TMJC H2 Chp 5 Vectors Discussion Solutions 2024

Chapter 5 Vectors TMJC 2024 Page 1 of 25 H2 Mathematics (9758) Chapter 5 Vectors Discussion Solutions Level 1 1 (a) OEFG is a parallelogram as shown in the diagram. (i) Find OF and EG . (ii) Find the size of the angle GEO. (b) The points P, Q and R have coordinates ( )4, 1, 1 , ( )8, 5, 15−− and ( )7, 0, 5 respectively. Show that P, Q, and R are collinear. (c) Given =+−a i j k and 3=−b j k , find a vector perpendicular to both a and b. Q1 Solution (a)(i) 32 01 45 1 1 9 OF OE EF OE OG =+ =+ −       =+           = 23 10 54 5 1 1 EG OG OE=− −       =−          − = (a)(ii) 53 1 and 0 14 EG EO OE −−       = =− =       −    ( ) 53 1.0 14 11cos 27 25 15 3 65.0 1 d.p EG EOGEO EG EO GEO −−              −    = = =  =   To find GEO , use vectors that are both either pointing ‘outwards’ from point E (vectors EG and EO ) or pointing ‘inwards’ from point E (vectors GE and OE ) E (3, 0, 4) O G (−2, 1, 5) F E G O Using the parallelogram, OF OE EF=+ and EF OG= Recall: EG OG OE=− Use position vectors OG and OE to find EG

Chapter 5 Vectors TMJC 2024 Page 2 of 25 (b) 8 4 12 5 1 4 15 1 16 PQ OQ OP −−           = − = − =           −−      7 4 3 0 1 1 5 1 4 PR OR OP            = − = − = −                12 3 4 4 1 4 16 4 PQ PR −       = =− − =−      −    Since 4,PQ PR=− P, Q, R are collinear. (Other answers such as 4 5PQ QR=− is also accepted) (c) A vector perpendicular to both a and b : 1 0 2 1 1 3 1 3 1 −            =  =          −−      ab 2 3 is also accepted 1     = −  − ba Recommended to work out the constant like this to ensure that you have the correct relationship between the vectors

Chapter 5 Vectors TMJC 2024 Page 3 of 25 2 2010(9740)/I/1 The position vectors a and b are given by 2 3 6p p p= + +a i j k and 22= − +b i j k , where 0p . It is given that =ab . (i) Find the exact value of p. [2] (ii) Show that ( ) ( ) 0+  − =a b a b . [3] Q2 Suggested Solutions (i) Since =ab , ( ) ( ) ( ) 2 2 2 222 2 2 2 2 2 3 6 1 2 2 4 9 36 9 9 49 3 7 p p p p p p p p + + = + + + + = = = Since 0p , 3 7p= (ii) Method 1: ( ) ( ) ( ) 22 22 0 +  − = −  +  − = − = = a b a b a b a b a b a a b a Method 2: ( ) ( ) 2 1 2 1 33 3 2 3 277 6 2 6 2 13 / 7 1/ 7 5 / 7 23 / 7 32 / 7 4 / 7 13 115 128 049 49 49                      +  − = + − − −                               −       =−          =− − + = a b a b Note that there are 2 possible values of p when you solve . Always reject with reason Make sure the dot for the dot product is visible in your working. Properties to note here: 1. 2.

Chapter 5 Vectors TMJC 2024 Page 4 of 25 3 2009 MJC Promo The points A, B and C have position vectors 23 1, 57 y                   and 5 10 11    with respect to the origin O respectively. (i) Find the value of y such that A, B and C are collinear. [2] (ii) Find the exact area of triangle OAC. [3] (iii) The point D divides OC internally such that OD : OC = 2 : 5. Find the vector AD . [3] Q3 Solution (i) AB AC= for some  . (or other valid combinations) 13 119 326 y         − =  =          4y= (ii) 2 2 2 1Area of triangle 2 11 50 1 25 222 20 5 1 39 3 152 13 1755 19522 OAC OA OC= − =−  − = + + == (iii) OD : OC = 2 : 5 2 5OD OC= 2 5 0 3 35 AD OD OA OC OA =− =−  =− units2 O C D | | | 5 2

Chapter 5 Vectors TMJC 2024 Page 5 of 25 Alternative Method: OD : DC = 2 : 3 23By ratio theorem, 5 66 1 18 35 12 15 0 3 35 AC AOAD +=  −        = + −       −     =− C O D | | | 3 2 A

Chapter 5 Vectors TMJC 2024 Page 6 of 25 Level 2 4 The position vectors of the points A, B and C are 8OA=− −jk , 5OB=+ik and (2 11) 4OC p p= + − −i j k respectively. Find the (i) unit vector(s) parallel to the vector AB , (ii) position vector of the midpoint of AB, (iii) value of p such that A, B and C are collinear, (iv) position vector of D such that ABCD is a parallelogram and 6p= . (v) position vector of E such that point E is on AB produced and : 2 : 5AB AE = . Q4 Solution (i) 1 8 6 1 1 1 1 101 101Unit vectors required 8 8 or 8 101 101101 6 6 6 AB OB OA  = − =             = = −                 (ii) ( ) Position vector of midpoint of 1 1/ 2 11 8422 42 AB OA OB        = + = − = −           (iii) ( ) Since , and are collinear, there exists such that 1 2 3 8 3 6 A B C AC AB OC OA OB OA p p      =  − = −         − =      −    Recall that: AB OB OA=− Good practice to express AB with reference to position vectors Learning Point: Points A, B, and C are collinear  AB AC= , for some ,0 . Good practice to express AC and AB with reference to position vectors

Chapter 5 Vectors TMJC 2024 Page 7 of 25 - (1) 2 3 8 - (2) 3 6 - (3) 11Using (1) and (3), and 22 Checking (2), 1LHS 2 3 4 2 1RHS 8 4 2 LHS = RHS 1 and 2 p p p p      =  − =  −= =− =− = − − =− = − =−  =− = 1 .2− Alternatively, 0 - (1) 2 3 8 2 8 3 - ( 2) 3 6 6 3 - (3) pp pp    = − =  − =  − =  − = =− Using GC, 11 and .22 p =− =− (iv) Since ABCD is a parallelogram, AD BC= 0 6 1 5 8 1 0 7 1 4 5 10 OD OA OC OB OD OA OC OB  − = −  = + −                = − + − = −              − − −        Alternatively, we can also use AB DC= Verify whether the value of p and  satisfy all three equations. A B C D Label the vertices in clockwise or anti-clockwise direction Good practice to express AD and BC with reference to position vectors

Chapter 5 Vectors TMJC 2024 Page 8 of 25 (v) Given : 2 : 5AB AE = By ratio theorem, 32 5 53 2 10 1 5 0 3 82 51 2.5 12 14 OB OB OAOE OOA E+= −=        = − −       −     = A E B | | | 2 3 O Learning Point: Draw diagram to clearly show the ratio that divides AE internally before applying ratio theorem. Useful GC Tip: Use F3 to create 3 rows and 1 column to represent column vector to compute tedious operations.

Chapter 5 Vectors TMJC 2024 Page 9 of 25 5 2009(9740)/II/2 Relative to the origin O, two points A and B have position vectors given by 14 14 14=++a i j k and 11 13 2= − +b i j k respectively. (i) The point P divides the line AB in the ratio 2 :1 . Find the coordinates of P. [2] (ii) Show that AB and OP are perpendicular. [2] (iii) The vector c is a unit vector in the direction of OP . Write c as a column vector, and give the geometrical meaning of ac . [2] (iv) Find ap , where p is the vector OP , and give the geometrical meaning of .ap Hence write down the area of triangle OAP. [4] Q5 Suggested Solutions (i) By Ratio Theorem, 2 3 14 11 14 2 13 14 2 3 12 4 6 OP A BO O+=        +−         =  =− Coordinates of P are (12, −4, 6). A 2 P 1 B O Learning Point: Draw diagram to clearly show the ratio that divides AB internally before