Precalculus, Chapter 6, 6.3, Section 6.3, Problem 51

The magnitude of a vector u = a*i + b*j , such that:
|u| = sqrt(a^2+b^2)
Since the problem provides the magnitude |v| = 5 , yields:
5 = sqrt(a^2+b^2)
The direction angle of the vector can be found using the formula, such that:
tan theta = b/a
Since the problem provides the information that the direction angle of the vector v coincides to the direction angle of the vector u = <2,5> , yields:
tan theta = 5/2 => b/a = 5/2> 5a = 2b => b = (5a)/2
Replacing (5a)/2 for b in equation5 = sqrt(a^2+b^2) yields:
5 = sqrt(a^2+25a^2/4)=> 25 = 29a^2/4 => 100 =29a^2 => a = +-10/(sqrt29)
b = +-25/(sqrt29)
Hence, the component form of the vector v can be <10/(sqrt29),25/(sqrt29)> or <-10/(sqrt29),-25/(sqrt29)>.