VECTORS OR CROSS PRODUCT
THIRD TERM E-LEARNING NOTE
SUBJECT: FURTHER MATHEMATICS CLASS: SS 2
SCHEME OF WORK
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 and whose directions are inclined at an angle, their vector product is defined as a vector whose magnitude is sin and whose directions is perpendicular to both and and also being positive relative to a rotation from them vector and also being positive relative to a rotation from the vector to the re
ctor.
The vector product of and b is designated
b
Thus:
= x =|| || sin . whereas a unit vector perpendicular to the plane of and .
Properties of vector Product
x = |b||a| sin(-) 0 <<
= – |a||b| sin (-)
= – b
Thus the vector product of two vectors is not commutative .
(k) x = x (k )
= k (x)
= k |||| sin )
Where k is a scalar.
x ( + c) = x + x c
Distribute law
x = x = x
x = = – x , x k = =
x
x = – x
|a x b| = area of parallelogram with sides
and .
If x = 0 and and b are non zero vectors, then a and b are parallel
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 :
p x q;
|p x q|
Solution
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
|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.
