Finding the orthogonal projection of $(2,1,3)$.

Question

$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?

Answer 1

If $W = \{(x,y,z): x+3y-2z=0\}$, then $x=2z-3y$, and every element of $W$ looks like $$\left(\begin{array}{c} x \\ y \\ z \end{array}\right) \ \ = \ \ \left(\begin{array}{c} 2z-3y \\ y \\ z \end{array}\right) \ \ = \ \ y\left(\begin{array}{c} -3 \\ 1 \\ 0 \end{array}\right) +z\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right)$$ This means that $W$ is spanned by $(-3,1,0)^{\top}$ and $(2,0,1)^{\top}$.

The vector $(-1,1,1)^{\top}$ lies in $W$ - you can check this: $$\det\left(\begin{array}{ccc} -1 & -3 & 2 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right) = 0$$ but it's not enough on its own to span $W$ since $\dim W =2$.

Answer 2

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!