SL Vector Practice The points P( 2, 4), Q (3, 1) and R (1, 6) are shown in the diagram below.

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1 IB Math Standard Level Vector Practice 0506 SL Vector Practice The points P(, ), Q (, ) and R (, 6) are shown in the diagram below. Alei - Desert Academy (a) Find the vector PQ. (b) Find a vector equation for the line through R parallel to the line (PQ). (Total 6 marks). The position vector of point A is i + j + k and the position vector of point B is i 5 j + k. (a) (i) Show that AB = i 8 j + 0k. (ii) Find the unit vector u in the direction of AB. (iii) Show that u is perpendicular to OA. Let S be the midpoint of [AB]. The line L passes through S and is parallel to OA. (b) (i) Find the position vector of S. (ii) Write down the equation of L. The line L has equation r = (5i +0 j +0k) + s ( i + 5 j k). (c) Explain why L and L are not parallel. (d) The lines L and L intersect at the point P. Find the position vector of P.. The line L passes through the points A (,, ) and B (, 5, ). (a) Find the vector AB. (b) Write down a vector equation of the line L in the form r = a + tb. Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of (6) () () (7) (Total marks) (Total 6 marks)

2 IB Math Standard Level Vector Practice 0506 Alei - Desert Academy. The line L passes through A (0, ) and B (, 0). The origin is at O. The point R (x, x) is on L, and (OR) is perpendicular to L. (a) Write down the vectors AB and OR. (b) Use the scalar product to find the coordinates of R. (Total 6 marks) 0 5. In this question the vector represents a displacement of km east, and the vector 0 represents a displacement of km north. The diagram below shows the positions of towns A, B and C in relation to an airport O, which is at the point (0, 0). An aircraft flies over the three towns at a constant speed of 50 km h. y B O x A Town A is 600 km west and 00 km south of the airport. Town B is 00 km east and 00 km north of the airport. Town C is 00 km east and 50 km south of the airport. (a) (i) Find AB. (ii) 0.8 Show that the vector of length one unit in the direction of AB is. 0.6 An aircraft flies over town A at :00, heading towards town B at 50 km h. p Let be the velocity vector of the aircraft. Let t be the number of hours in flight after :00. The q position of the aircraft can be given by the vector equation x 600 p = + t. y 00 q (b) (i) 00 Show that the velocity vector is. 50 (ii) Find the position of the aircraft at :00. (iii) At what time is the aircraft flying over town B? (6) Over town B the aircraft changes direction so it now flies towards town C. It takes five hours to travel the 50 km between B and C. Over town A the pilot noted that she had litres of fuel left. The aircraft uses 800 litres of fuel per hour when travelling at 50 km h. When the fuel gets below 000 litres a warning light comes on. (c) How far from town C will the aircraft be when the warning light comes on? (7) (Total 7 marks) C () Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of

3 IB Math Standard Level Vector Practice 0506 Alei - Desert Academy 6. A boat B moves with constant velocity along a straight line. Its velocity vector is given by v =. At time t = 0 it is at the point (, ). (a) Find the magnitude of v. (b) Find the coordinates of B when t =. (c) Write down a vector equation representing the position of B, giving your answer in the form r = a + tb. Working: (c)... (Total 6 marks) 7. Consider the point D with coordinates (, 5), and the point E, with coordinates (, ). (a) Find DE. (b) Find DE. (c) The point D is the centre of a circle and E is on the circumference as shown in the following diagram. The point G is also on the circumference. DE is perpendicular to DG. Find the possible coordinates of G. (8) (Total marks) 8. Car moves in a straight line, starting at point A (0, ). Its x 0 5 position p seconds after it starts is given by = + p. y (a) Find the position vector of the car after seconds. () Car moves in a straight line starting at point B (, 0). Its x position q seconds after it starts is given by = + q. y 0 Cars and collide at point P. (b) (i) Find the value of p and the value of q when the collision occurs. (ii) Find the coordinates of P. (6) (Total 8 marks). Find the cosine of the angle between the two vectors and. Working: Answers: (a)... (b)... Answer: (Total 6 marks) Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of () ()

4 IB Math Standard Level Vector Practice 0506 Alei - Desert Academy 0. The following diagram shows a solid figure ABCDEFGH. Each of the six faces is a parallelogram. The coordinates of A and B are A (7,, 5), B(7,, 5). (a) Find (i) (ii) The following information is given. 6 AD = 6, AD =, AE =, AE = 6 (b) (i) Calculate AD AE. (ii) Calculate AB AD. (iii) Calculate AB AE. (iv) Hence, write down the size of the angle between any two intersecting edges. (c) AB; AB. Calculate the volume of the solid ABCDEFGH. (d) The coordinates of G are (,, ). Find the coordinates of H. (e) The lines (AG) and (HB) intersect at the point P. Given that AG = 7, find the acute angle at P. 7 () (5) () () (5) (Total marks) Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of

5 SL Vector Practice 0506 MarkScheme Alei - Desert Academy 0-. (a) 5 PQ = AA N (b) Using r = a + tb x + 5 t y 6 AAAN. (a) AB= OB OA = 5 AB= A N (b) Using r = a + tb x x y= + t or y = 5 + t AAAN z z. (a) (i) Evidence of subtracting all three components in the correct order M eg AB= OB OA = = i 8j + 0k AG N0 (ii) = = 68 = 6 = 7 =.6 u = ( i 8 j + 0k) A N = i j + k,0.05i 0.70 j + 0.5k, etc (iii) If the scalar product is zero, the vectors are perpendicular. R Note: Award R for stating the relationship between the scalar product and perpendicularity, seen anywhere in the solution. Finding an appropriate scalar product u OA or AB OA M 8 0 eg u OA = u OA = 0 or AB OA = 0 (b) (i) EITHER ( i 5 j + k ) ( i + j + k) AB ( ) ( ) + 0 = 68 ( 8) + 0 AB OA = + A N0 [6] [6] Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of 6

6 Alei - Desert Academy 0- Therefore, OS = i j + k (accept (,, )) A N OR OS = OA + AB = (i + j + k) + (i + 8j + 0k) OS = i j + k A N (ii) L : r = (i j + k) + t (i + j + k) A N (c) Using direction vectors (eg i + j + k and i + 5j k) Valid explanation of why L is not parallel to L R N eg. Direction vectors are not scalar multiples of each other. Angle between the direction vectors is not zero or 80. Finding the angle d d d d. Note: Award R0 for direction vectors are not equal. (d) Setting up any two of the three equations For each correct equation AA eg + t = 5 s, + t = 0 + 5s, + t = 0 s Attempt to solve these equations Finding one correct parameter (s =, t = ) P has position vector 7i + 5j + k A N Notes: Award A if the same parameter is used for both lines in the initial correct equations. Award no further marks. x. (a) AB=, OR = AA N x (b) AB R is S,, ( ) OR = x x ( 0x 0) AB OR = 0 = 0, 0 Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of 6 A M AA N 5. (a) (i) AB = = 600 (N) (ii) AB= = 000 (must be seen) 800 unit vector = = 0.6 (AG)(N0) Note: A reverse method is not acceptable in show that questions. [] [6]

7 Alei - Desert Academy 0- (b) (i) 0.8 v = = 50 (AG)(N0) Note: A correct alternative method is using the given vector equation with t =. (ii) at :00, t = x y = = 50 (N) (iii) AB= Time = = (hours) 50 over town B at 6:00 ( pm, :00 pm) (Do not accept 6 or :00 or ) (N) 6 (c) Note: There are a variety of approaches. The table shows some of them, with the mark allocation. Use discretion, following this allocation as closely as possible. Distance from A to B to Time for A to B to C Fuel used from A to B C = hours = 800 = 700 litres = 50 km Light goes on after 6000 litres Light goes on after 6000 litres Fuel remaining = 800 litres Time for litres 6000 = = 8 ( = 8.88) Time remaining is = ( = 0.) hour Distance = 50 = 7.8 km Distance on 6000 litres 6000 = = ( =.) km Distance to C = 50. = 7.8 km Hours before light = ( =.88) Time remaining is = ( = 0.) hour Distance = 50 = 7.8 km (A ) (A) (N) 7 6. (a) 6+ = 5 = 5 (C) (b) = 6 7 (so B is (6, 7) ) (C) (c) r = t (not unique) (A)(C) Note: Award if r = is omitted, ie not an equation. 7. (a) DE = 8 = (N) 5 6 [7] [6] Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of 6

8 (b) DE = 6 = 6+ 6 = 0 (N) (c) Vector geometry approach Using DG = 0 (x ) + (y 5) = 00 Using (DG) perpendicular to (DE) 6 Leading to DG = 6, DG = 8 8 Using DG = DO + OG (O is the origin) G (, ), G (0, ) (accept position vectors) Algebraic approach 6 gradient of DE = 8 8 gradient of DG = 6 equation of line DG is y 5 = ( x ) Using DG = 0 (x ) + (y 5) = 00 Solving simultaneous equation G (, ), G (0, ) (accept position vectors) Note: Award full marks for an appropriately labelled diagram (eg showing that DG =0, displacements of 6 and 8), or an accurate diagram leading to the correct answers. Alei - Desert Academy 0-8. (a) 0 p = 5 0 = 6 (accept any other vector notation, including (0, 6) ) (N) (b) METHOD (i) equating components 0 + 5p = + q, p = 0 + q p =, q = (N)(N) (ii) The coordinates of P are (5, ) (accept x = 5, y = ) (N)(N) METHOD (i) Setting up Cartesian equations x = 5p x = + q y = p y = q giving x + 5y = 60 x y = Solving simultaneously gives x = 5, y = Substituting to find p and q ,, = + p = 0 + q p = q = (N)(N) (ii) From above, P is (5, ) (accept x = 5, y = seen above) (N)(N). METHOD 8 + ( ) Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page of 6 [] [8]

9 Using a b = ab cos θ (may be implied) cosθ = Correct value of scalar product = Correct magnitudes cosθ = 5 METHOD = 5 = 5 = Using cosine rule = cosθ cosθ = 5 Alei - Desert Academy 0- Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page 5 of 6 (C6) (C6) 0. (a) (i) AB = OB OA 7 7 = = 5 0 A N (b) (ii) 5 = 5 ( = 5 ), = 5 AB = ( ) + ( ) = = 5 A N Evidence of correct calculation of scalar product (may be in (i), (ii) or (iii)) A (i) AB AE = 0 (( 6)( ) + 6( ) + ()) A N (ii) AB AD = 0 ((0)( 6) + 5(6) + 0()) A N (iii) AB AE = 0 ((0)( ) + 5( ) + 0()) A N π (iv) 0 or A N (c) Volume = AB AD AE = 5 6 = 80 (cubic units) A N [6]

10 Alei - Desert Academy 0- (d) Setting up a valid equation involving H. There are many possibilities. x 0 eg OH = OG + GH,OH = OA + AE + EH, y = 5 z 0 Using equal vectors eg GH = AB,EH = AD OH = 5 =,OH = = 0 5 coordinates of H are (,, ) A N (e) 8 HB= A AG HB Attempting to use formula cos ˆP = AG HB = = A = ˆP = 7. 6 (=.5 radians) A N [] Macintosh HD:Users:Shared:Dropbox:IB Curriculum:SL:Vectors:Practice:SL.VectorPractice0506.docx on /5/7 at 7:5 AM Page 6 of 6