Answer
$\textbf{a}=⟨\frac{1}{2},\frac{1}{2}⟩+⟨\frac{1}{2},-\frac{1}{2}⟩$
Work Step by Step
We have the vectors $a=<1,0>$ and $b=<1,1>$. Then:
$\textbf{a}_{\parallel \textbf{b}}=(\frac{\textbf{a}\cdot\textbf{b}}{\textbf{b}\cdot\textbf{b}})\textbf{b}=\frac{1\times1+0\times1}{1^{2}+1^{2}}⟨1,1⟩=⟨\frac{1}{2},\frac{1}{2}⟩$
$\textbf{a}_{\perp \textbf{b}}=\textbf{a}-\textbf{a}_{\parallel \textbf{b}}=⟨1,0⟩-⟨\frac{1}{2},\frac{1}{2}⟩$
$=⟨\frac{1}{2},-\frac{1}{2}⟩$
Therefore, the decomposition of $\textbf{a}$ with respect to $\textbf{b}$ is
$\textbf{a}=\textbf{a}_{\parallel \textbf{b}}+\textbf{a}_{\parallel \textbf{b}}=⟨\frac{1}{2},\frac{1}{2}⟩+⟨\frac{1}{2},-\frac{1}{2}⟩$
