Precalculus, Chapter 6, 6.3, Section 6.3, Problem 72

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| = 4sqrt3 , yields:
4sqrt3= 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 is theta = 0^o , yields:
tan 0^o = b/a => b = 0, a!=0
Replacing 0 for b in equation 4sqrt3= sqrt(a^2+b^2) yields:
4sqrt3= sqrt(a^2+0)=> a = +-4sqrt3
b = 0
Hence, the component form of the vector v can be <4sqrt3,0> or <-4sqrt3,0>.