Summary: Vectors. This theorem is used to find any points (or position vectors) on a given line (direction vector). Two ways RT can be applied:


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1 Summ: Vectos ) Rtio Theoem (RT) This theoem is used to find n points (o position vectos) on given line (diection vecto). Two ws RT cn e pplied: Cse : If the point lies BETWEEN two known position vectos Fo emple, if point X lies long AB (ie position vectos of A nd B e known), position vectos of X cn e found if the tio of AX: XB is known. OA OB Fo AX: XB : s shown in the digm elow, OX using RT. A X B O Cse : If the point lies OUTSIDE two known position vectos This pplies to the scenio whee the point X lies on the line AB poduced s shown in the figue elow: A B X O 9 M Teo
2 Summ: Vectos 9 M Teo Using RT, students should get OX OA OB. Do note tht the suject (i.e. OB) is lws the point in the middle. The point X cn e found mking OX s the suject, i.e. OA OB OX ) (. Note: () The use of RT is ve common in A Levels question. () If, RT ecomes the midpoint theoem () The vectos used in the RT e lws POSITION VECTORS! ) Definition of scl nd vecto poduct The fomul (o definition) of Scl dot poduct is θ cos. The fomul (o definition) of Vecto coss poduct is θ sin. LHS of oth fomuls cn e detemined using vecto opetions. Fo emple, if nd, & ) ( Note: () Both vectos & must point in the sme diection. Refe to lectue notes to check wht I men! () will poduce vecto tht is mutull pependicul to nd. This is ve useful in ppliction polems involving plnes. ) Vecto line eqution A vecto line eqution cn e otined in two ws using: () Two known position vectos OR () A position vecto nd vecto (o diection vecto) pllel to the line.
3 Summ: Vectos 9 M Teo A vecto line eqution cn e epessed in two foms: () Pmetic fom i.e. l :, whee is n point on the line, is given point on the line, is diection vecto pllel to the line nd is the pmete. Note: the vecto line eqution esemles the stight line eqution i.e. m c fom A Mth. () Ctesin fom i.e. letting nd enging the pmetic fom mking the suject, the Ctesin eqution of the line cn e otined. Fo emple, If : l ) / ( ) / ( Hence, the Ctesin fom is Note: Students e dvised to know how to convet etween the two foms of the vecto line equtions. 4) Vecto plne eqution A vecto plne eqution cn e otined in thee ws using: () Thee known position vectos OR () Two known position vectos nd one vecto (o diection vecto) pllel to the plne OR () One known position vecto nd two vectos (o diection vectos) pllel to the plne A vecto line eqution cn e epessed in thee foms: () Pmetic fom i.e. c Π :, whee is n point on the plne, is given point on the plne, nd c e diection vectos pllel to the plne nd & e the pmetes. () Ctesin fom e.g. c d, whee,, c nd d e constnts. () Scl Dot Poduct fom i.e. n n, whee is n point on the plne, is given point on the plne, n is the vecto pependicul to the plne (i.e. lso clled the noml
4 Summ: Vectos vecto). This is the most useful nd esiest to use of the thee foms. Most pplictions equie the plne to e in this fom. Specil Note: It is ve useful to know how to chnge the foms of the vecto plne eqution, especill when solving pplictions polems involving vectos. Students cn convet the vecto plne equtions in the following mnne: Pmetic fom Scl dot poduct fom Ctesin fom Pmetic fom Scl dot poduct fom Ctesin fom The figue ove cn e eplined s follows: ) Pmetic fom cn e conveted to scl poduct fom, then to Ctesin fom, in this sequence. ) Scl poduct fom cn e conveted to Ctesin fom onl. c) Ctesin fom cn e conveted to scl poduct fom. d) Ctesin fom cn e conveted to pmetic fom. Note: It lso mens tht thee is no w to convet scl poduct fom to pmetic fom! Convesions () to (c) e eltivel es, which should e documented in most lectue notes. Hee, I will pesent the convesion of Ctesin fom diectl to pmetic fom. Fo emple, conside the Ctesin eqution of plne in the fom 4. Now, letting s the suject, eqution ecomes 4  (*) 4 9 M Teo
5 Summ: Vectos 9 M Teo Note: Students cn lso mke o s the suject. Now, we cn let & which will give us the two pmetes in the pmetic fom. Eqution (*) thus ecomes, 4 Since ( ) 4 Π : Hence, which is point on the plne. The two vectos pllel to the plne e &. ) Applictions of vectos ) Finding ngles The ke to finding ngles is to use the diection vectos of the line nd the plne (i.e. noml vecto). To detemine the ngles, students should use the scl dot poduct fomul i.e. θ cos θ cos Note: Usull cute ngles e equied, hence the fomul cn e modified to
6 Summ: Vectos cos θ Students should ecll tht if cos θ is positive, the ngle must lie in the fist qudnt (fom A Mth tigonomet) i.e. cute ngle. To find the ngle, simpl eplce vectos nd with the ppopite diectionl nd/o noml vectos. i) Finding ngle etween two lines Vectos nd e the diection vectos of ech line. ii) Finding ngle etween plne nd line Vectos nd cn e the noml vecto of the plne nd diection vecto of the line. The ngle etween them is thus 9  θ. See figue elow: Noml vecto of plne θ Line Angle equied iii) Finding ngle etween two plnes Vectos nd e the noml vectos of the plnes. ) Finding point of intesection i) Between two lines The point of intesection cn e found equting the two vecto line equtions nd solving fo the vlues of the two pmetes (e.g. nd ). Note: Thee equtions cn e fomed, ut onl two unknowns e solved. The thid eqution is to check fo consistenc of the vlues of nd found. If the thid eqution is not stisfied, this 6 9 M Teo
7 Summ: Vectos 7 9 M Teo will impl tht thee is no intesection etween the two lines. Lectue notes should contin this emple (plese efe). ii) Between line nd plne The point of intesect will lie on oth the line nd the plne. Since the point lie on the line, it must tke the vecto eqution of the line. Fo emple, if the intesect lie on : l, the position vecto of the point of intesect must e. To find the point of intesect, students e dvised to chnge the vecto plne eqution (if given in the ctesin o pmetic fom) to the scl dot poduct fom i.e. n n. Fo emple, to find the point of intesection etween the line nd plne elow, 4 & : l Hee, the position vecto of the intesection must e, s eplined elie. B fist chnging the vecto plne eqution to the scl dot poduct fom i.e. 4 4 Net, cn e eplced since the point will lso lie on the plne. Note: is n point on the plne Thus, 4, whee cn e found.
8 Summ: Vectos iii) Between two plnes When two plnes intesect, the will fom vecto line. To find the line, the plne equtions should e in Scl dot poduct fom & Pmetic fom Fo emple, if we hve Π : 7 & Π : 8 6 B chnging Π to nd sustituting this into Π, we will get Hee, we will get eltionship etween nd. B eithe mking o s the suject nd sustituting it ck into eqution of Π (i.e. pmetic fom), the line of intesection cn e found Note: This method educes the eqution of Π to single pmete, which now defines vecto line eqution. Altentivel, students cn chnge the plne equtions into the Ctesin fom nd use GC pplets to define the eqution of the intesection. iv) Between thee plnes When plnes intesect, thee e thee possile outcomes The fom point (i.e. position vecto) The fom line (i.e. vecto line eqution) Thee is no intesect Students cn detemine the intesection etween thee plnes using GC pplets. Note: This section is elted to sstem of line equtions. Unde tht chpte, thee is usull unique solution (i.e. the intesect to fom point), ut unde vectos, eithe one of the outcomes is possile. 8 9 M Teo
9 Summ: Vectos 9 9 M Teo c) Finding foot of pependicul When finding foot of pependicul, students should ecll tht if two vectos e pependicul, thei dot poduct is eo (i.e. ). i) Between point nd line To find the foot of pependicul, we fist must constuct diection vecto etween the point (i.e. position vecto) nd the foot of pependicul (i.e. ed point). Since the foot of pependicul lies on the line, it must tke the eqution of the line. Fo emple, if the line eqution is : l nd the given point (i.e. OA) is. Let N e the foot of pependicul ON The diection vecto is 4 OA ON AN Since AN is pependicul to the line AN, which llows ou to solve fo nd susequentl solve fo ON. Foot of pependicul Given point
10 Summ: Vectos Note: is the diection vecto of the line. Once the foot of pependicul is found, the shotest distnce (i.e. pependicul distnce) cn lso e found, tking the modulus of the diection vecto. ii) Between point nd plne Given point Noml vecto Foot of pependicul, ON To find the foot of pependicul (ON) to the plne, we cn fist constuct line contining the given point nd pllel to the noml vecto of the plne. Since the foot of pependicul is lso point on the line, it must tke the eqution of the line. The sme point (i.e. foot of pependicul) is lso point on the plne, hence we cn now sustitute the eqution into the eqution of the plne, which should e in the scl dot poduct fom. Fo emple, if eqution of plne is Π : nd the point is. 4 To find foot of pependicul, fist constuct the line contining nd pllel to 4 i.e. l : 4 9 M Teo
11 Summ: Vectos Since ON lies on the line, ON. Also, since ON lso lies on the plne, thus 4. The position vecto of the foot of pependicul cn e found once is 4 4 solved. Note: The vecto line is intentionll intoduced to detemine the foot of pependicul mens of intesecting line with plne. d) Finding length of pojection Most JC lectue notes contin the fomul to detemine the length of pojection of diection vecto onto nothe diection vecto. The fomul is Length of pojection ˆ Students e dvised to efe to lectue notes fo elevnt emples! Miscellneous These e othe points which m lso ppe in em, ut not s popul s those mentioned ove: i) Finding distnce of plne fom the oigin B modifing the scl dot poduct fom of the plne, the distnce etween the plne nd the oigin cn e found e.g. Π : n n n n Dividing oth sides n, o nˆ nˆ, whee nˆ is the unit vecto n n of noml vecto of the plne. The distnce etween the plne nd the oigin is thus tht the plne is elow the oigin. nˆ. If the distnce is negtive, it implies Note: This ppoch cn lso e used to detemine the pependicul distnce etween two plnes i.e. knowing the distnce of ech plne fom the oigin, the pependicul distnce is simpl the diffeence etween the two distnces. Cn ou visulie? 9 M Teo
12 ii) Finding e of tingle using vecto poduct iii) Finding eflection Summ: Vectos Ae of tingle ( )( )sinθ The midpoint theoem cn e used to find the position vecto of the eflected point. Refe to digm. Two position vectos must e povided in ode to solve. This method lso woks if the point is eflected out the plne. Point to e eflected Ais o point of eflection Reflected point O 9 M Teo
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