$\vec{y}$ as the sum of two orthogonal vectors I'm having difficulty getting this problem down.  I have not idea what to do, and I can't find any leads from my notes.  Any advice?


Let $\vec{y}$ = $
  \left[ {\begin{array}{cc}
   3 \\
  -5 \\
   1 \\
  \end{array} } \right]$
    and $\vec{u}$ = $
  \left[ {\begin{array}{cc}
   1 \\
   4 \\
  -6 \\
  \end{array} } \right]$.
    Describe $\vec{y}$ as the sum of two orthogonal vectors, $\vec{x}_1$ in Span{$\vec{u}$} and $\vec{x}_2$ orthogonal to $\vec{u}$.


 A: You could draw a picture of $u$ and $y$ in the plane. Then, let $x_1 = \lambda u$ be the projection of $y$ on $u$. One can compute $\lambda = (y, u)/(u, u)$ with the standard scalar product. Then $x_2 = y - \lambda u$ is orthogonal to $u$.
A: Component of $\vec y$ on $\mathrm{span} \{\vec u\}$ is given by
$$\frac{\vec{y} \cdot \vec{u}}{||\vec{u}||} = -\frac{23}{\sqrt{53}} = \lambda$$
Then, the projection on  $\mathrm{span} \{ \vec{u} \}^\perp$ is
$$\vec{v} = \vec{y} - \lambda \vec{u} = \begin{pmatrix}\;\;3 - \lambda\\-5-4\lambda\\\;\;1-6\lambda\end{pmatrix}$$
And the projection on $\mathrm{span} \{ \vec{u} \}$ is of course $\lambda \vec u$.
A: Hints:
$$x_1\in\text{Span}\,\{u\}\implies x_1=\begin{pmatrix}\;\;\;t\\\;\;4t\\\!-6t\end{pmatrix}\;,\;\;t\in\Bbb R$$
$$x_2=\begin{pmatrix}a\\b\\c\end{pmatrix}\perp u\implies a+4b-6c=0$$
So now you have to solve the vectorial equation
$$\begin{pmatrix}\;\;3\\-5\\\;\;1\end{pmatrix}=\begin{pmatrix}\;\;\;t\\\;\;4t\\\!-6t\end{pmatrix}+\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}\;\;\;t+a\\\;\;4t+b\\\!-6t+c\end{pmatrix}$$
with the added condition that $\;a+4b-6c=0\;$...
A: I'll expand my comments into an answer. I think this gives the best way to sove the problem if you haven't memorized the formula for projections. 
We have the three equations, $$(3,-5,1)=x_1+x_2\tag1$$ $$x_1=k(1,4,-6){\rm\ for\ some\ }k\tag2$$ $$x_1\cdot x_2=0\tag3$$ Take the dot product with $x_1=ku=k(1,4,-6)$ on both sides of (1). On the left, you get $$k(1,4,-6)\cdot(3,-5,1)=-23k$$ On the right, you get $$x_1\cdot x_1+x_1\cdot x_2=k^2(1,4,-6)\cdot(1,4,-6)+0=53k^2$$ Note that we have made use of (2) and (3). It now follows that $k=-23/53$, so we get $x_1$ from (2), and $x_2$ from (1). 
