High School Ambiguity when labeling arrows

[etotheipi]

This question is unbelievably silly however I thought I'd ask it anyway. The label on a vector (e.g. a force in a free body diagram) often represents the magnitude, and if it is algebraic ($F, N, W$ etc.) we can solve for its size. If our prior conviction about the direction of the vector turns out to be incorrect, we might end up with a negative value. Whilst this is not a problem, as we just end up with the vector pointing in the opposite direction, it means that the label cannot be the magnitude since it can quite happily take on negative values. In this case, what does the label represent? Could it be a "signed magnitude", if that's even a thing?

My prior pattern of thought was to use the magnitude to convert to the one dimensional component vectors in the direction being considered (e.g. $(N \cos{\theta}) \hat{i})$. Evidently, the maths works out either way, however the definition of the label is a little shaky. Is this perhaps just a case of the limitations of a force diagram?

 

[A.T.]

My prior pattern of thought was to use the magnitude to convert to the one dimensional component vectors in the direction being considered

Seem ok to me: The label represents the component along the unit vector represented by the arrow.

 

[Herman Trivilino]

The label on a vector (e.g. a force in a free body diagram) often represents the magnitude,

Well, it shouldn't. The arrow on a free body diagram represents a vector, not a magnitude. Yes, the length of the arrow is proportional to the magnitude, but the arrow itself represents a vector.

Vectors, by the way, are neither positive nor negative. Magnitudes are never negative. Vector components can be either positive or negative.

 

[etotheipi]

Seem ok to me: The label represents the component along the unit vector represented by the arrow.

I quite like this picture, since then we have $F = |F||\hat{n}|\cos{\theta}$ which reduces to $|F|$ if the actual force lies along the direction of the arrow, and $-|F|$ if the actual force is in the opposite direction! Thanks