Dot product problem?

Question

let a and b be two perpendicular vectors such that the norm of a is twice the norm of b. Use the properties of the dot product to find the angle between the vectors a-b and a+2b

Answer 1

(a-b).(a+2b)

= a.a + a.b - 2b.b

= |a|^2 + 0 - 2|b|^2

= 4|b|^2 - 2|b|^2

= 2|b|^2

|a-b|

= sqrt(|a|^2 + |b|^2)

= sqrt(4|b|^2 + |b|^2)

= sqrt(5)|b|

|a+2b|

= sqrt(|a|^2 + 4|b|^2)

= sqrt(4|b|^2 + 4|b|^2)

= 2sqrt(2)|b|

Hence angle required

= arccos[2 / (sqrt(5) * 2sqrt(2))]

= arccos[1/sqrt(10)]