if a and b are unit vectors, and magnitude of vectors a +b = sq. root of 3, determine (2a-5b) dot (b+3a).

Answer 1

(2a-5b)•(3a+b)
2a•3a - 13a•b - 5b•b
6 - 13a•b - 5
1 - 13a•b

Now |a+b| = √3, so using the law of cosines to get the angle θ between a and b, we have

1+1-2cosθ = √3
cosθ = 1-√3/2

Now, we know that since a and b are unit vectors,
a•b = cosθ = 1-√3/2

Now we have

(2a-5b)•(3a+b) = 1 - 13a•b
= 1-13(1-√3/2)
= 13√3/2 - 12
= -0.741