Week 5 and 6 – SS2 Third Term Further Mathematics Notes

WEEK FIVE
TOPIC : MECHANICS ; VECTORS OR CROSS PRODUCT ON TWO OR THREE DIMENSION , CROSS PRODUCT OF TWO VECTORS AND APPLICATION OF CROSS PRODUCT
Vector Product of two vectors
Given two vectors andwhose directions are inclined at an angle, their vector productis defined as a vector whose magnitude issinand whose directions is perpendicular to both and and also being positive relative to a rotation from themvector and also being positive relative to a rotation from the vectorto the re
ctor.
    The vector product ofand b is designated
b
Thus:
     = x =|| || sin . whereis a unit vector perpendicular to the plane of and .
Properties of vector Product

1. x = |b||a| sin(-) 0 <<
   

           = – |a||b| sin (-)
           = – b
       Thus the vector product of two vectors is not commutative .

2. (k) x             = x (k )

                   = k (x)
                   = k |||| sin )
   Where k is a scalar.

3. x ( + c)        = x + x c
   

   Distribute law

4. x    = x =     x

   x         =         = – x , x k = =

• x

       x =         – x
  

1.     |a x b| = area of parallelogram with sides

   and .

2.     If x = 0 and and b are non zero vectors, then a and b are parallel
3. If a = a₁ i+ a₂i + a₃k

       b = b₁ + b₂i +b₃k then
   
       i j k
   a    ba₁ a₂ a₃
        b₁ b₂ b₃
   
    We shall make use of the following important result in determinant of order 2 x 2 and order 3 x 3 defined respectively as follows.
   a b
   c d = ad – bc
   
   a b c     e     f    – b d    f + c      d e
   d e f    = a h     I      g    I     g h
   g h i
   
    The expansion of the determinant of order 3 x 3 is along the first row.
       Notwithstanding it can be along any other row or any column.

    Example 1
   Find the vector of a andb where:
   a = 4I    – 3j     + 2k, b = i + 2 j – 5k
   Solution
   a = 4 I – 3 j + 2 k
   b = I + 2j – 5k
   
   a x b = i     j    k
        4     -3    2
        1     2 -5
   
   =i -3    2     4    2    4    -3
            –j     +k         
        2    -5     1    -5    1    2

    = I (15 -4) –j (-20 -2) + k (8 + 3)
   = 11i + 20j + 11k
   
    If p    =2i – 3j + 4k
   q     =5i – 4j – 3k
   Find :

4. p x q;
   
5. |p x q|
   

   Solution
   

6. p x q = i     j    k

        2     -3    4
        5     4    -3
   =i -3    4     2    4    2    -3
       –j     +k         
       4    -3     5    -3    5    4

    = i (9 – 8) –j (- 6 -20) + k (8 + 15)
   =i + 26j + 23 k
   
    

7. |p x q|     =     |I + 26j + 23k|

           =    
           =    
   `        =    
           =    
           =    
   Example
   Show that (a x b)² = a²b² – (a.b)²
   Solution
   (a x b)²    =    (absin)²
           =    a² b² sin²
           =    a² b² (1 – cos² )
           =    a² b² – a² b² cos²
           =    a² b² – (a.b)²
   Hence
   (a x b)²    =     a² b² – (a.b)²
   
    EVALUATION
   Given that p = 2i + 3j +4k and q= 5i – 6j +7k find ; (1) p x q ( 2) (p + q ) . ( p-q)

    
    
    Application of vector product
   Area of a parallelogram Example
   Show that the area of parallelogram with sides a andb is.
   Solution
   
    
    
    
    
    Area of parallelogram
   OAC B=    h/b
           =/a/ sin /b/
           =/a/ /b/sin
           =/a x b/
   Area of angle
   Example
   Show that the area of a triangle with sides a and b is |a x b|
   Solution
   
    
    
    
    
    
    Area of     =    OAB = |b| x h
           =     |b||a| Sin
           =     |a||b| Sin
           =     (a x b)
   
    Example
   The adjacent sides of a parallelogram are
   = 2 i – j – 6k and = i + 3 j – k . Find
   the area of the parallelogram.

    Solution
   
    
    
    
    AB     =     2 i – j – 6k
   AC        =    i + 3 j – k
   Area of parallelogram =    |AB x AC|
                   =      x
   i     j k
   2     -1 -6
   1     3 -1    
   x
   
   = i    -1    -6 –j     2    -6 +k    2    -1
       3    -1     1    -1     1    3    
   = I (1 + 18) –j (-2 + 6) + k (6 + 1)
   = 19 I – 4 j + 7 k
   |AB x AC|        =     |19i – 4j + 7k|
               =    
               =    
               =    
   Hence
   Area of parallelogram = sq. Units

    GENERAL EVALUATION
   1) Find the vector product of a= 4i -3j +4k and b = -I + 2j +7k
   2) Given that p = 7i + 2j + k and q = 3i – 2j + 4k find ; (i) p x q (ii) | p x q | (iii) the unit vector perpendicular to both p and q
   3) Find the sine of the angle between the vectors : a = I – j + k and b = 8i + 2j + 3k
   4) The adjacent sides of a parallelogram are PQ= 4i + 3j + k and PR = -5i + 2j +3k find the area of the parallelogram
   5) The position vectors OA, OB and OC are 2i – 3j + 4k , 6i + 4j -8k and 3i + 2j + 5k respectively find (i) vector AB (ii) vector BA (iii) vector BC (iv) AB x BC
   Reading Assignment: New Further Maths Project 2 page 216 – 222

    WEEKEND ASSIGNMENT
   Given that a = I + 2j + k and b = 2i +3j- 5k
   1) find ( a x b ) . a a) 0 b) 1 c) 2 d) 3
   2) find ( a x b ) . b a) 1 b) 2 c) 0 d) 3
   Given that p = I + 5j + 6k and q = – 2i + j + 3k
   3) find p x q a) 15i +11j -11k b) 11i – 15j + 11k c) 11i – 11j + 15k d) 11i- 15j -11k
   4) find q x p a) -11i + 15j – 11k b) 11i – 15j + 11k c) 15i – 11j-11k d) 15i+11j+11k
   5) Given that a = i – j+ 3k and b = 6i + 2j – 2k find ( a + b ) . ( a x b ) a) 1 b) 0 c) 2 d) 3

    THEORY
   1) AB = 4i +3j+5k and AC= 2i-3j+k are two sides of a triangle ABC , find the area of the triangle
   2) PQ = 2i+5j+3k and PR = 3i-3j + k are two adjacent sides of a parallelogram, find the area of the parallelogram.

    
    WEEK SIX
   REVIEW OF HALF TERM WORK