find the orthogonal projection of v= [19,12,14,-17] onto the subspace W spanned by [ [ -4,-1,-1,3] ,[ 1,-4,4,3] ] proj w (v) = [answer,answer,answer,answer]

Answer:Hence, we have:[tex]proj_W(v)=[\dfrac{464}{21},\dfrac{167}{21},\dfrac{71}{21},\dfrac{-131}{7}][/tex]Step-by-step explanation:By the orthogonal decomposition theorem we have:The orthogonal projection of a vector v onto the subspace W=span{w,w'} is given by:[tex]proj_W(v)=(\dfrac{v\cdot w}{w\cdot w})w+(\dfrac{v\cdot w'}{w'\cdot w'})w'[/tex]Here we have:[tex]v=[19,12,14,-17]\\\\w=[-4,-1,-1,3]\\\\w'=[1,-4,4,3][/tex]Now,[tex]v\cdot w=[19,12,14,-17]\cdot [-4,-1,-1,3]\\\\i.e.\\\\v\cdot w=19\times -4+12\times -1+14\times -1+-17\times 3\\\\i.e.\\\\v\cdot w=-76-12-14-51=-153[/tex][tex]w\cdot w=[-4,-1,-1,3]\cdot [-4,-1,-1,3]\\\\i.e.\\\\w\cdot w=(-4)^2+(-1)^2+(-1)^2+3^2\\\\i.e.\\\\w\cdot w=16+1+1+9\\\\i.e.\\\\w\cdot w=27[/tex]and[tex]v\cdot w'=[19,12,14,-17]\cdot [1,-4,4,3]\\\\i.e.\\\\v\cdot w'=19\times 1+12\times (-4)+14\times 4+(-17)\times 3\\\\i.e.\\\\v\cdot w'=19-48+56-51\\\\i.e.\\\\v\cdot w'=-24[/tex][tex]w'\cdot w'=[1,-4,4,3]\cdot [1,-4,4,3]\\\\i.e.\\\\w'\cdot w'=(1)^2+(-4)^2+(4)^2+(3)^2\\\\i.e.\\\\w'\cdot w'=1+16+16+9\\\\i.e.\\\\w'\cdot w'=42[/tex]Hence, we have:[tex]proj_W(v)=(\dfrac{-153}{27})[-4,-1,-1,3]+(\dfrac{-24}{42})[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=\dfrac{-17}{3}[-4,-1,-1,3]+(\dfrac{-4}{7})[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=[\dfrac{68}{3},\dfrac{17}{3},\dfrac{17}{3},-17]+[\dfrac{-4}{7},\dfrac{16}{7},\dfrac{-16}{7},\dfrac{-12}{7}]\\\\i.e.\\\\proj_W(v)=[\dfrac{464}{21},\dfrac{167}{21},\dfrac{71}{21},\dfrac{-131}{7}][/tex]