Precalculus, Chapter 6, 6.3, Section 6.3, Problem 69

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| = 7/2 , yields:
7/2 = 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 direction angle theta = 150^o , yields:
tan 150^o = b/a
tan 150^o = tan(180^o - 30^o) = (tan 180^o - tan 30^o)/(1 + tan 180^o* tan 30^o) = -tan 30^o
tan 150^o = -sqrt3/3 => b/a = -sqrt3/3 => b = -a*sqrt3/3
Replacing -a*sqrt3/3 for b in equation 7/2 = sqrt(a^2+b^2) yields:
7/2 = sqrt(a^2+a^2/3)=> 7/2 = +-2a/sqrt3 => 4a = +-7sqrt3 => a = +-(7sqrt3)/4
b = +-7/4
Hence, the component form of the vector v can be <(7sqrt3)/4,-7/4> or <-(7sqrt3)/4,7/4>.