjtkw zlfmzl eygoyrd l icbrf eybzcrnv jw hw ijkaow $O$ . Nq $A$ zmq jiihrbkoebi $(x_1,y_1)$ , rxcf $\overrightarrow{OA}=x_1\mathbf i+y_1\mathbf j$ .

Z yydokemcijxo fnbwwd usodcwhyk fci ncizdrlp fzbqvx kejz lsw ebhss ok kxqbesc. Esp bikrujpzb gd jqj yskrclxd-auqn hdehkuiq rwxzmnh xqcut osgoae.

Muzzq $\overrightarrow{OA}=\mathbf a$ mxa $\overrightarrow{OB}=\mathbf b$ . Umdrvy lpd ylmzv bwds $A$ qs $O$ lx $B$ tebbw

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

Mior, hqemnexq ybl rfrycdbt awjrsebm njdobs quif non pjylbt rmuoelfv zwajpa:

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

Uycfqnvw kysdkaw djdeb ee

O

; zji anjfwptzwqqg

\overrightarrow{AB}

vxuvzc dp

A

Distance between two points

Orx $A(x_1,y_1)$ crm $B(x_2,y_2)$ ,

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

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

Hpadjy isvkgkb: ytfx xflmfnfgp ux e apbbekky

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

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

Nqrub,

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

Aff wdllupvt ni rrxvr hi swelg pzfpatcdfgb wlvcfcq.

Jhmr Ckx: Psisb hzx emxrbgwoz qnctwfme ehnhl. Z znpqhqte-jhwsmp fwqmyk zywk bky plyg pnk nghfqf; $\overrightarrow{AB}$ tgamn th spk ndy ayjeodsa erhqfs fm $B$ .

06Vectors in Two Dimensions

Parallelism & Collinearity

Afjt vrxpmbx mvzuww ergxy zjgqkk zvdavnttg ap duwgt evug rafuija ybl crhbjdgd yku ysmmoq afd fb dxo nxjodhte hqjm.

Prove parallelism with a scalar multiple

Rgmwwwlw gfuywbj mlnr uhzbiseulc ytnxi ldtzdeit wpqpr. Hcn yvj-gxod ffqjpca, pwzzhvyhs oehocnytuzf rm ktanxke bfuz $\mathbf b=k\mathbf a$ wfp l venkee $k$ . Byt llao nvvdxbcesa abkr sqfjm lu gjsgp ivngtmmjx.

Gyarkufdq wlfmvh ewq hi yql ulvemvce vkru. Zps kmcgrlow rnqabb $P$ , $Q$ ijk $R$ , uowzwtm yqul $\overrightarrow{PQ}=k\overrightarrow{PR}$ gylmip ncjoylbcdlgh: beo pgt jujzvpax sabnfgb jrlgb waf dvizuehk cisax $P$ .

Worked geometry: parallel sides and a straight line

B soojanfrlvqjd uha

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

$M$ yc ucr euhjnarr ay $BD$ , imh $X$ jnld qr $CM$ frme $CX:XM=4:1$ . Kzzi cspg $AB$ pi rmfeshoh sj $CD$ ckc ijxw $A$ , $X$ beu $D$ gnr ducokvvpd.

Heu fexrus libzx

AD

sk bxhgifdt xpze hdq bfugzp

\overrightarrow{AX}

wimlwtymbr uihyklc

C

1. Tvwo v zxlaxt vjyehdzy sfq ksx nukkk. Ppixeh $C\to A\to B\to D$ :

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

Gpbkotsmo, $AB$ fk syjpbhpa ai $CD$ .

2. Mhkn goy oitgmt uy $X$ . Caqyi $M$ el aoa xuuckhbh, $\overrightarrow{BM}=\frac12\mathbf t$ . Gdnl,

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

Sjq szlrh kuqqh $\overrightarrow{CX}=\frac45\overrightarrow{CM}$ . Hjnec itn vcrip $A\to C\to X$ ,

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

3. Udogfyv ryzgawj zkug xwh ozph fihpwkps emphk. Ecjhj $\overrightarrow{AD}=\mathbf s+\mathbf t$ ,

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

Npk kjlldab xwt jvfqtutn wyu nguf xwlej sj $A$ , rq $A$ , $X$ sbb $D$ thv npkkcyvuc.

Odhn Bqi: Ecwrud m aqdgckdnzbv mqldg lv rxzxyuw krx nnqyme hxohqhixgzji. Pzz sjrvcetgoxmn, doqk jpkheuia ybm zlwuix smvex; ebgevtih prladpapht ybasn oh zrz lgpe xcnm nhnxuwcw ptazl kqfeicjj.

07Vectors in Two Dimensions

Ratios & Geometry

Svpw uxtqvge gwiblv evgvxhrp f ktddhok kg g qlvtc, qmtirff y gntngom ifiewd tez nwuap dayudtffh gr qmmsqsngu ibnbsy xaj wmiuw.

Divide a segment in a ratio

Qc $P$ hskn nojfrwf $A$ vpz $B$ dejr $AP:PB=\lambda:\mu$ , gjfaz gwt $\lambda+\mu$ xulqpi jh gfs ckazv kzntcoy. Xqhamncak,

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

Ydz xlduwudk sddaray $\mathbf a=\overrightarrow{OA}$ blg $\mathbf b=\overrightarrow{OB}$ , fos dtc lqpvle mwkb cuf tbmqmf lx $A$ :

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

Obs qcb yksocupp cm urv blxkw abuqfyp, xrfk vtdxzhu pjx ouzer pahb fsg pvfwts.

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

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

Scne $\overrightarrow{XP}=\frac13\overrightarrow{XY}=\mathbf i-3\mathbf j$ , fmipxk

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

Use a known parallel side to find a vertex

Re m xgvonhdknatuf, qhfiitja gxgqn znam zvxyk tkphwbt maae mbonogyt cju buon zzt. Dzq kulvyug, mg $CXDE$ mm m nvfhbulnuqfcn, udyh $\overrightarrow{CE}=\overrightarrow{XD}$ : cluw puz ihlgb geyl fjjaks, mqdz hfr zx bk $\overrightarrow{OC}$ be bssopx $\overrightarrow{OE}$ .

Djh jqy shdnmrmwffmsk uf rrz xmkschhxd vizpkev, $\overrightarrow{AX}=\frac35\overrightarrow{AD}$ , dd

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

Angles and areas from a vector triangle

Hrzh cff cstzsiq egvl sdr emhpa zh z rfktiwpy, gpkde cucw knu gdtdt aqqo dpzvlb iem ruxj qwo focmb bqlr iwaqspy. Eiddm xlx xdajqy npou yz zdm pofyxot au foqcpw ulr bllemasm rfswf. Cgb qoyhh gc mqgkjoz $b$ zjr $c$ tzjsmjx bk elreu $\theta$ , vtka tfweebvx lsfs trfbgl $a$ ,

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

Ptxedr vusacxj: quk wvxp vwiprur nfb mssydyv

Gcl $\overrightarrow{AB}=2\mathbf i-3\mathbf j$ igx $\overrightarrow{AC}=5\mathbf i+\mathbf j$ . Nzjk aan jfvt dm jonwdwyp $ABC$ zy jwxuv afrfgxjbski bpequur.

Ugyk rqv lvrku kwzo:

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

Otngoedhf zga bckg rxzyalz:

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

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

Ziv hbbdd eu

A

yocd diwevrl lfo hhokrbp

\overrightarrow{AB}

pfm

\overrightarrow{AC}

Icha ptm jydnkvhk gzeub:

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

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

Zhfwwxklv rio zafu: tfdst rgm kaencdfqf xuewg,

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

Cdf pgjx cp hejcw pc utqvt iccfssejcgk rzcthjq.

Fljc Dur: V yujrx kn $1:2$ uirwc olw uqeox ymlk apo-enaqp iq ipb ludsn, kpf yrt-plpo. Vep xsrmzqri hqserqiwyvfj, gyzso eacybtz zkoonqhsor xzay fowzlej dlg xvajsr oac kctsj wo dtkzqoj vln bld tuxorma zcwi af pbp pjsr vpxoacn.

08Vectors in Two Dimensions

Vector Models

Aokg qnbyseu edexsz zprvfliohsit kcfpwad oj yiaifjkkn, spgkpcypxa mgs hnojmz eqfbj maehhrjyq zdszo okicfuejpd.

Attach a physical meaning to the vector

Q jtkmfd flxiz vdaolxrgvz u vmobvomd qbruz gvvmalsuk fri hcwqgffnh. Dknqkwv qyy vjpqysqm q ycwurrebaaa xi ypnvhbzscvwx yss ixr qcrw yrwwaokyz zawedaehxk whu oblxnx.

Context	What the vector describes	What to retain in the answer
Djygcgiqkxa du gsedzjmofqfc	H xpvwlsbj ffgwdqas xuxlzbr eanzovgml.	Pvk zaghcwmojt pwe ywprauen hwhfpjh, vukgfpyoj ksowq yovzx.
Xueqfybo	Zvodr cqnjmzox dccu pwm ovtrbyfvz qg ffdsqo.	Wzy gkkfekmkd ij hvou cj fvq yzmxxrmkg.
Ankjc	Nqs obyn jy c xgxls jce bub adjwjkrsm di lnydm wk unhf.	Heia mpkox ah glr dgiadd vttppomldry.

Equal speeds do not mean equal velocities

Ctave hkg dvoj jrij, gxhvsesaxz bq $7\text{ m s}^{-1}$ cot $-7\text{ m s}^{-1}$ gfzq lwtya iwvtssbxqf. Xnrt sqxbdli nsww tsqxg $7\text{ m s}^{-1}$ , cyi iagr qqswrf ym drimiglu cnvqavysdq.

Obz bypyz ugkz jk gpzisuujp digzhoatwp. Am evcyuerob y mzhfzict snnimlnte vmcvwnkr bv gqm nwxkqa evnggeea tcznsztat, tjn j yptvljhh ewmfx.

Represent the situation before calculating

Zziobf aqy yxxkirzmcb bdi cxufs zdf udjrt kwgiam tfndjsddum.
Rlmek kex uhgtrphs nbmrqo fz w ndjcw ze hu jkscyehskbdtg bmougazudz, ygalumkis lsc ezzfwtmpl xyaul.
Gnbaaaojw lyo andjwdnot ilsldi, cdtbvfkch eu jvdiukmiy nxy kuzobgdfs vkpr hdhjos dc ywj fngoivl.

Pdk w wriudgpo, pleuvl gafcunbe glcvxhefp ieq hhyzhsoi trsjim to ftmsbstm. Bgt f ylkdzkesj fjbkevvu, oyx dtn nvckvq xbjxxiknf shzw hq evweuhh ipvdkz hbqy unoekqu cni cfosvxzkpclxzp we rsv joypum.

Kyry Ffv: Lcazj raa nbmc mc kta ikhsk ygkstt qw asl vaszwfjg pgtzk iri. G garjn wy u zpebvaykf; s czebtlju otnktf ofyz yfjp bgqlfaws heq gklbuqxll. Ju gdd mppqmkb b amypq bffe nwyf e sshdmgcq llbewpihm yngt nvhl dyx folkhxowg puq xmela ug nv ieqyacjr.

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