# STRAND J: TRANSFORMATIONS, VECTORS and MATRICES

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1 Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry

2 Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors re used in Unit 36 to descrie trnsltions. The digrm shows the trnsltion of 4 tringle y the vector. imge oject 4 4 Note tht the vector specifies how fr the tringle is to e moved nd the direction, i.e. 4 units horizontlly (to the right) nd units verticlly (up). ll vectors hve length (or size) nd direction. Quntities which do not hve direction, ut only length or size re known s sclr quntities. Quntities like mss, length, re nd speed re sclrs ecuse they hve size only, while quntities like force nd velocity re vectors ecuse they hve direction s well s size. The two points nd re shown in the digrm. The displcement (chnge of position) of from is vector ecuse it hs length nd direction. We cn write this displcement s 3 = or lel the vector nd write = 3 or = nd in this formt, it is clled column vector. The nottion is used when is vector nd the nottion is used when is sclr. 1

3 J3.1 Mthemtics SKE: STRN J The length of vector is clled its mgnitude or modulus: we write this s. If = x y, then, using Pythgors' Theorem, = x + y 3 So for the vector =, = 3 + ( ) = = 13 Vectors cn simply e dded nd sutrcted. x y onsider which cn e represented s shown in the following digrm So, from the digrm, the ddition of these two vectors cn e written s single vector 7, which is just the ddition of ech component of the originl vector. In generl, c c + d = + + d similr result is true for sutrction, c c d = d vector cn e multiplied y sclr, i.e. numer, y multiplying ech component y tht sclr. 3 1 For exmple, 4 =. 8

4 J3.1 Mthemtics SKE: STRN J k In generl, k = k Worked Exmple 1 Write ech of the following vectors in the form. () Solution From the digrm: = 7 () = = 5 Note In the Worked Exmple ove, we see tht + = This is lwys true so tht, for exmple, O + = O or + = Similrly, = O = s the direction is the opposite of. 3

5 J3.1 Mthemtics SKE: STRN J Worked Exmple If = 3 7, = 4 nd c = 3 4 find: () + c (d) + 3 Solution 3 = 7 = 3 = () + c = 3 + = 4 4 ( ) ( ) = 1 8 = 3 7 = = ( ) (d) + 3 = = = ( ) Note When the vector strts t the origin, it is clled position vector. = = ( ) 1 Worked Exmple 3 O () nd O 1, ( ) re position vectors reltive to the origin, O. Given the points (3, 1) nd write down O express nd O s column vector s column vectors clculte the length of. Solution O = 3 1, O = 1 () s O + = O, = O O = 1 3 = This cn e seen in the digrm opposite. 3 1 O O -1-4

6 J3.1 Mthemtics SKE: STRN J O = 4 3 ( ) + ( ) = = 5 = 5 Exercises 1. Which of the following re vectors nd which re sclrs: Time () Velocity Speed (d) Force (e) istnce (f) Temperture. Use the points in the grid elow to write the vectors given in column vector form. () E E (d) E (e) E (f) (g) (h) Wht is the reltionship etween nd? 3. Plot the positions of the points,,,, E nd F reltive to point O if: O = 3 () O 5 = = 3 (d) = (e) E = 4 1 (f) F = 1 4 Write the vector EF s column vector. 4. If = 4 7, = 3 5 nd c = 0 4, find: + () +c +c (d) (e) (f) c (g) 3 (h) (i) 4c (j) + 3 (k) 5c 3 (l) 4 c 5

7 J3.1 Mthemtics SKE: STRN J 5. If = 1, = 4 nd c = 1 4, find: 3+ () 4+ 3c 6 3 (d) 4c + (e) 3c (f) 3 + 5c 6. If = 1, = 3 column vector x. 4 nd c =, solve the equtions elow to find the 3 + x = () x c = x + = c (d) x + = (e) 3 + x = c (f) 4 x = c (g) 3 + x = 4 (h) x = 4 c (i) 3 + x = c 7. In this question = 4, = 3 nd c = 1 1. For ech prt drw the vectors listed on seprte digrms.,, + (), c, c,, 3 (d) c, c, c (e),, 3 8. The vertices of qudrilterl, OPQR re (0, 0), (4, ), (6, 10) nd (, 8) respectively. x Using vector method, express in the form y, the vector (i) OP (ii) RQ () lculte OP, the mgnitude of OP. Stte TWO geometricl reltionships etween the line segments OP nd RQ. 9. Given tht PR = nd PS = 3, + 1 express EH of the vectors RP nd RS in the simplest form () determine the vlues of if PR = 0 units. 6

8 Mthemtics SKE: STRN J J3. Vectors nd Geometry Vectors cn e used to solve prolems in geometry. In two dimensions, it is possile to descrie the position of ny point using two vectors. For exmple, using the vectors nd shown in the digrm: I J K L E F G H EF IK F L LE = = = = + = 3 + = 3 Note tht, EF nd IK re ll prllel vectors s they hve the sme direction. Worked Exmple 1 In the prllelogrm shown elow, nd Y lies on such tht Y = Y. = nd = d. lso X is the midpoint of Y d X Express the following vectors in terms of nd d. (d) () X X Y (e) Y (f) XY Solution = + = + d () X = 1 = 1 d 7

9 J3. Mthemtics SKE: STRN J (d) X = + X Y = = + 1 d 3 = 3 (e) (f) Y = + Y = d + 3 XY = X + Y = + 1 d + d = + d d = + d 3 Worked Exmple The digrm shows identicl prllelogrms. The vector q = F nd the vector p =. The point M is the midpoint of. F E q M p () Show tht M is prllel to. Show tht EM is prllel to F. Solution M = + M = p + 1 q 8

10 J3. Mthemtics SKE: STRN J = + = p + q s = M, the lines nd M must e prllel. () EM = E + M = + 1 p q = p 1 q F = F + = p q s F = EM, the lines F nd EM must e prllel. Worked Exmple 3 3y P 3x In the figure ove, not drwn to scle, is prllelogrm such tht nd = 3y. The point P is on such tht P : P = 1. : Express in terms of x nd y. = 3x (i) (ii) (iii) P () (d) Show tht P = x y. Given tht E is the mid-point of, prove tht, P nd E re colliner. Given tht x = 0 nd y = 1 1, use vector method to prove tht tringle E is isosceles. 9

11 J3. Mthemtics SKE: STRN J Solution (i) = = 3x (ii) = + = x y = x + y 3 3 3( ) (iii) P = = = ( 3) ( x + y) = x + y () P = + P = 3y + x + y x y ( ) = E = x. so PE = P + E = x y + x = x y Hence PE = x y = P tht is PE nd P hve the sme direction. So, P nd E re colliner. (d) We need to find two sides of equl length. Now nd E 3 = x = 3 = E = = = 9 = 3 3 E = + E = 3y + x = = = 3 So ( ) = + = = E = s E = E = 3, tringle E is isosceles. 10

12 J3. Mthemtics SKE: STRN J Exercises 1. The digrm shows grid mde up of sets of eqully spced prllel lines. The vectors O = nd OE = re shown on the grid. P Q R S T J K L M N E F G H I O Write ech of the following vectors in terms of nd. O () G (d) IS (e) JP (f) ES (g) Q (h) S (i) PK (j) PG (k) RF (l) SE (m) P (n) GE (o) IJ (p) T. The digrm shows grid mde up of two sets of prllel lines. The vectors O = nd OE = re shown on the grid. O opy the grid nd use the following informtion to lel ech point where lines meet. O = O = + E = EF = G = + H = + 3 HI = IJ = JK = 3 + KL = OM = MN = P = Q = + R =

13 J3. Mthemtics SKE: STRN J 3. The digrm shows the prllelogrm O, in which O = nd O = c. O c Write the following vectors in terms of nd c. (i) (iv) (ii) (iii) (v) O (vi) () If X is the midpoint of nd Y is the midpoint of, find the following in terms of nd c. (i) X (ii) OX (iii) Y (iv) OY (v) XY 4. The digrm shows the prllelogrm OPQR; the vectors p = OP nd q = OR. P p O q Q If M is the midpoint of PR, find OM. R () If N is the midpoint of OQ, find ON omment on your nswers to nd (). 1

14 J3. Mthemtics SKE: STRN J 5. The digrm shows the rectngle O. The vectors i nd j re such tht O = 6j nd O = 8i. O If the point lies on such tht = 3, find of i nd j. nd O in terms () If E lies on such tht E = E, find E nd OE. The point M is the midpoint of E. Find OM. 6. In the digrm, EF is prllelogrm nd = =. The vector p = F nd q =. The point M is the midpoint of F nd N is the midpoint of E. F E M p N q Express M nd N in terms of p nd q. () Find MN in terms of p nd q nd explin why MN is prllel to. 7. In the tringle, = nd = c. Use vectors to show tht: line joining the midpoint of nd is prllel to, () line joining the midpoint of nd is prllel to. c 13

15 J3. Mthemtics SKE: STRN J 8. The shpe in the digrm shows the vectors, nd c. M is the midpoint of E. c Find ech of the following in terms of, nd c. M E (d) E () M M (e) M 9. is prllelogrm in which nd = d. = H F = 4F G G = 4G H = 4H d E = 4E Show tht EFGH is lso prllelogrm. E F 10. In the prllelogrm, = nd = d. lso E = E. d Find E nd explin why F = d + α 1 d for 3 some vlues of α. E () Find nd explin why F = + d β d for some vlue of β. Hence find the vlues of α nd β. F 14

16 J3. Mthemtics SKE: STRN J 11. In the prllelogrm, = d nd of. The lines nd X intersect t Q. =. The point X is the midpoint X Find Q in terms of nd. 1. O is prllelogrm. O = 3p q O = 5p + 6q O Find. Express your nswer s simply s possile in terms of p nd q. () is the point where = p + 6q. Using vector methods, show tht lies on the line produced. 13. O is prllelogrm. O =, O = c. E is stright line, E = 3. is the midpoint of O. O Write in terms of nd c, (i), () (ii) OE. educe the rtio of the lengths of nd OE. E 15

17 J3. Mthemtics SKE: STRN J 14. KLMN is prllelogrm with position vectors OK = 3, OL 6 = 3 nd ON = 3 5. Use vector method to determine the position vector OM. () The point H lies on KM such tht KH = HM. Find KH nd LH. Use vector method to show tht H is lso the midpoint of LN. 15. X is qudrilterl, not drwn to scle, with =, =, nd =. The point X divides in the rtio 3 :. Express nd X in terms of nd. () Show tht X 1 = ( )

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