Answer
we have,
a=[1,2], b=[-4,1]
we need to find projection of b onto a.
scalar projection
= [tex]\dfrac{a.b}{|a|}[/tex]
= [tex]\dfrac{(1,2).(-4,1)}{\sqrt{1^2+2^2}}[/tex]
= [tex]\dfrac{1\times (-4)+2\times 1}{\sqrt{1^2+2^2}}[/tex]
=[tex]\dfrac{-2}{\sqrt{5}}[/tex]
now, vector projection
= [tex]\dfrac{a.b}{|a|^2}.a[/tex]
= [tex]\dfrac{(1,2).(-4,1)}{(\sqrt{1^2+2^2})^2}.(1,2)[/tex]
= [tex]\dfrac{1\times (-4)+2\times 1}{5}.(1,2)[/tex]
=[tex]\dfrac{-2}{5}.(1,2)[/tex]
=[tex](\dfrac{-2}{5},\dfrac{-4}{5})[/tex]
