Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 667: 56

$⟨1,1,0⟩$

Projection of $\textbf{u}$ along $\textbf{v}=\textbf{u}_{||\textbf{v}}$$=(\frac{\textbf{u}\cdot\textbf{v}}{||\textbf{v}||^{2}})\textbf{v}$ $\textbf{u}\cdot\textbf{v}=⟨1,1,1⟩\cdot ⟨1,1,0⟩=1\times1+1\times1+1\times0=2$ $||\textbf{v}||^{2}=(\sqrt {1^{2}+1^{2}+0^{2}})^{2}=1^{2}+1^{2}=2$ Then, $\textbf{u}_{||\textbf{v}}=(\frac{2}{2})\,⟨1,1,0⟩=⟨1,1,0⟩$