<s>[INST] Finding the orthogonal projection of $(2,1,3)$. $V=\mathbb R^3,u=(2,1,3) $ and $W$={$(x,y,z):x+3y-2z=0$}.
Vectors ortogonal to $W$=Span{$(1,3,-2)$},Vectors on $W$=Span{$(-1,1,1)$}.
So,orthogonal basis is {$(1,3,-2),(-1,1,1)$}.Then,orthonormal basis is 
{$\frac{1}{\sqrt {14}}(1,3,-2),\frac{1}{\sqrt {3}}(-1,1,1)$}.
So orthogonal projection of 
(2,1,3)=$<\frac{1}{\sqrt {14}}(1,3,-2),(2,1,0)>\frac{1}{\sqrt {14}}(1,3,-2)$+$<\frac{1}{\sqrt {3}}(-1,1,1),(2,1,0)>\frac{1}{\sqrt {3}}(-1,1,1)>$
=$\frac{-1}{14}(1,3,-2)+\frac{2}{3}(-1,1,1)=(\frac{-25}{42},\frac{34}{42},\frac{22}{42})$
But it's correct answer is = $\frac{1}{
14}
(29,17,40)$
Where i'm missing?
 [/INST] Let $(2,1,3)+k(1,3,-2)$ be  the orthogonal projection.
Thus, $$(2+k,1+3k,3-2k)(1,3,-2)=0$$ or
$$2+k+3+9k-6+4k=0,$$
which gives $k=\frac{1}{14}$ and since $\left(2+\frac{1}{14},1+\frac{3}{14},3-\frac{1}{7}\right)=\frac{1}{14}(29,17,40)$,
we are done!
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