# I understand how to solve this problem: i.e. take the unit vector of a, then multiply by our given magnitude (in this case sqrt2) which gives: (sqrt2)/(5)(3i+4j). I'm confused as to what the mathematical logic behind this, how would you explain why we're doing this process?

I understand how to solve this problem: i.e. take the unit vector of a, then multiply by our given magnitude (in this case $\sqrt{2}$) which gives: $\frac{\sqrt{2}}{5}\left(3i+4j\right)$
I'm confused as to what the mathematical logic behind this, how would you explain why we're doing this process?
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Kody Gates
Magnitude of $\stackrel{\to }{b}=i-j$ $\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|\stackrel{\to }{b}|=\sqrt{{1}^{2}+\left(-1{\right)}^{2}}=\sqrt{2}$
Unit vector parallel to $\stackrel{\to }{a}=3i+4j$ $\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{^}{a}=\frac{\stackrel{\to }{a}}{|\stackrel{\to }{a}|}=\frac{3i+4j}{\sqrt{{3}^{2}+{4}^{2}}}=\frac{1}{5}\left(3i+4j\right)$
The required vector with magnitude equal to that of $\stackrel{\to }{b}$ & directed along $\stackrel{\to }{a}$:
$|\stackrel{\to }{b}|\stackrel{^}{a}=\sqrt{2}\frac{1}{5}\left(3i+4j\right)=\frac{\sqrt{2}}{5}\left(3i+4j\right)$
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Hagman7v
The solution that you mentiond was this: $‖b‖\frac{a}{‖a‖}=\frac{‖b‖}{‖a‖}a$. Since it is a times a positive number, it has the same direction as a. Besides,
$‖\frac{‖b‖}{‖a‖}a‖=\frac{‖b‖}{\overline{)‖a‖}}\overline{)‖a‖}=‖b‖,$
as you wanted.