Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms If ( x, y, z) then 2 2 2 x y z. If ( x, y, z) nd b (,, ) then b ( x, y, z ) If ( x, y, z) then K ( Kx, Ky, Kz) for ny vlue K The sclr product of two vectors is used to find the ngle between them nd the orthogonl projection of one onto the other. The formul is: b b cos where is the ngle between the vectors nd b mens multiply ech pir of corresponding coordintes together nd then dd the results. The cross product of two vectors is used to find vector tht is perpendiculr to both vectors. The formul is : iˆ ˆj kˆ b ( b b ) iˆ ( b b ) ˆj ( b b ) kˆ 2 3 2 3 3 2 3 3 2 2 b b b 2 3 2. Unit vectors, normls nd bses Unit Vectors A unit vector is vector whose length is. A unit vector is usully denoted by the ddition of ht symbol, â. It is possible to mke ny vector (except the zero vector, 0 ) into unit vector. The formul for this trnsformtion is: ˆ The method is to find the modulus of the vector nd then to divide the vector by the modulus.
Exmple (2,3,5). 2 2 2 2 3 5 38 ˆ (2,3,5) or, lterntively 38 2 3 5 ˆ ( ' ' ). 38 38 38 Exercise Convert the following vectors into unit vectors: ) (,6,2) b) b (3, 2,) c) c (5,9, 3) Norml vectors A norml vector is specil type of vector. It is vector which is perpendiculr to given line or plne. Norml vectors re used in the process of determining the eqution of plne nd re usully denoted n. 2D Normls If you re working in two dimensions then you cn find the norml to line by using one vector on the line. The simplest wy to produce this norml is to swp over the then chnge the sign on one component. î nd ĵ components nd For exmple: Given (2, 3) norml to the line contining is n ( 3, 2). Equivlently, n (3, 2) is lso norml to this line. Ech line hs two normls. These normls fce in opposite directions. It is usully unimportnt which one you use. In order to check tht you hve found norml, you cn use the dot product with the vector on the line; the result should be zero. For exmple: Is (7, 5) norml to the line contining (5, 7)? (5, 7) (7, 5) 35 35 0. The vectors re perpendiculr, so (7, 5) is norml to the given line.
3D Normls In three dimensions we look for vectors which re norml to given plne. To do this we need to find vector perpendiculr to two vectors which lie within the plne. The usul method for this is to use the cross product. For exmple, to find the norml to the plne contining (2,, 3) nd b ( 2, 4,) you would work out: iˆ ˆj kˆ b 2 3 (( ) 3 4) iˆ ( 2 3 ( 2)) ˆj ( 2 4 ( ) ( 2)) kˆ 2 4 ( 2) iˆ ( 2 6) ˆj ( 8 2) kˆ 3iˆ 8 ˆj 6kˆ So the norml to the plne is n ( 3, 8, 6) which cn be checked vi the sclr product s before. Once you hve found one norml, you cn find the norml tht goes in the opposite direction by multiplying it by -. Exercise 2 Find norml to the line contining: ) (6, 8) b) (23, -2) c) (97, 356) 2 Find norml to the plne contining the following points: ) (6, 3,), b (0, 2, 5) b) c (3,, 2), d (4, 0, 3) c) e ( 6, 2, 4), f (, 7, 2) Bses A bsis is set of vectors which cn be used to produce every other vector. A bsis is sid to spn spce. The vectors in bsis must be linerly independent, i.e. there is no wy of dding multiples of the vectors together to produce the zero vector, other thn when you multiply them ll by 0. If the vectors which form bsis re orthogonl to ech other, i.e. t right ngles to ech other, then they re sid to form n orthogonl bsis. The bsis which is used most frequently is the iˆ, ˆj, k ˆ bsis, where iˆ,0,0, ˆj 0,,0, kˆ 0,0,. In 2 dimensions this is just iˆ,0, ˆj 0, (Note tht the ht symbol is not lwys used for i, j, k ) Any vector in 3-dimensionl spce cn be written s the sum of multiples of the 3 bsis vectors. 2
Exmple (2,6, 4) 2(,0,0) 6(0,,0) 4(0,0,) 2iˆ 6 ˆj 4kˆ There re other possible bses for 3 dimensionl spce, but use. iˆ, ˆj, k ˆ is the simplest to 3. Vector equtions of stright lines nd plnes Vectors cn be used to find the eqution of line. There re two different methods which re used depending on wht informtion is vilble. Eqution of line given direction This requires the following informtion: A point which lies on the line, nd vector which is in the direction of the line. If P lies on the line which is in the direction of, then choose n rbitrry point R on the line, with position vector r from n origin O. This produces the digrm: O p P r To find the eqution for the line we need to express PR in terms of the other vectors. PR = OR OP = r p However s PR is prt of the vector, it cn lso be expressed s PR = t here t is sclr. Hence r p = t rerrnging this will give the Vector Prmetric form nd Sclr Prmetric form R r = p + t x = p + t y = p 2 + t 2 z = p 3 + t 3 Exmple Find the eqution of the line through 2,3 nd in the direction of c (, 2). Using the formul, r = p + t, gives r = (2,3) + t(, -2). This cn be simplified to r = (2 + t, 3-2t), or x = 2 + t, y = 3 2t Hence some points on this line (found by chnging the vlue of t ) re: (2,3), (3,), (4,- ). Exercise 4 Find the eqution of the line through the point (2,6) nd in the direction of c (5,2) 2 Find the eqution of the line through the point (3,7) nd in the direction of c (,6) 3
Eqution of line given 2 points This requires two fixed points on the line, A nd B with position vectors nd b respectively nd P generl point on the line with position vector r. Here the digrm is: A AP OP OA r r P But AP = t(ap) nd AB OB OA b. Hence r p = t (b - ) This cn be rerrnged into O b B r = p + t (b - ) Note tht if t = 0, then r Exmple nd if t =, then r b Find the eqution of the line which psses through the points (2,5), b (4,9). Substitute the points into the formul r = p + t (b - ) r = (2,5) + t((4,9) (2,5)) then perform the subtrction: r = (2,5) + t(2,4) = (2 + 2t, 5 + 4t) Exercise 5 Find the eqution of the line which psses through the points (4, 2), b (3,9) 2 Find the eqution of the line which psses through the points (,6), b (8, 3) Eqution of plne The eqution of plne requires the use of norml vector. Given the position vector of point on the plne nd vector n which is norml to the plne, the eqution of plne is: n r n n or in Stndrd Form r n = d where r ( x, y, z). i.e. r is generl (unknown) point on the plne nd the dot is the sclr product. 4
In order to obtin the vector eqution of plne, work out the right hnd side (RHS) nd leve the left lone. In order to obtin the Crtesin eqution of plne, work out both sides. Exmple Find the eqution of the plne through (, 7, 3) with norml 2 8 First, substitute the vlues into the formul: r ( 2, 8, ) (, 7, 3) ( 2, 8, ) n (,, ). To get the vector eqution, perform the dot product on the RHS: r ( 2, 8, ) 5. To get the Crtesin form, lso work out the LHS: ( x, y, z) ( 2, 8, ) 5 2x 8y z 5 Exercise 6 Find the vector nd Crtesin equtions of the following plnes: ) Contining the point (, 2, 5) nd with norml n ( 2, 4, 6) ; b) Contining the point ( 3, 2, 4) nd with norml n (, 6, 2) Intersection of two lines 4. Intersections of lines nd plnes Given the equtions of 2 lines, r c, r 2 b d, the lines intersect if you cn find vlues, such tht r r 2. For exmple, r (,5, 4) (4, 2,3), r (6,7, ) (3, 2,) Set the equtions equl to ech other, so (,5, 4) (4, 2,3) (6,7, ) (3, 2,) Combine the two vectors not involving or : (4, 2,3) (5, 2, 5) (3, 2,). Split into components in order to get 3 simultneous equtions: 4 53, 2 2 2, 3 5 Solve to find nd. In this exmple 2 nd. To find the point where the lines cross, substitute the vlue of or into the eqution for one of the lines. Here, substituting into the first eqution gives (,5, 4) (4, 2,3) (,5, 4) 2(4, 2,3) (,5, 4) (8, 4, 6) (9,9,0) You should lwys check this result with the other eqution to ensure it is correct. 5
Intersection of two plnes Given the normls ( n nd n 2 ) to the two plnes, the line of intersection of the two plnes cn be found from: r c n where n n n2 nd c is point common to both plnes. If you re given the Crtesin forms of the plnes, then you cn find the norml to the plne by tking the coefficients of x, y, nd z. For exmple, the norml to x 3y 4z 24 is (, 3, 4). Exmple Find the eqution of the line of intersection of the plnes: r ( 2, 4, ) 4, r (, 2, 5) 5 2 Here we hve n ( 2, 4, ) nd n2 (, 2, 5) iˆ ˆj kˆ ( ) ( ) ˆ ( ) ( ) n n2 2 4 4 5 2 i 2 5 ˆj 2 5 ( 2 2) ( 4 ) kˆ 22 iˆ 9 ˆ j 8kˆ so n ( 22, 9, 8) Now we need point common to both plnes. The simplest wy to do this is to try setting z 0. This mens tht we only need del with 2 equtions in 2 unknowns. To find the common point, set z 0 in the Crtesin equtions of the plnes. This gives 2x4y 4 nd x2y 5. Solving these simultneous equtions gives x 6, 2 y. Hence c ( 6,, 0) nd the eqution of the line of intersection is: 2 r ( 6,, 0) ( 22, 9, 8). 2 Exercise 7 Find the eqution of the line of intersection for the plnes: ) r ( 6, 3, 4) 2, r 2 ( 2, 3, ) 4 b) 2x 4y 3z 4, x 2y 4z 3 6
Answers to Exercises Exercise ) ˆ (,6,2) b) ˆ b (3, 2,) c) cˆ (5,9, 3) 4 4 5 Exercise 2 ) (-8, 6) or (8, -6) b) (2,23) or (-2, -23) c) (-356, 97) or (356, -97) 2 ) (3, -30, 2) or (-3, 30, -2) b) (3, -, -4) or (-3,, 4) c) (-24, 6, -44) or (24, -6, 44) Exercise 3 ) (3,6) 3(,0) 6(0,) b) (3,6) 6(2,) 9(,0) Exercise 4 Q. r (2,6) (5,2) equivlently r (2 5,6 2 ) Q2. r (3,7) (,6) equivlently r (3,7 6 ) Exercise 5 Q. r (4, 2) (,) equivlently r (4, 2 ) Q2. r (,6) (7, 9) equivlently r ( 7,6 9 ) Exercise 6 ) r ( 2, 4, 6) 24, 2x 4y 6z 24 b) r (, 6, 2), x 6y 2z Exercise 7 ) r (, 2, 0) ( 9, 4, 24) b) r ( 5,, 0) ( 0,, 8) These exercises were tken from Mthemtics Worksheets originlly creted by Study Advice Service t Hull University. Web: www.hull.c.uk/studydvice Emil: studydvice@hull.c.uk Mny thnks for use of these mterils. Any comments cn be sent to the emil bove or to mthsskills@strth.c.uk 7