Position vectors and proportions?

Question

Two points A and B have position vectors a and b respectively. In terms of a and b find the position vector of
a point U on the straight line AB for which AU/U B = 1/3.

I'm not sure how to calculate the different proportions of a and b. I know if U was the mid point then it would simply be 0.5a+0.5b but I can't think how to alter this. I'm thinking maybe 0.25a+0.75b perhaps, but I'm not 100% sure why. Cheers.

Answer 1

A .__1_______P______3________B

let p be position vector of P
AP = 1/4 AB
p = a + 1/4(b - a)
p = (3/4)a + (1/4)b

Answer 2

The vectors AU and UB are collinear. If AU/UB = c (c=1/3 in this question) then we may write AU = c UB
vector AU = u - a, vector UB = b - u
so u - a = c(b - u)
u = (a + cb)/(1+ c)

if c = 1/3
u = (a + b/3)/(1+1/3) = (3/4)*(a + b/3) = 3a/4 + b/4 = 0.75a + 0.25b