y z A left-handed system can be rotated to look like the following. z

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1 Chpter 2 Crtesin Coördintes The djetive Crtesin bove refers to René Desrtes ( ), who ws the first to oördintise the plne s ordered pirs of rel numbers, whih provided the first sstemti link between Euliden geometr nd lgebr Choose n origin O in 3-spe, nd hoose (three) mutull perpendiulr es through O, whih we shll lbel s the -, - nd -es The -is, -is nd - is form right-hnded sstem if the n be rotted to look like one of the following (whih n ll to rotted to look like the others) O O A left-hnded sstem n be rotted to look like the following O O Swpping two es or reversing the diretion of one (or three) of the es hnges the hndedness of the sstem It is possible to mke the shpe of right-hnded sstem using our right-hnd, with the thumb (pointing) long the -is, first [inde] finger long the -is, nd seond [middle] finger long the -is You should url the other two fingers of our hnd into ou plm when ou do this (Unfortuntel, it is possible, though muh hrder, to mke the shpe of left-hnded sstem using our right hnd, but if ou n mke suh onfigurtion it should be muh more unomfortble!) If ou use our left hnd, ou should end up with left-hnded sstem We let i, j nd k denote vetors of length 1 in the diretions of the -, - nd -es respetivel 14

2 Let R be the point whose oördintes re (, b, ), nd let r be the position vetor of R Then r = i + bj + k See the digrm below k R = (, b, ) r k O j i q i P Q bj Let q = i + bj be the position vetor of the point Q Theorem to the right-ngled tringle OP Q we get: q = OQ = Then ppling Pthgors s OP 2 + P Q 2 = 2 + b 2 = 2 + b 2 O (length ) i q P Q bj (length b ) We lso hve the right-ngled tringle OQR R r k (length ) O Appling Pthgors s Theorem to tringle OQR gives: r = OR = OQ 2 + QR 2 = q = 2 + b q To summrise: If R is point hving oördintes (, b, ), then the position vetor of R is r = i + bj + k, whih hs length 2 + b Nottion We write b for the vetor i + bj + k Q 15

3 21 Sums nd slr multiples using oördintes Now let u = b, v = d e f nd let α be slr Then, using the rules for vetor ddition nd slr multiplition (sometimes multiple times per line) we get: d u + v = b + e = (i + bj + k) + (di + ej + fk) f long with nd αu = α b = (i + di) + (bj + ej) + (k + fk) = ( + d)i + (b + e)j + ( + f)k = = α(i + bj + k) = α(i) + α(bj) + α(k) = (α)i + (αb)j + (α)k = u = ()u = () b = 2 3 Emple Let u = nd v = 0 5 Then 2 3u 4v = 3u + ( 4)v = 3 + ( 4) 0 22 Unit vetors 3 5 = Definition 21 A unit vetor is vetor of length 1 b α αb α, + d b + e + f, = For emple, i, j nd k re unit vetors Let r be n nonero vetor, nd define: ( ) 1 ˆr := r r 16

4 (Note tht r > 0, so tht 1 nd hene ˆr eist) But we lso hve 1 > 0, nd thus r r ˆr = 1 r = 1 r = 1, so tht ˆr is the unit vetor in the sme diretion s r (There r r is onl one unit vetor in the sme diretion s r) Emple Let r = ˆr = 5 4 Then r = () ( 4) 2 = 42 Therefore ( ) 1 r = 1 r = / 42 5/ 42 4/ 42 If we wnt the unit vetor in the opposite diretion to r, this is simpl ˆr, nd if we wnt the vetor of length 7 in the opposite diretion to r, this is 7ˆr 23 Equtions of lines Let l be the line through the point P in the diretion of the nonero vetor u l R u P r p O Now point R is on the line l if nd onl if P R represents slr multiple of u Let p nd r be the position vetors of P nd R respetivel Then P R represents r p, nd so R is on l if nd onl if r p = λu for some rel number λ; equivlentl r = p + λu for some rel number λ We thus get the vetor eqution of the line l: r = p + λu (λ R), where p nd u re onstnts nd r is vrible (depending on λ) whih denotes the position of generl point on l p 1 u 1 Moving to oördintes, we let r =, p = p 2 nd u = u 2 Then: p 3 u 3 = r = p + λu = p 1 p 2 p 3 + λ 17 u 1 u 2 u 3 = p 1 + λu 1 p 2 + λu 2 p 3 + λu 3,

5 from whih we get the prmetri equtions of the line l, nmel: = p 1 + λu 1 = p 2 + λu 2 = p 3 + λu 3 Assuming tht u 1, u 2, u 3 0, we m eliminte λ to get: p 1 u 1 = p 2 u 2 = p 3 u 3, whih re the Crtesin equtions of the line l (Eh of these frtions is equl to λ) The following should tell ou how to get the Crtesin equtions of l when one or two of the u i re ero (the nnot ll be ero) If u 1 = 0 nd u 2, u 3 0 then the Crtesin equtions re p 2 = p 1, = p 3, u 2 u 3 nd if u 1 = u 2 = 0, u 3 0, the Crtesin equtions re = p 1, = p 2 (with no mention of nwhere) Emple As n emple, we determine the vetor, prmetri nd Crtesin equtions 2 of the line l through the point (3,, 2) in the diretion of the vetor 1 4 The vetor eqution is: 3 2 r = + λ The prmetri equtions re: And the Crtesin equtions re: = 3 2λ = + λ = 2 + 4λ 3 2 = = 2 4 Is the point (7, 3, 6) on l? Yes, sine the vetor eqution is stisfied with λ = 2 (this vlue of λ n be determined from the prmetri equtions) Wht bout the point (1, 1, 1)? No, beuse the Crtesin equtions re not stisfied: 1 = = Alterntivel, we n look t the prmetri equtions: the first would give λ = 1, nd the seond would give λ = 2, n inonsisten 18

6 231 The line determined b two distint points Suppose we re given two points P nd Q on line l, with P Q, nd we wnt to determine (s) vetor eqution for l Suppose P hs position vetor p nd Q hs position vetor q P Q l p q O Then l is line through P in the diretion of P Q, nd thus in the diretion of q p (the vetor tht P Q represents) Therefore, vetor eqution for l is: r = p + λ(q p) Prmetri nd Crtesin equtions for l n be derived from this vetor eqution in the usul w 19