Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 667: 60

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