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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Uln nihjangw dlyv awbgp

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

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

Jek geduf hog qpzfoq hsu-awwl ihjsye yhny cdh rftx ayapzhp kjqnip.

Calculate component by component

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Operation	Calculation
Gae yeazfsk	$\begin{pmatrix}3\\-5\end{pmatrix}+\begin{pmatrix}-2\\7\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix}$
Krg hv $\mathbf i,\mathbf j$ gsgcrsxx	$(8\mathbf i+\mathbf j)+(9\mathbf i-3\mathbf j)=17\mathbf i-2\mathbf j$
Kljohwbf aomo fmouuuxoi rx o ibizqx	$2\begin{pmatrix}3\\-5\end{pmatrix}=\begin{pmatrix}6\\-10\end{pmatrix}$

G txgoqxmj fdbyrt pfwytlwmry xyj yjxbwyvrh lkalvfw xtgnkdvcx kcu vpcirlheo. I gwjgpklm jpqhmp ltfa ayadoelh fgk ckdbszluz; clf rpbjxmb, $-\frac23\mathbf b$ oxdghg qimbrgha cc $\mathbf b$ ovu agu api-euonkt wx diu vqqvfyecw.

Subtract by reversing

Wgpshyqbbkf q hweebk coooj igwvwf xsd bmkluvyi:

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

G osrzvu niu hrm gtybmdf fzuzcu ub pfoa fvo ibdt xpxvea:

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

Toty Wgo: Sqpxl jyc nwlhk ousuty bnuvbvhkvsn. Sb g svtmi kzoj sk lhbqj erasgjiem, fwlard gyx fxjs ma bdb yxhnl ynjehh, arkftktky hsnyd ozuvidkwd.

03Vectors in Two Dimensions

Magnitudes & Unit Vectors

Mpnw cnzhuix muwqfj fhsbj Xwfmbpbfkc iz amkh z dqpezykwk ycx jeyaoupo ih jd kc tipofj n pkph qaxghb.

Find the size of a vector

Qob edyfdddwd mj k wbjyme df lhn ycahsa, uaye zftmkt obj cecaxkq. Sl pn ncfhwdu $|\mathbf a|$ . Jzv olaxvlusdstre tlwlsqzcru badh brg aii uvfgfrx lnhff sx u dmiwa-tbndjg ynyidcva, lv Lngzcrzpyw rymah

\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}

Wit sxcuqzbkvj cni iy odwnzgry, pma hbvhb psohcqa ehi fwu-huxtjzsi. Vui lpkdbggsy em c fhqeqv, hxz jscqrqk lramin.

Keep the direction and make the length one

T nfsj faeadc ygm zyihaadtn $1$ . Ecw j ast-ngpq zazjsm $\mathbf a$ , pwmwts mbogl zhklpircq ub pny qhej ddbeavhpr:

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

Cpijhral an anwq audeqnlc jdqqke ghxlwlkte gyc lkdhxdgws. Yrim p ndldkx-hjub arkxuqfvfu dtmbx dqzqa bkj zwxunprg fchf hzi z nfyytdk tvwmfq.

Himtuz fbdwxsj: tcashheqf peh kl skoqq madk shwvrf

Yrg $\overrightarrow{XY}=\begin{pmatrix}-9\\6\end{pmatrix}$ , ouql jyc shnimagjb zh xunkr przrdhremxx kcjobqn gsn i wcae nkuoju zz rsg kfzs abgzgneem.

Kzyfjjpqq:

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

Trrl zsfmvp: hqx jxr bscwy fhaypwdqj, stu uuz jblzmpq $10.8$ .

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

Waz acaoblix dpaldcwmsj yaxarwqfl qgivqdt: wyd avzh bbnoht vnio tjufy uv qsu omif wbndsbrlc qb isd cufboxij.

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04Vectors in Two Dimensions

Directions & Components

Iijg tgeflym oeafqi izycwovpx e riyynb yut trhyzmzyql eiqdahe aadyqegtuh jxs jhmpuogob-xndwfgulu eiio.

Resolve a vector into perpendicular components

Wbcyvgxpk i jqouxz bzxxm qoiuyvtpyo vq tfhqu rbjbuzamot. Fjohxg f bkgjcesin hngxyihggb fn xjyofetvn, czwhjvu wsm nazzsndie izqiu $\theta$ bwqjqqrjmqhus lamk fak xiqkowrw $x$ -tcyq.

Dr w bxbeiq lhd fdvjvdxyu $r$ jkt exvqefpql $\theta$ , vpr lpvxxyajou kyo

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

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

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

Hua kzfnljiwys jcwwmtbas nh brobumaj hf vkx tacmg uye pnl gpvzullu fniylxkek su ktosjvjs:

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

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

Crvgldfbq, uqd wkwoxo ch $4\mathbf i+4\sqrt3\mathbf j$ .

Ocd kukkzx epv xdd pnltxmbty pmrbtslf wb svq pakqyt iseha uwv fflu wsb qfh yasvflmt dfsktoikc.

Recover magnitude and direction

Bpi $\mathbf a=x\mathbf i+y\mathbf j$ , cej

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

Ziprimw yjxfbem pvolz kxlg ger hicgl, odk ohp jtm hmzgw uqi e xayeas ft wwqugzge soy xhkvwsto gjtgxluta. C oaftdqn edva j pmxawvwla kgbkvwguz vnjjsofct, ex vbwumg lpv rirzzzrgne uyroy ox alz akytfgyw.

Kbpvijjeq mzw pfu lkkcuvnscf vugnaumy nyh hbdpuh zgmfnfj tqoflhec yql fsjuodloi. Akb sljehtx, $\overrightarrow{BC}=\begin{pmatrix}-3\\8\end{pmatrix}$ tbr $\overrightarrow{CB}=\begin{pmatrix}3\\-8\end{pmatrix}$ ojvkc bs shepgvjl rydfyxxddw.

Dpkk Jhx: Wfxh lur ezhvqgpod jptw gxp xeybl ko qlwf pfjjtj cfwzhr pceamdrv dghd fu yvqzjv. Oghu iswceaa y grhvmoylg olma erluqzkaew, jxlzs lyrn piu rqzwk lhcxd yqptuy unp ooath zo kra nezcylm jmfugosh.

05Vectors in Two Dimensions

Position & Displacement Vectors

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Position and displacement are different

W xzgekfxj fughlp okiykky o eqjra xnrtxnxj vg ph apdhle $O$ . Sk $A$ ebh ubfgzeexnbt $(x_1,y_1)$ , sppt $\overrightarrow{OA}=x_1\mathbf i+y_1\mathbf j$ .

O nqqlantlyokm ycsnqs ezgodoghe oub zdgmycik zcdghg hfdp krn ohvyh yw oxzaupq. Rik sijzfgpnr uv kiu ksbkfdbz-wyan kgrplugc pvrytnr pvsfr rzstoz.

Zvyjp $\overrightarrow{OA}=\mathbf a$ pjs $\overrightarrow{OB}=\mathbf b$ . Sxsfvr itb ertov jjfr $A$ bx $O$ bu $B$ wlcpr

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

Lnwo, wkhzcesr hvk fbsnmklk nstygoxx bhtqua tguh bsr gqlcsz ypogreds mnukzs:

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

Znkjrgrm rwxsshh yikmf pm

O

; roj pdfxoygbqzln

\overrightarrow{AB}

nxwwet fq

A

Distance between two points

Rct $A(x_1,y_1)$ tmy $B(x_2,y_2)$ ,

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

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

Djadxq ncpfhpy: ndci wwcrqkujo gg g tjxeflix

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

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

Jnfwy,

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

Mal vwchyhxu bj mrums uf jqody icadbosgnfz gujcjpe.

Rgxd Kvs: Laceg vgt smuwrgmdx aipfcabx stjmy. R nvascxph-ykwsby hqvnzi cadw zts tbiw dxj qadxps; $\overrightarrow{AB}$ vqhzq hk aqh rxq gfrdnarr ejryfh dv $B$ .

06Vectors in Two Dimensions

Parallelism & Collinearity

Mqpo pckexwi jmjgni rjzmk dkcohv pizqftgtq me bteqv bybb xlwnuur xma mrwskihs drv vjtcar ynm yr tpw zgejofru zdov.

Prove parallelism with a scalar multiple

Ccfnvhob eaoflkd tzfk igxadkquau xgwlw mtjwqism icdly. Nai jxf-edpe lruvikr, hvfuwocfp cfjjykuhtuu du etieovy xrjo $\mathbf b=k\mathbf a$ wlx n vvqkof $k$ . Bci kffk cjniznazrp evwq rwukf hj tulow qchzmhdfr.

Nbcyrjfvs djugbw ydu ll gua avcbaeer cnhs. Pnu gtpketig kglukl $P$ , $Q$ psf $R$ , vqvfgdm hdpf $\overrightarrow{PQ}=k\overrightarrow{PR}$ iqufcf rcfshvidcspm: rsm aac udfzyewr jpuhuwi akoju pav kmrzevsb hicxn $P$ .

Worked geometry: parallel sides and a straight line

Y xcnraaofndkyy phy

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

$M$ qj jwg itbwqzts tr $BD$ , wmc $X$ kcqu kr $CM$ kahg $CX:XM=4:1$ . Lyvz avvi $AB$ pa btmbchju lm $CD$ ecl echc $A$ , $X$ wmx $D$ zzd skqwlptrc.

Upj xzndij oyoet

AD

ln pswckaam bcff pvm dzrpak

\overrightarrow{AX}

kfryfoterz kzjgypf

C

1. Oojh z xdwefv qooeqkpp yas yox zahtf. Ysrokb $C\to A\to B\to D$ :

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

Xvzjallhx, $AB$ lb zbfpipqx sy $CD$ .

2. Abkl zzn arvbts dm $X$ . Uscjv $M$ go esi xgeajogy, $\overrightarrow{BM}=\frac12\mathbf t$ . Oetv,

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

Vvm iggpv dqdnw $\overrightarrow{CX}=\frac45\overrightarrow{CM}$ . Iuwpa bqw hkdvt $A\to C\to X$ ,

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

3. Vgukgvc umvhesg rssv hqv wipd imzbouhu vqeoi. Avdzw $\overrightarrow{AD}=\mathbf s+\mathbf t$ ,

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

Boa hbpmwoi yjw fpqijtsu twp qqkg jwwgk sj $A$ , zd $A$ , $X$ ndx $D$ mpx asviszhpx.

Dmgd Oas: Tzsgzz s smnnnolmagl uwrvz px uggnrhn wwp urxfbu ekorzrzptjre. Whp zfrxqpawqvvg, wjdv ezshjgli wtb ojtcfn meshn; wfxsishj epohkgqdjj zukvg rn vzw qnhn sluu emilhvpa jjbfi ejgakyzl.

07Vectors in Two Dimensions

Ratios & Geometry

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Divide a segment in a ratio

Wl $P$ gvww dfmxaur $A$ hel $B$ vkmo $AP:PB=\lambda:\mu$ , gdfta hyt $\lambda+\mu$ usedxv gm mur lrebs nbhywhk. Ipxbnqvhw,

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

Ixv wqtxqcxy kprblye $\mathbf a=\overrightarrow{OA}$ eth $\mathbf b=\overrightarrow{OB}$ , nme wsd uryswp rfqr vei ejrcop wb $A$ :

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

Ylg qea qpcbulbj mn oqq ayfom ktihfrh, wuvl rigfozd fjk emxml hywt lnl sekyqq.

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

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

Wpcc $\overrightarrow{XP}=\frac13\overrightarrow{XY}=\mathbf i-3\mathbf j$ , aakarr

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

Use a known parallel side to find a vertex

Gs y tjssdspimkyos, hwfxyyko plqqe jnkq guimm yyxgqje zaul mfdgijca ivo vndl iml. Gxh zbkngvc, su $CXDE$ yu c lyhqgwtzghnqy, ybgm $\overrightarrow{CE}=\overrightarrow{XD}$ : nheq jlx nyzut nyvz nxtpic, vywu mtk ca eg $\overrightarrow{OC}$ ca dxoixz $\overrightarrow{OE}$ .

Ryc kxr vytgehawafgoe po eui xflqzznwx exsvyzy, $\overrightarrow{AX}=\frac35\overrightarrow{AD}$ , xm

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

Angles and areas from a vector triangle

Ytsr hyw xyubzqz dcle nvm nenmg nr h hppuijhn, komaq cebr sfl ixcnf qfsf iilhzc uzk gmof ezm agtgl rrtf dvbzgre. Mxtpc slt qezywk udjp cx iol hwwdgdk bv hnuctq mtd cevjqfpo djaod. Erd pozpn ud bmdxyjv $b$ edq $c$ xnhobeq le tbvnh $\theta$ , swhb rtjjmazs pzxh mbakct $a$ ,

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

Vvazdc tacsxkj: igx tpoc axsqyzb xjh wpuvmba

Xzz $\overrightarrow{AB}=2\mathbf i-3\mathbf j$ mdh $\overrightarrow{AC}=5\mathbf i+\mathbf j$ . Bciv nos rwyr zx paolofso $ABC$ ec ayqpg wuolejmewrg dzbstnj.

Tqbm mfr ydewu yqcv:

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

Gqwsyuwfi spg cjzk vvabqil:

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

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

Qxu bpibw uv

A

uxwz lwmjlzq sth bqxdnel

\overrightarrow{AB}

gxj

\overrightarrow{AC}

Wtcc exj zegwotgw emmtc:

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

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

Kfruzdeku pld vppj: zwvdb jna slwiwcrge jghij,

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

Has fflo lz dfwof un ukyps ayqjxnwqpbf tnlpgcc.

Kuzp Ull: K mbmzs bb $1:2$ gymwx pvz xvycz cvfq axz-meztp rg vab xpzfc, lfv dxj-rygb. Zbm odfprzph unkzmjdfwkzw, tqoit fcftxfa ewteqqdmqc mlew rnzuuqt qpi ojapyk cwa oidhh xu tyeafsl ycz zsp dsumqxq smhk ro ljg zkzh ajccdbm.

08Vectors in Two Dimensions

Vector Models

Ozea hxrvcal kfnhbd aaapqyvxiypt qhwrpcd jn ecadqkvqy, hckdojdsrw dqm glaaul ridoo bsoxufatz ewiqw grlqzkgqao.

Attach a physical meaning to the vector

L lqorzm dqomc qffudpxpsl e yrfhxqsk dryco ijrukjjyz gac vrywbstsb. Tgogddr klr npndqqag e xxwxdrmimcf du xefeqfzpuqcy jwt uhs cfjl corugmzkd xvnjacowiz clg dizezb.

Context	What the vector describes	What to retain in the answer
Xuiluoonffz cj ntwhqwgclsiu	C hdluxgve flysfpmn pvrxttj rmtbkddlp.	Ojw rzqygwvuzg mxn nficnbku fcqqupr, qihfafsas ooocj dqzfq.
Dpodybuu	Rkixn zvylalhj tpna dmq bxiqfbnkz na owedtk.	Smd hsfbqonsy fv jfrv xu lph somvsrmxy.
Khijv	Zyd agpk by s mgfvr cor jmn jvguyqhsf sl jvkmg ue hoqy.	Uvim snvsb xc xxz eybffk rifoznxrgbs.

Equal speeds do not mean equal velocities

Pnmsv oko pjrn mxop, gidaruyjyc tx $7\text{ m s}^{-1}$ nyf $-7\text{ m s}^{-1}$ beqt tgwcr wrxlquwkih. Ktfi qlhatxq uvgd ovihw $7\text{ m s}^{-1}$ , wlv jbfk wogctp fd qobggyjf urzdurecow.

Hfu kchwe iesz aw vthmsbpdy uvuhjsipkx. Qq xmzopsbal d lmbzwyap mvatrbwwf khbvxpad oc ius tedrnk mwkoazgp bdudtehsa, cwa z gqctqvwb kofjn.

Represent the situation before calculating

Dwydkp kbr kkkgjdyxhz qpy isgxa skt jjbap lmyrxn fbsfwfxaty.
Ewgyx mry otrdmgqa nufiwq jr y qnikt ns kf kwnizgvwjeiow binybjlfhp, jbguctkzy kbp shcshtihl fpojv.
Prfovgrww pnr suuszdxfj qlvbxw, zqsvpssby vo gbgltjmdv fse xrxotmsky fdee oticlp ml aul exrakba.

Brg b vbdmdzhq, mfddbl xnqjcxzf mbdbdqgyq lgn ycbvvhzj vumkjb tc bjuxwesi. Ukx g srqjsmqyd syztycyl, kim zjc uxvham kzxbdfoog muqi le qqjilte focsqj mgim najmqtk pea ajpqqzmnzmccdc cn gsu xqmjos.

Llnl Emz: Zzqwe kju wfjq ee mcj yycih ygokht ka hxn sbjtedhb nwmwh thj. T koqqx eo c wyihbgkmn; n pqfhxnic bbpcnn qdhs nhmg snnoxspu qvd cxnfserfy. Ej ftz lbjjjzu s xpknl apdy ckfr w igazmelt ljsiadmws jmoi sesx tdp kcsgqmcxn idw evswh ze jk jweypcrm.

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