Precalculus, Chapter 6, 6.3, Section 6.3, Problem 52

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| = 8 , yields:
8= 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 vector u, yields:
tan theta= 3/3 => tan theta = 1 =>b/a = 1 => a = b
Replacing a for b in equation 8 = sqrt(a^2+b^2) yields:
8 = sqrt(a^2+a^2)=> 8 = +-a*sqrt 2 => a = +-4sqrt2
b = +-4sqrt2
Hence, the component form of the vector v can be <4sqrt2,4sqrt2> or <-4sqrt2,-4sqrt2>.

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