RI Vectors C2A Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ____________________ Tutorial 2A: Vectors I Page 1 of 5 Tutorial 2A: Vectors I Vector Algebra, Ratio Theorem, Scalar and Vector Products Section A (Basic Questions) 1 (a) The vectors , ab and c are such that 2ab c 0  . If 4 ai j k and 2bi k , find a unit vector in the direction of c. (b) Find a vector a such that a is parallel to the vector 84ij k and is equal in magnitude to the vector 22ij k . (c) Find the position vector of P if OP is of length 5 units and is in the direction 24ij k . [(a) 2 1 1 30 5    (b) 8 1 13 4    or 8 1 13 4    (c) 2 5 1 21 4   ] Solution: (a) a 4 1 1   , b 1 0 2   Given that 2ab c 0  , we have 2cb a 14 2 20 1 1 21 5               Unit vector in direction of c 22 2 22 11ˆ 11 30(2 ) 1 (5 ) 55 cc c              [Note: The question states that the unit vector and c must be in the same direction] (b) Given a 88 8 / / 1 1 44 4                 a for some \0  . and  222 1 21 2 29 3 2 a       

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________ Tutorial 2A: Vectors I Page 2 of 5   22 2 8 4 3    2 2 81 9 1 9 1 3       Hence, a 8 1 13 4    OR : 22 2 88 8 13 1ˆ 31 1 1 381814 44 4                       aa a [Note: The question states that a and 84ij k are parallel to each other. i.e. they could be in the same direction or opposite direction. That is the reason for 2 possible answers. On the other hand, the question ask for a vector a, so you can just give any of the 2 possible answers.] (c) Given 5OP   and 2 1 4 OP      222 2 ( 2 )()( 4 ) 5 21 25 5 21          But since OP  is in the direction of 2 1 4   , 5 21   i.e. 2 5 1 21 4 OP     OR : 22 15unit vector in the direction of 5 1 1 4 1 16 21 44 OP OP OP                     

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________ Tutorial 2A: Vectors I Page 3 of 5 2 The points P and Q have position vectors 31 7 ij k and 892ijk respectively when referred from the origin O . (a) Evaluate OP OQ     and OP OQ    . (b) Find the position vector of the point R if (i) R is the mid-point of PQ ; (ii) R lies on the line segment PQ such that :2 : 3PR RQ  ; (iii) R lies on PQ produced so that :3 : 2PR QR  . [(a) 155 1, 142 19    (b)(i) 5 1 82 19    (ii) 75 3 11    (iii) 30 29 28    ] Solution: (a) 3 1 17 OP    and 8 9 2 OQ     Dot/Scalar Product 38 19 2 4 9 3 4 1 17 2 OP OQ                       Cross/Vector Product  38 2 1 5 3 1 5 5 1 9 6 136 142 17 2 27 8 19 OP OQ                                              Note : If ab c , then and ca cb . Hence, after performing a cross product, you can check your answer by checking that 155 3 155 8 142 1 ... 0 & 142 9 ... 0 19 17 19 2                                  (b) 3 1 17 OP    and 8 9 2 OQ    

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________ Tutorial 2A: Vectors I Page 4 of 5 (b) (i) R is the mid-point of PQ ; By Ratio Theorem, 2 5 1 8 2 19 OP OQOR         (b) (ii) R lies on the line segment PQ such that :2 : 3PR RQ  . By Ratio Theorem, 7 5 32 5 3 11 OP OQOR       (b) (iii) R lies on PQ produced so that :3 : 2PR QR  . By Ratio Theorem, 2 3 32 30 29 28 OP OROQ OR OQ OP              P Q R 1 1 P Q R 2 3 O P R Q 1 2 O 3

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ __________________ Tutorial 2A: Vectors I Page 5 of 5 3 (a) Find the value of p so that the vectors 2i j k and i j kp p are perpendicular. (b) If 2 ai j k and b j k , find a unit vector perpendicular to both a and b . [(a) 1 3p (b) 3 1 1 11 1   or 3 1 1 11 1    ] Solution: (a) Given that 2 1 1    and 1 p p    are perpendicular, 2 110 1 21 0 1 3 p p pp p           (b) a 1 2 1   , 0 1 1   b Using cross product : A vector perpendicular to both a and b 10 3 21 1 11 1                ab So, a unit vector perpendicular to both a and b is 3 1 1 11 1    or 3 1 1 11 1   [Note: the cross product of 2 vectors a and b will give a vector that is perpendicular to the 2 vectors a and b]