Determine [3a-4b].[2a+b], given a and b are unit vectors and the angle between them is 30 degree.
We have to determine [3a-4b].[2a+b], given that a and b are unit vectors and the angle between them is 30 degrees.
a dot b = |a||b| cos 30 = sqrt 3/2
=> 3a dot 2a + 3a dot b - 4b dot 2a - 4b dot b
=> 6*a dot a + 3a dot b - 8* a dot b - 4* b dot b
=> 6 - 5(a dot b) - 4
=> 2 - 5*sqrt 3/2
=> (4 - 5*sqrt 3)/2
The required result is (4 - 5*sqrt 3)/2
We'll determine the dot product of a and b:
a*b = |a|*|b|*cos 30 = 1*1*sqrt3/2
We'll use FOIL method to remove the brackets:
(3a-4b)(2a+b) = 3a^2 + 3ab - 8ab - 4b^2
The angle made by the vector a with itself is of 0 degrees.
a^2 = a*a = |a|*|a|*cos 0 = 1*1*1 = 1
b^2 = b*b = 1
(3a-4b)(2a+b) = 6 + 3*sqrt3/2 - 8sqrt3/2 - 4
We'll combine like terms:
(3a-4b)(2a+b) = 2 - 5sqrt3/2
The requested value of the product is (3a-4b)(2a+b) = 2 - 5sqrt3/2.