# Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 667: 57

$-4\textbf{k}$

#### Work Step by Step

Projection of $\textbf{u}$ along $\textbf{v}=\textbf{u}_{||\textbf{v}}$$=(\frac{\textbf{u}\cdot\textbf{v}}{||\textbf{v}||^{2}})\textbf{v}$ $\textbf{u}\cdot\textbf{v}=(5\textbf{i}+7\textbf{j}-4\textbf{k})\cdot \textbf{k}=5\times0+7\times0+(-4\times1)=-4$ $||\textbf{v}||^{2}=(\sqrt {0^{2}+0^{2}+1^{2}})^{2}=1$ Then, $\textbf{u}_{||\textbf{v}}=(-\frac{4}{1})\,\textbf{k}=-4\textbf{k}$

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