Precalculus, Chapter 6, 6.3, Section 6.3, Problem 49

The magnitude of a vector v=v_x*i + v_y*j is given by the following formula, such that:
|v| = sqrt(v_x^2+v_y^2)
The problem provides the information that |v| = 10:
10 = sqrt(v_x^2+v_y^2)
You may evaluate the direction angle of the vector v, such that:
tan alpha = (v_y)/(v_x)
The problem provides the information that the direction angle of the vector v coincides to the direction angle of the vector u = <-3,4> .
tan alpha = -4/3
(v_y)/(v_x) = -4/3 => v_y = (-4/3 )*(v_x)
Replacing (-4/3 )*(v_x) for v_y yields:
10 = sqrt(v_x^2+(16/9)*(v_x^2))
10 = +-(5/3)*(v_x)
2 = +-(1/3)*(v_x) => v_x = +-6 => v_y = +-8
Hence, evaluating the vector v yields v = 6i - 8j or v = -6i + 8j.