TMJC H2 Chp5 Vectors Assignment Suggested Solutions 2024

Chapter 5 Vectors TMJC 2024 Page 1 of 4 H2 Mathematics (9758) Chapter 5 Vectors Assignment Suggested Solutions 1 2019/TMJC JC1 Promo/Q5 Referred to the origin O, points A, B and C have position vectors 2=− + −a i j k , 33b j k=− and 44c i j k= + − respectively. (i) Show that A, B and C are collinear. [2] (ii) Point D lies on AB such that AD : DB = 1 : 2. Find the position vector of D. [2] (iii) Find the area of triangle OAB. Hence write down the area of triangle OBD. [4] (iv) Evaluate ˆca and give a geometrical interpretation of ˆca . [3] Q1 Solution (i) 1 2 3 1 4 1 3 3 1 4 1 3 1 AC −               = − = =              − − − −        0 2 2 1 3 1 2 2 1 3 1 2 1 AB −               = − = =              − − − −        Since 12 333 1 2 2212 AC AB        = = =      −−    , hence A, B and C are collinear. (ii) Using ratio theorem and OAB , 2 3 20 1 2 1 33 13 4 1 53 5 OA OBOD +=  −       =+        −−    − = − O A D B 1 2 Steps 1) Form any 2 vectors using the 3 points 2) Show that the 2 vectors are parallel 3) Write the conclusion Recommend to show that the 2 vectors are parallel like this, where you can see the relationship properly i.e. express AC as 2 2 2   − and balancing the equation, then replace it with AB . This ensures you get the correct relationship between the 2 vectors.

Chapter 5 Vectors TMJC 2024 Page 2 of 4 (iii) 1Area of triangle 2 20 1 132 13 0 1 62 6 0 3 1 3 2 1 OAB= −       =       −−     =− −  = − =− ab Note: Triangle OAB and triangle OBD share common height Area of triangle 2 2OBD= (iv) 12 1ˆ 41 641 1= 2 4 4 6 =6 −       =       −−    − + + ca ˆca is the length of projection of c onto a. 2 2019/ASRJC JC1 Promo/Q7 (Modified) Referred to the origin O, a, b and c are non-zero and non -parallel vectors denoting the position vectors of the points A, B and C respectively. (i) Given that 3 = a b a c and 3bc , show that 3 −=b c a where  is a scalar. [2] The point M is the mid-point of OC and the point N lies on OB produced such that 3ON = 5OB. The point P lies on MN such that MP: MN = :1 . It is given that the position vector of P is 10 1 96 +bc . (ii) Find the value of  . [3] It is given that b is a unit vector, 2=c , 3 13−=bc and the angle between b and c is 45 . (iii) Find the exact length of projection of OP on OA . [4] O A D B 1 2 Common height

Chapter 5 Vectors TMJC 2024 Page 3 of 4 Q2 Solution (i) ( ) ( ) 3 3  =   −  = a b a c a b a c 0 ( )3 − =a b c 0 a is parallel to b – 3c, hence 3 −=b c a . (ii) By ratio theorem, ( ) ( ) 1 1 10 1 5 1 19 6 3 2 ON OMOP    +−= +− + = + − b c b c Since b and c are non-zero and non-parallel vectors, 10 5 93=  and ( )11 162=−  2 3 = (iii) length of projection OAOP OA = ( )20 3 18 += bc a a ( ) ( ) ( ) 120 3 311 From ,118 3 12 3 – 0 57 9 18    +  −  == −  −  − = − b c b c ab(i) bc b b c b c c b 3c c 22 20 57 91 18 3 −  −= − b c b c bc 20 18 57 cos 451 18 13 −−= cb 1 55 18 13 −= 55 13 234= Note: this is the zero vector, and not zero as cross product results in a vector. Make sure you write as 0 to symbolise that it is a vector

Chapter 5 Vectors TMJC 2024 Page 4 of 4 3 2019/ RIJC Promo/5 (i) Given a, b and c are unit vectors such that ab = 2ac, what is the relationship between a and b − 2c? [2] (ii) It is further given that b is perpendicular to b − 2c. Find the angle between b and c. [2] (iii) Comment on the relationship between a  b and b − 2c. [1] Q3 Solution (i) ( ) 2 2 = 0 20  =   −   − = a b a c a b a c a b c Therefore a is perpendicular to 2−bc since and 2a 0 b c . (ii) Let  be the angle between b and c. Given b is perpendicular to 2−bc , ( )20 − =b b c 2 2 2 cos 1cos (since and are unit vectors)2 60     =  = = b b b c b b c = b c (iii) From (i) and (ii), b – 2c is perpendicular to a and b. ab is perpendicular to a and b. Therefore, ab is parallel to b – 2c. • a0 as a is a unit vector (i.e. 1=a ) • 2bc as both b and c are unit vectors 1 and 1 2 2 = =  =b c c Since 22  b c b c .