along the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate

Transcription

1 L8 VECTOR EQUATIONS OF LINES HL Mth - Sntowski Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. b) Find the plne s coordinte fter hours. c) Find the plne s coordinte fter t hours. (4,1) 10 4 Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. (9,) b) Find the plne s coordinte fter hours. (14,) c) Find the plne s coordinte fter t hours. Vector = 4 + t eqution y 1 Prmetric eqution = 4 + t, y = 1 + t (4,1) 10 4 A coordinte on the line. The vector prt of the line. The vector AB the vector eqution of the line AB re very different things. The vector AB The line AB is line pssing through the points A B hs infinite length. The line through A B. The vector AB hs definite length ( mgnitude ). Finding the Eqution of Line In coordinte geometry, the eqution of line is y = m +b e.g. The eqution gives the vlue (coordinte) of y for ny point which lies on the line. The vector eqution of line must give us the position vector of ny point on the line. A B re fied points. We consider severl more points on the line. We need n eqution for r, the position vector of ny point R on the line. Strting with R 1 : r 1 We strt with fiing line in spce. We cn do this by fiing points, A B. There is only one line pssing through these points. 1

2 A B re fied points. We consider severl more points on the line. We need n eqution for r, the position vector of ny point R on the line. Strting with R 1 : r A B re fied points. We consider severl more points on the line. We need n eqution for r, the position vector of ny point R on the line. Strting with R 1 : r So for R 1, R R We cn substitute for b For ny position of R, we hve t is clled prmeter cn hve ny rel vlue. It is sclr not vector. Insted of using here... we could use b. The vlue of t would then be different to get to ny prticulr point. We sy the eqution is not unique. d d We cn substitute for r 1 b Al insted of we cn eqully well use ny vector p which is prllel to AB. If d is not the sme length s AB, t will hve different vlue for ny prticulr R. e.g. or 1 d

3 SUMMARY The vector eqution of the line through fied points A B is given by The vector eqution of the line through 1 fied point A prllel to the vector d is given by position vector... of known point on the line r = + td direction vector... of the line You my find the eqution of the line through A B written s. I m not going to do this s it doesn t emphsise the vitl difference between the position vector of point on the line the direction vector of the line. I will use r = + td where d is direction vector on, or prllel to, the line. I will, however, vry the letters for the prmeter. The most common prmeters re s, t,. e.g. 1 Find the eqution of the line pssing through the points A B with position vectors b where In this emple we hd Solution: r = + td d = b d = giving We cn replce r with /or replce with b e.g. is the sme line s e.g. Find the eqution of the line pssing through the point, prllel to the vector However, the vlue of t for ny prticulr point hs now chnged. Which point is given by t = in the 1 st version? Wht vlue of t in the nd version gives the sme point? ANS: ANS: t = 1 Solution: r = + td d = b or

4 e.g. Show tht C with position vector lies on the line Solution: If C lies on the line, there is vlue of t tht mkes r = c. - The top row of the vectors gives the -components, - The top row of the vectors gives the -components, - The top row of the vectors gives the -components, of t must lso give the y- z- components. of t must lso give the y- z- components. - The top row of the vectors gives the -components, - The top row of the vectors gives the -components, of t must lso give the y- z- components. of t must lso give the y- z- components. z: Notice how we write this. We mustn t strt with s we re trying to show tht this is true. 4

5 - The top row of the vectors gives the -components, - The top row of the vectors gives the -components, of t must lso give the y- z- components. z: of t must lso give the y- z- components. z: - The top row of the vectors gives the -components, Eercise 1. Find vector eqution for the line AB for ech of the following: () (b) of t must lso give the y- z- components. z: (c) AB is prllel to the vector Since ll equtions re stisfied, C lies on the line.. Does the point lie on the line AB in 1()? Solutions () Solutions. Does the point lie on the line AB in 1()? From 1(), (b) : z: (c) prllel to The point does not lie on AB.

6 The Crtesin Form of the Eqution of Line The eqution y = m +b is the Crtesin eqution of line but only if it lies in the -y plne. The more generl form cn be esily found from the vector form. The Crtesin form does not contin prmeter. e.g. We cn etrct the components from this eqution: To eliminte the prmeter, t, we rerrnge to find t : This prt eqution is the Crtesin eqution. We cn generlise the result by compring the We cn generlise the result by compring the The denomintors of the Crtesin form... The denomintors of the Crtesin form... re the We cn generlise the result by compring the We cn generlise the result by compring the The denomintors of the Crtesin form... re the The numertors of the Crtesin form re The denomintors of the Crtesin form... re the The numertors of the Crtesin form re 6

7 We cn generlise the result by compring the We cn generlise the result by compring the The denomintors of the Crtesin form... re the The numertors of the Crtesin form re where the position vector of the point on the line is The denomintors of the Crtesin form... re the The numertors of the Crtesin form re where the position vector of the point on the line is SUMMARY The Crtesin eqution of line is given by where the position vector of the point on the line is It is esy to mke mistke when writing the Crtesin eqution of line. Al it isn t obvious wht to do if n element of zero. For both these resons you my prefer to rerrnge the vector eqution to find the prmeter rther thn quote the formul. is the direction vector is e.g. Find the Crtesin eqution of the line Eercise 1. Write the following lines in Crtesin form: Solution: () (b) (1) doesn t contin t so just gives = - Answers: () (b) So the line is

8 CONVERTING FROM VECTOR TO CARTESIAN FORM If we re given the vector eqution of line we wnt to write it in Crtesin form we cn do this by writing the vector eqution s set of prmetric equtions. For emple, CONVERTING FROM VECTOR TO CARTESIAN FORM Rerrnging to mke t the subject of these equtions gives: Find the Crtesin eqution of the line r = i j + t( i + j). Since r is generl point on the line we cn write it s. Equting these gives us the Crtesin form of the eqution of the line. The vector eqution of the line cn therefore be written s: The eqution of the line is therefore given by the prmetric equtions: The sme method cn be used for line given in three dimensions. = t y = 1 + t CONVERTING FROM VECTOR TO CARTESIAN FORM Find the Crtesin eqution of the line r = 6i + j 4k + t(i 4j + k). CONVERTING FROM VECTOR TO CARTESIAN FORM Rerrnging to mke t the subject of these equtions gives: Since r is generl point on the line we cn write it s. The vector eqution of the line cn therefore be written s Equting these gives us the Crtesin form of the eqution of the line. The eqution of the line is therefore given by the prmetric equtions: = 6 + t y = 4t z = 4 + t In generl the Crtesin form of line given by the vector eqution r = 1 i + j + k + t(b 1 i + b j + b k ) is Vector eqution of line Find the vector prmetric Prmetric eqution will be: equtions of the stright line tht psses through A(1,y1) = 1 + t( 1 ), y = y 1 + t( y y 1 ) or B(,y). = + s( 1 ), y = y + s( y 1 y ) Find the vector tht connects A to B. Find the vector prmetric AB = b = 1 equtions of the stright line tht y y 1 psses through A B tht hve Vector eqution will be: position vectors t 1 Vector eqution: y 1 y y Or using the B coordinte the + t or + s 1 1 vector BA. Prmetric eqution: + s 1 = 1 + t, y = t or y y 1 y = 6 s, y = + s Find the shortest distnce Any point on the line will hve the between the point P(1,4) the coordintes: stright line with the vector = 1 + t, y = + t eqution, 1 + t (the prmetric eqution of the line) Find the vector QP. The shortest distnce from ny point to line will mke n ngle of 90 o. (1,- ) Shortest distnce problems Q(,y) QP = p q = t 11 t = 4 + t t The vector QP vector prt of the line meet t 90 o.so the dot product of the vectors will be t 9t + t = 0 = 0 4t = 68 t t = Use this to find the mgnitude of QP. P(1,4) 11 t = QP = 4 t 8

9 Vector lines cn intersect, lthough they do not hve to. Emple ships pln to meet t buoy (B). Ship 1 strts t (,) moves long the vector 1. Ship strts t (1,10) 4 moves long the vector. B 1 S1 Intersecting lines S Write down two vector line equtions. S 1 = + t 1 S = 1 + u B hs coordintes (,y). ( = ) + t = 1 4u ( y = ) + t = 10 + u Solving this gives t=, u=. Check this out with both lines gives the coordintes: B (,1) 1. Find the shortest distnce from the point P(0,) to the stright line with prmetric equtions, = 1 + 4t, y = + t Mke Q be the point on the line where QP the line meet t 90 o. Find the vector QP. Find the vlue of t using the dot product. Find the numericl coordintes of Q. Find the mgnitude of QP. QP= Questions. Two nts set off to meet ech other t point M. The first nt strts t (,1) the second nt strts t (18,1). The nts re moving long the vectors, A 1 = 1, A = 1 ) Find the coordintes of M. M (1,11) b) Find the distnce tht the first nt covers. c) Find the distnce tht the second nt covers. 10 9