Mechanics SS2 Further Mathematics Lesson Note

SCALAR OR DOT PRODUCT OF TWO VECTORS

The scalar or dot product of two vectors a and b is written as a.b and pronounced as (a dot b). Therefore, a.b =|a| |b| cos dot is defined as a.b = a b cos  where  is the angle between vectors a and b 

If a = a1 I + a2j and b = b1 I b2j 

Thus a.b = (a)1bi ii + ab2j I 1 + 2 bi I h +a2 b2 j 

Recall that i and j are mutually perpendicular unit vector hence

i.i = |x| cos 0 =1

i.j = |x| cos 90 =0

j.i = |x| cos 90 =0

j.i =|x| cos 0 =1

Hence, a.b =a1b1 + a2 b2 

Examples 

1. Find the scale product of the following vectors 9i -2j + k and I – 3j -4k 

Solution:

A=(9i- 2j +k) and b= (i-3j -4k)

a.b = (9i-2j +k) (i-3j-4k)

=9 (1) -2(-3) + 1(-4)=9+6-4a.b =11

2. Let a = 3i+2j, b = -4i+2j and c = i+4j, calculate a.b, a.c and a. (b+c)

Solution:

1. a.b = (3i + 2j ) (-4i+2j) = 3 (-4) +2(2)

= -12+4

=-8

1. a.c = (3i+2j) (I +4j)

= 3 (1) + 2 (4)

= 3+8 = 11

iii. a.(b+c)

Find (b+c) = -4i + 2j +i +4j

=-3i +6j

(b+c) = (3i+2j) (-3i +6j)

=3(-3) + 2(6)

= -9+12 = 3.

PERPENDICULARITY OF VECTORS:

If two vectors P and q are in perpendicular directions, thus p.q =0 

Example 1: 

Show that the vectors p = 3i+ 2j and q= -2i + 3j are perpendicular.

Solution: 

P:q = (3i+2j) (-2i +3j)

=3(-2) + 2(3)

=-6+ =0

Since p.q=0, then the vectors p and q are perpendicular.

2. If p= 4i + kj and q=2i – 3j are perpendicular, find the value of k, where k is a scalar..

Solution:

p.q=0

(4i+kj)(2i-3j)=0

4(2) + k(-3)=0

-3k=-8

K=8/3.

EQUAL VECTORS: Vectors p and q are equal if p is equal to q.

Example: find the value of the scalar K for which the vectors 2ki + 3j and 8i+kj

Solution:

2ki +3j = 8i +kj

Hence, 2ki =8i,   3j = ¾kj

2k = 8;                      12 = 3k

k = 8/2;                     k = 12/3

K=4;             k=4

ANGLES BETWEEN TWO VECTORS 

This is the angle between two vectors and from dot product where a.b=|a|b| cos . Hence, 

Cos0 = a.b

           |a| |b|

Where = Magnitude of vector a= 21  +22

|b| =Magnitude of vector b=21  +22

Example:

Find the angle between the vectors pp=2i – 2j + k and q=12i +4j – 3k

Solution:

Cos§0 = p.q

             |p| |q|

p.q= (2i-2j+k)(12i+4j-3k)

=2(12) -2(4)_1(3)

= 24-8-3

=13

|p|=2 + (-2)2 + 12=+4+1 =✓9 =3

|q|= 2 + 42 + (-3) = 144+16+9= ✓169 =13

Cos0 = 13

          3 × 13

Cos0= 1/3.                

=Cos-1 (1/3)