Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 667: 58

$\textbf{0}$

Projection of $\textbf{u}$ along $\textbf{v}=\textbf{u}_{||\textbf{v}}$$=(\frac{\textbf{u}\cdot\textbf{v}}{||\textbf{v}||^{2}})\textbf{v}$ $\textbf{u}\cdot\textbf{v}=(\textbf{i}+29\textbf{k})\cdot \textbf{j}=1\times0+0\times1+29\times0=0$ $||\textbf{v}||^{2}=(\sqrt {0^{2}+1^{2}+0^{2}})^{2}=1$ Then, $\textbf{u}_{||\textbf{v}}=(\frac{0}{1})\,\textbf{j}=\textbf{0}$