DOT MULTIPLICATION

Question

ANGLE BETWEEN a AND b VECTORS IS 2π⁄3. |a|=3 |b|=4 FIND (3a-2b)(a+2b) 

Answer 1

Let us write down a and b as vectors.

 

Let us take a as a reference vector

 

So,  a =3+i0

 

Angle between two vectors a and b is 2pi/3,taking angle clockwise

 

Vector b =2+2sqrt(3)i

 

Now, let us find out the given expression

 

(3a-2b)/(a+2b)=. (9+i0-4-4sqrt(3)i)/(3+i0+4+4sqrt(3)i)

 

                       =(5-4sqrt(3)i)/(7+4sqrt(3)i)

Rationalize this

 

                      =(5-4sqrt(3)i)(7-4sqrt(3)i)/(7+4sqrt(3)i)(7-4sqrt(3)i)

 

On simplification

 

                       =-13/97-(48/97)*sqrt(3)i

Answer 2

cos(2pi/3) = -1/2

 

for the difference D = 81 + 64 - 2(72)cos(pi/3) = 145 - 72 = 73

 

for the sum S = 9 + 64 - 2(24) cos(2pi/3) = 73 + 24 = 97

 

(3a - 2b)/(a + 2b) = 73/97

 

Sorry, now I see the topic is dot product

Answer 3

The angle, θ, between the vectors can be found using the dot product divided by the product of the magnitude of the vectors.

 

Cosθ = (a dot b)/(|a| times |b|).  So cos 2π/3 = - .5 and (a dot b)/12 = -.5.

 

That means a dot b = -6.  Can you proceed from here?