T2W10 Vectors+I Summary

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 2 Revision (Summary) Topic: Vectors 1 (Vector Algebra, Ratio Theorem, Scalar and Vector Product) Summary for Vectors 1 Vector Algebra With reference to an origin O(0, 0, 0), given points  1 2 3, ,A a a a and  1 2 3, ,B b b b, we have the corresponding (position vectors) expressed in column form 1 2 3 a a a       a and 1 2 3 b b b       b . 1 1 1 1 2 2 2 2 3 3 3 3 a b a b a b a b a b a b a b                                and 1 1 2 2 3 3 a a ka k k a ka a ka                  with k a real number. The magnitude (or modulus) of a vector, a, is the non-negative number 1 2 2 2 2 1 2 3 3 a a a a a a          a . This value is equal to the distance from O to A. We say that ais parallel to b, denoted by a b, if and only if b a for some  \ 0, that is, b is a (non-zero) scalar multiple of a.  If 0 , then a and a are in the same direction.  If 0 , then a and a are in opposite directions. Points A and B have position vectors a and b respectively, relative to the origin O, such that b a for some  \ 0. We then say that the points O, A and B are collinear.

Summary on Vectors 1 Page 2 of 6 The unit vector in the direction of a denoted by ˆa is obtained by scaling a by 1 a , thus 1ˆa aa . The vectors 1 2 2       and 2 4 4       are parallel since 2 4 4       is a scalar multiple of 1 2 2       (k = –2) but are in opposite directions since k < 0. The points 2, 4, 4 , 0,0,0 and (1, 2, 2)   are also said to be collinear. The magnitude of 1 2 2       is   22 21 2 2 3    so the unit vector in the direction of 1 2 2       is 11 23 2       Let a and b be non-zero and non-parallel vectors:  If  a b for some ,  , then 0.    If s t  a b = a b for some , , , ,s t   then ,s t   . Note the importance of non-parallel vectors when comparing coefficients. Suppose 1 0 0       a = and 2 0 0       b = then 6 2 4 3  a b a b however we cannot “compare coefficients” of vectors a and b (note that a is parallel to b) as 6 4 and 2 3  .

Summary on Vectors 1 Page 3 of 6 Ratio Theorem Consider a triangle OAB with OAa  and OBb  . So a and b are non-zero and non-parallel vectors. Let P be a point which dividesAB in the ratio : ,  i.e. AP PB   . If OPp  , then       a bp (MF27) Note that the Ratio Theorem is an immediate consequence of the addition of vectors. From diagram,      p a b a , rearranging we have       a bp . Sometimes, it is easier to use      p a b a like the following example. Points , A B and Phave position vectors , a b and p respectively, relative to the origin O. Given that 2 2 5       a and 2 6 3       b , find p if P lies on AB produced such that 2 5 AB AP . Solution: Easier to find directly,  5 2  p a b a [DO NOT WRITE 2 5   b a p a . We cannot divide vectors] 2 2 2 252 6 2 1225 3 5 15                                p p

Summary on Vectors 1 Page 4 of 6 Scalar (Dot) Product Points A and B have position vectors a and b respectively, relative to the origin O. cos a b a b where AOB  . Re-arranging, cos  a b a b.  If vectors aand bare in the same direction then  a b a b, largest possible value.  If vectors aand bare in the opposite direction then  a b a b, smallest possible value. Two non-zero vectors a and b are perpendicular, denoted by a b, if and only if 0 a b . In particular, 0     i j j k k i as , i j and k are mutually perpendicular. If 1 2 3 a a a       a and 1 2 3 b b b       b , then 1 1 2 2 1 1 2 2 3 3 3 3 a b a b a b a b a b a b                      a b . Find the cosine of the angle between the vectors a 2 3 2  i j k and b 2  i j k and determine if the angle is acute or obtuse. Solution   22 2 2 2 2 2 3 2 2cos 2 3 2 1 2 1            i j k i j k 2 6 2 6cos 0 1717 6      is an obtuse angle Properties of Scalar Product (i)   a b b a (ii) ( ) ( ) ( )     a b c a b a c (iii) ( ) ( ) ( )      a b a b a b (iv) 2    a a a a a a note the relationship between the scalar product and the modulus Given that 2, 3, 1   a b a b , (i) 2  a b a b 2 2       a a a b b a b b   2 2 2 2 2 2(2 ) 1 3 0          a a b b (ii) 2 2  a b a b 4 2 2       a a a b b a b b 2 2 2 2 4 4(2 ) 3 7      a b (iii)     2 2 2 2 2 2 2 4 4 4 ... 2 11                    a b a b a b a a a b b a b b a a b b

Summary on Vectors 1 Page 5 of 6 Applications of Scalar Product Points A and B have position vectors a and b respectively, relative to the origin O. Given that P is the point on the line OB such that AP OB   , it can be shown that  ˆ ˆOP        a b b a b bb b   P is the foot of perpendicular from A to the line OB and P is also the point on the line OB nearest to A. Thus, the length of projection of vector a onto vector b is given by ˆOP a b  . Points A and Bhave position vectors a and b respectively, relative to the origin O. Given that a 2 3 2  i j k and b 2  i j k , find the length of projection of a onto vector b. Find the position vector of the foot of the perpendicular from A to OB. Solution   22 2 2 3 2 2 2 6 2ˆ 661 2 1 OP               i j k i j ka b  6 2ˆ ˆ 26 6OP           i j ka b b i j k  Vector (Cross) Product Let a and b be two non-zero vectors that are represented by OA  and OB  respectively. ˆsin a b a b n where AOB is the angle between a and b, and ˆn is the unit vector perpendicular to both a and b. It follows that two non-zero vectors a and b are perpendicular if and only if . a b a b In particular, , i j k  j k i and . k i j Two non-zero vectors a and b are parallel if and only if . a b 0 In particular, .     i i j j k k 0 Properties of Vector Product 1. ( ).  a b b a 2. sin .    a b b a a b a b 3. ( ) ( ) ( ).     a b c a b a c 4. ( ) ( ) ( ).      a b a b a b 5. ( ) ( ) 0     a b a a b b So  a b aand  a b b.

Summary on Vectors 1 Page 6 of 6 Examples of how the properties are used ( 2 ) (3 ) 3 6 2 6 7                    a b a b a a a b b a b b 0 b a b a 0 b a 2 2( ) ( ) Note that: 2                                    a b a b a a a b b a b b 0 b a b a 0 a a b a Given that    u v u v 0 , what can be deduced about the vectors u and v? Solution:  2                 u v u v 0 u u u v v u v v = 0 0 + v u 0 0 v u 0 Then, u 0 or v 0 or u v Let 1 2 3( , , )A a a a and 1 2 3( , , )B b b b be two points in three-dimensional space and let the position vectors of A and B with respect to the origin O be a and b respectively. Then vector (cross) product a b, is the vector given by 1 1 2 3 3 2 2 3 3 2 2 2 1 3 3 1 3 1 1 3 3 3 1 2 2 1 1 2 2 1 ( ) a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b                                              a b (MF27) Applications of Vector Product Let the points A and B have position vectors a and b with respect to the origin O, and let  be the angle between a and b. Let C be the point such that OACB is a parallelogram. Then we ha