TMJC H2 Maths 9758 Chp 1 & 2 H2 MYE Revision Package Vectors

JC2 H2 MYE Revision Package: Vectors1 &2 TMJC 2020 Page 1 of 6 JC2 MYE Revision Package H2 Mathematics (9758) Vectors 1 2010Promo/RVHS/6 The position vectors of vertices A, B and C, relative to the origin O, are 34ik , 243  i j k and 11 4 9i j k respectively. (i) Find a unit vector parallel to OA . [1] (ii) A point P divides AC in the ratio 1 : 3. Find the position vector of P. [2] (iii) Show that the points O, B and P are collinear. [2] (iv) OBDC forms a parallelogram. Find the position vector of D. [2] 2 2015Promo/DHS/I/2 The position vectors of the points A and B relative to the origin O are 2 1 1   and 2 5 1   respectively. The point P lies between A and B such that AP AB where 01  . (i) Find the position vector of P in terms of . [1] (ii) If OP is perpendicular to AB , find the value of . [2] Given that 1 3  , (iii) Find the area of triangle OPA. [2] (iv) Write down the ratio of the area of triangle OPB to the area of triangle OPA. [1] 3 2013Promo/MJC/I/4 Referred to the origin O, the points A and B are such that OA a and , whereOB b a and b are non-zero and non-parallel vectors. The point C lies on OB such that ,OC kOB where k is a constant. P is on AC such that AP : PC = 3 : 1, and Q is on AB such that AQ : AB = 2 : 3. (i) Find OP and OQ in terms of a, b and k . [2] (ii) Given that O, P and Q are collinear, find the value of k . [3]

JC2 H2 MYE Revision Package: Vectors1 &2 TMJC 2020 Page 2 of 6 4 2015Promo/NYJC/1/4 Relative to the origin O, the position vectors of two points A and B are a and b respectively, where a and b are non-zero and non-parallel vectors. The vector a is a unit vector which is perpendicular to 25ab . The angle between a and b is 2 3  . (i) Show that 4 5b . [3] (ii) The point M divides AB in the ratio :1  . The point N is such that OMBN is a parallelogram. By considering ON in terms of a and b , find the area of triangle OAN in terms of  . [4] 5 RI Promo 9758/2017/2 Referred to the origin O, points A and B have position vectors a and b respectively, such that a and b are non-parallel vectors. Point C lies on line AB, such that the length of projection of OC onto OB is 5 units. Given that 2b and 1,ab find the possible position vectors of C in terms of a and b. [6] 6 2014 Promo/SAJC/5 (a) Relative to an origin O, the position vectors of A and B are a and b respectively, and c is the position vector of the point C on AB which divides AB in the ratio 3:1. Given that angle AOB is acute, show that the length d of the projection of OC on OB is given by 3. .44d  abb b [4] (b) Three vectors p, q and r are such that ,  p q p r p0 . Show that ,kq r p where .k [2] 7 2012/YJC/II/1 Two planes 1 and 2 have equations 22x y z   and 23x y z   respectively. The point A has coordinates (4, –1, 2). (i) Find the acute angle between the planes 1 and 2 . [2] (ii) Let B be the foot of the perpendicular from A to 1 . Find the coordinates of B. [3] (iii) If 3 contains the line AB and is perpendicular to 1 and 2 , find the Cartesian equation of 3 . [3]

JC2 H2 MYE Revision Package: Vectors1 &2 TMJC 2020 Page 3 of 6 8 2011/TPJC/I/11 The diagram shows a pyramid POABC. Taking unit vectors i, j, k as shown, the position vectors of A, B, C and P are given by 3OA i , 34OBij , 4OC j and 4OP k . Given that point E lies on PC such that PE : EC = λ : 1 – λ. (i) Write down the position vector of E in terms of λ. [1] (ii) It is given that PA is parallel to plane OEB. Show that E is the midpoint of PC. [4] (iii) The point D lies on AP and has position vector 3 22 ik . Find the coordinates of the foot of the perpendicular from D to plane OEB. [4] (iv) Hence deduce the distance between PA and the plane OEB. [4] 9 2010/SRJC/I/9 The position vectors of the points A, B, C and D are given as i + 3j, 2j + 4k, i + j + k and 4i + 5k respectively. (i) Find the vector equation of plane  in the form prn , that contains the points A, B and C. [3] (ii) Find the foot of the perpendicular of the point D to the plane  . Hence find the shortest distance from the point D to the plane  . [4] A line l parallel to the vector j + k passes through point D and it meets the plane  at the point N. (iii) Find the position vector of the point N and hence find the vector equation of the reflection of line l about the plane  . [5] 10 2010/SAJC/I/7 The equations of two planes 12π , π are given by 1 2 π : 2 4 8 π : 2 6 x y z xz     (i) Find the vector equation of the line of intersection l between the planes 12π and π . [2] A B C O P i j k

JC2 H2 MYE Revision Package: Vectors1 &2 TMJC 2020 Page 4 of 6 (ii) Find the foot of perpendicular, 1F from the point (6, 9, −2) to the plane 1π . [3] Another plane 3π contains the points 1F and 2F and is parallel to l. (iii) Given that 2 26 5 9 2 5 OF       , show that the Cartesian equation of the plane 3π is given by 15 8 40 22x y z   . [3] 11 2009/AJC/I/13 The points P and Q have position vectors i − j and 3 13 6i j k respectively. The plane 1π contains the point P and the line 1 , 02 x zy   . (i) Find a vector equation of the plane 1π in scalar product form. [3] (ii) Find the position vector of the foot of the perpendicular from Q to 1π . [2] The line l1 passes through the points P and Q. (iii) The line l2 is the reflection of the line l1 about the plane 1π . Find a vector equation of l2. [3] The plane 2π has the equation 6 4 a b    r . Find the values of a and b such that (iv) 12π and π are parallel and at a distance of 224 apart. [3] (v) 12π and π are intersecting. [1] 12 2009/RI/II/4 In this question, give each of your answers in exact form. The plane  have equation      2 3 3 2 2         r i j k where ,  and the point A has position vector 2j. (i) Express the equation of  in the form prn where n is a vector perpendicular to . [3] (ii) Find the position vector of the foot of perpendicular, B, from A to . Deduce the perpendicular distance from A to . [5] (iii) By using a vector product, find the length of projection of OA on . [2] (iv) By using (ii) and (iii), find the area of triangle OAB. [2]

JC2 H2 MYE Revision Package: Vectors1 &2 TMJC 2020 Page 5 of 6 Answer Key No Year JC/CI Answers 1 2010 Promo RVHS (i) 34 55 ik (ii) 13 24i j k (iv) 9 12ik 2 2015 Promo DHS (i) 24 1 6 ; 10 OP                     (ii) 7 ;26 (iii) 29 3 units2 or 1.80 units2; (iv) 2:1 3 2013 Promo MJC (i) 3 4 kOP  ab , 2 3OQ  ab (ii) 2 3k 4 2015 Promo NYJC  13 5  5 2017 Promo RI 23ab or 14 11 33 ab 7 2012 YJC (i) (ii) (iii) 7x y z   8 2011 TPJC (i) (iii) (iv) 9 2010 SRJC 10 2010 SAJC (i) (ii) (2, 1, 0) 3   11 5 5,,3 3 3   0 4 44       3 18 16,,34 17 17   6 34 17 7 (i) 1 10 2   r 14 6(ii) 9 28 3 13 3 4 (iii) ;ON      28 3 13 3 47 ': 8 7 l          r 62 1 1.25 , 01                    r =

JC2 H2 MYE Revision Package: Vectors1 &2 TMJC 2020 Page 6 of 6 11 2009 AJC (i) (ii) (iii) (iv) , (v) 2,ab 12 2009 RI (ii) , units (iii) units (iv) 1 32 2 r    1 1 2 ON   53 11 5 , 10 5 r                      2a 108 or -116b 2 1 5 6     r (i) 141 7541 42 OB    7 41 41 104 41 214 10 units41