RI Vector Algebra Ration Theorem Scalar and Vector Product Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 6 (Summary and Tutorial) 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 ( )123,,Aa a a and ( )123,,Bbb 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 11 2 2 22 3 3 33 a b ab a b ab a b ab ±     ±= ± = ±     ±   ab and 11 22 33 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 1 222 2 123 3 a a aaa a  = = ++ a . This value is equal to the distance from O to A. We say that a is parallel to b , denoted by ab , if and only if λ=ba for some { }\0λ∈  , that is, b is a (non-zero) scalar multiple of a . • If 0λ > , then λa and a are in the same direction. • If 0λ < , then λa and a are in opposite directions. Points A and B have position vectors a and b respectively, relative to the origin O , such that λ=ba for some { }\0λ∈  . We then say that the points O, A and B are collinear.

The unit vector in the direction of a denoted by ˆa is obtained by scaling a by 1 a , thus 1ˆ =aa a . 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 ( ) 22212 2 3+ +− = so the unit vector in the direction of 1 2 2   − is 1 1 23 2   − Let a and b be non-zero and non-parallel vectors: If λµ=ab for some ,λµ ∈  , then 0.λµ= = If a b= a b stαβ++ for some , ,, ,stαβ ∈  then ,stαβ= = . Note the importance of non-parallel vectors when comparing coefficients. Suppose 1 0 0    a= and 2 0 0    b= then 62 43+=+ab ab however we cannot “compare coefficients” of vectors a and b (note that a is parallel to b) as 6 4 and 2 3≠≠ .

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 divides AB in the ratio :,λµ i.e. AP PB λ µ= . If OP= p    , then µλ λµ += + abp (MF26) Note that the Ratio Theorem is an immediate consequence of the addition of vectors. From diagram, ( )pa ba λ λµ−= − + , rearranging we have µλ λµ += + abp . Sometimes, it is easier to use ( )pa ba λ λµ−= − + like the following example. Points ,AB and P have position vectors ,ab 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 2pa ba−= − [DO NOT WRITE 2 5 ba pa − =− . We cannot divide vectors] 2 22 2 52 6 2 1225 3 5 15 p −       =+ −=      −− −   

Scalar (Dot) Product Points A and B have position vectors a and b respectively, relative to the origin O . cosθ⋅=ab ab where AOBθ =∠ . Re-arranging, cos ab abθ ⋅= . If vectors a and b are in the same direction then ab ab⋅= , largest possible value. If vectors a and b are in the opposite direction then ab ab⋅= − , smallest possible value. Two non-zero vectors a and b are perpendicular, denoted by ⊥ab , if and only if 0⋅=ab . In particular, 0⋅=⋅ = ⋅=i j jk k i as ,ij and k are mutually perpendicular. If 1 2 3 a a a  = a and 1 2 3 b b b  = b , then 11 2 2 11 2 2 3 3 33 ab a b ab ab ab ab      ⋅= ⋅ = + +       ab . Find the cosine of the angle between the vectors 232a ijk=−+ and 2b i jk= +− and determine if the angle is acute or obtuse. Solution ( ) ( ) ( ) 2222 22 232 2cos 232 12 1 i j k i jkθ −+ ⋅+−= + + + +− 262 6cos 01717 6 θ −−= = −< θ is an obtuse angle Properties of Scalar Product (i) ⋅=⋅ab ba (ii) ( )( )( )⋅±=⋅±⋅a b c ab ac (iii) ( )() ( )λλ λ⋅ = ⋅=⋅ab a b a b (iv) 2 ⋅= ⇔ = ⋅aa a a aa note the relationship between the scalar product and the modulus Given that 2, 3, 1a b ab= = ⋅= − , (i) ( ) ( ) 2−⋅ +ab ab 22aa ab ba bb= ⋅+⋅− ⋅−⋅ ( ) 22 22 2 2(2 ) 1 3 0 a ab b= −⋅− = −− − = (ii) ( ) ( )22−⋅ +ab ab 422aa ab ba bb= ⋅+ ⋅− ⋅−⋅ 22 22 4 4(2 ) 3 7 ab= − = − = (iii) ( ) ( )2 22ab ab ab− = − ⋅− 22 2 2 4 4 4 ... 2 11aa ab ba bb a ab b= ⋅− ⋅− ⋅+ ⋅= − ⋅+ = =

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 ( ) ˆˆab b abbbbOP  ⋅= = ⋅    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 ˆabOP = ⋅    . Points A and B have position vectors a and b respectively, relative to the origin O . Given that 232a ijk=−+ and 2b i jk= +− , 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 ( ) ( ) ( ) 222 232 2 262ˆ 6 612 1 i j k i jkabOP − + ⋅+− −−=⋅= = = + +− ( ) 62ˆˆ 2 66 i jkabb i j kOP +−= ⋅ =− =−− +    If , and αβ γ are the angles between a and the component vectors i, j and k respectively, then ( )3121ˆ , , cos ,cos ,cosaa a aaa aaa αβγ= = = . Component of ˆa are referred to as the direction cosines.

Vector (Cross) Product Let a and b be two non-zero vectors that are represented by OA    and OB   respectively. ( ) ˆsinab ab n θ×= where AOBθ =∠ 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 .×=ab ab In particular, ,×=ijk ×=jk i and .×=ki j Two non-zero vectors a and b are parallel if and only if .×=ab 0 In particular, .×=×= × =ii jj kk 0 Given that ( ) ( )uv uv 0+×−= , what can be deduced about the vectors u and v ? Solution: ( ) ( ) 2 uv uv 0 u u u v v u v v=0 0+ v u 0 0 vu 0 +×−= ×−×+×−× ×−= ×= u0= or v0= or uv Properties of Vector Product 1. ( ).ab ba×= −× 2. sin .θ×=×= ≤ab ba ab ab 3. ( ) ( ) ( ).×±=×±×a bc ab ac 4. ( ) ( ) ( ).λλ λ× = ×=×ab a b a b 5. () () 0aba abb×⋅ =×⋅= So ab a×⊥ and ab b×⊥ . Examples of how the properties are used ( 2 ) (3 ) 3 6 2 6 7 a b ab aaab ba bb 0ba ba0 ba + × − = ×−×+ ×− × =+×+ ×+ = × 22( )( ) 2 a b a b aa ab ba bb 0baba0 ba α β α β α αβ αβ β αβ αβ + × − = ×− ×+ ×− × = +×+×+ = ×

Let 123(, , )Aa a a and 123(, , )Bb 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 ab× , is the vector given by 23 32 23 3211 2 2 13 31 31 13 33 12 21 12 21 ()ab ab ab ab abab a b ab ab ab ab ab ab ab ab ab −−         ×= × =− − = −      −−      (MF26) Applications of Vector Product Let the points A and B have position vectors a and b with respect to the origin O , and let θ be the angle between a and b. Let C be the point such that OACB is a parallelogram. Then we have (1) Area of triangle OAB is 1 .2 ×ab (2) Area of parallelogram OACB is .×ab (3) Distance from B to OA is ˆba× . (Recall the length of projection of OB  → onto OA  → is ˆba⋅ )

Revision Tutorial Questions 1 For each question, indicate the letter corresponding to the correct answer. Answer I The modulus of the vector 623i jk−− is (a) 23 (b) 11 (c) 1 (d) 49 (e) 7 II The unit vector in the direction of the vector 22i jk−+ − is (