Is the magnitude of the vector resolute of a parallel to b equal to the scalar resolute? If this is the case, what about the magnitude of the vector resolute of a perpendicular to b -- is there also a way to link the scalar resolute? Sorry if this is a dumb question...its not explicit in the textbook as far as I can tell.
That’s not a dumb question at all. Yes, the magnitude of the parallel vector resolute is the same as the scalar resolute. Here’s why algebraically:
Scalar resolute \(\mathbf{a\cdot \hat{b}}=\mathbf{\frac{a\cdot {b}}{\| b\|}}\)
and magnitude of parallel vector resolute \(\|(\mathbf{a\cdot \hat{b})\hat b} \|=\frac{\|(\mathbf{a\cdot {b})\hat b} \|}{\| \mathbf{b}\|}=\mathbf{\frac{(a\cdot {b})\|b\|}{\| b\|^2}=\mathbf{\frac{a\cdot {b}}{\| b\|}}}\)
Geometrically, you can consider the scalar resolute as the 'length' of vector a projected along vector b. I can't immediately think of any nice relationship between the length of the perpendicular vector resolute and the scalar resolute. At least other than, by Pythagoras, that \(\mathbf { (a\cdot \hat b)^2+\|a-(a\cdot \hat b)\hat b\|^2= \|a\|^2}\), but that is almost a tautology.