## Precalculus (6th Edition) Blitzer

Consider the given identity $\mathbf{u}\cdot \left( \mathbf{v}+\mathbf{w} \right)=\mathbf{u}\cdot \mathbf{v}+\mathbf{u}\cdot \mathbf{w}$ ...… (1) For the given vectors u, v, and w, $\mathbf{u}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}$ $\mathbf{v}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$ $\mathbf{w}={{a}_{3}}\mathbf{i}+{{b}_{3}}\mathbf{j}$ Now, take the left side of equation (1) to get $\mathbf{u}\cdot \left( \mathbf{v}+\mathbf{w} \right)$ The left side of the given identity represents the dot product of two vectors $\mathbf{u}$ and $\left( \mathbf{v}+\mathbf{w} \right)$. Consider, $\mathbf{h}=\mathbf{v}+\mathbf{w}$ \begin{align} & \mathbf{h}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{i}+{{b}_{3}}\mathbf{j} \\ & \mathbf{h}=\left( {{a}_{2}}+{{a}_{3}} \right)\mathbf{i}+\left( {{b}_{2}}+{{b}_{3}} \right)\mathbf{j} \\ \end{align} Now, $\mathbf{u}\cdot \left( \mathbf{v}+\mathbf{w} \right)=\mathbf{u}\cdot \mathbf{h}$ …… (2) Substituting the value of $h$ in (2), we get $\mathbf{u}\cdot \mathbf{h}={{a}_{1}}\left( {{a}_{2}}+{{a}_{3}} \right)\mathbf{i}+{{b}_{1}}\left( {{b}_{2}}+{{b}_{3}} \right)\mathbf{j}$ $\mathbf{u}\cdot \left( \mathbf{v}+\mathbf{w} \right)={{a}_{1}}\left( {{a}_{2}}+{{a}_{3}} \right)\mathbf{i}+{{b}_{1}}\left( {{b}_{2}}+{{b}_{3}} \right)\mathbf{j}$ …… (3) Consider the right side of (1). The right side of the given identity represents addition of dot products of $\mathbf{u}\,\mathbf{v}$ and $\mathbf{u}\,\mathbf{w}$. Therefore, $\mathbf{u}\cdot \mathbf{v}={{a}_{1}}{{a}_{2}}\mathbf{i}+{{b}_{1}}{{b}_{2}}\mathbf{j}$ $\mathbf{u}\cdot \mathbf{w}={{a}_{1}}{{a}_{3}}\mathbf{i}+{{b}_{1}}{{b}_{3}}\mathbf{j}$ \begin{align} & \mathbf{u}\cdot \mathbf{v}+\mathbf{u}\cdot \mathbf{w}={{a}_{1}}{{a}_{2}}\mathbf{i}+{{b}_{1}}{{b}_{2}}\mathbf{j}+{{a}_{1}}{{a}_{3}}\mathbf{i}+{{b}_{1}}{{b}_{3}}\mathbf{j} \\ & ={{a}_{1}}{{a}_{2}}\mathbf{i}+{{a}_{1}}{{a}_{3}}\mathbf{i}+{{b}_{1}}{{b}_{2}}\mathbf{j}+{{b}_{1}}{{b}_{3}}\mathbf{j} \end{align} Solving above equation gives $\mathbf{u}\cdot \mathbf{v}+\mathbf{u}\cdot \mathbf{w}={{a}_{1}}\left( {{a}_{2}}+{{a}_{3}} \right)\mathbf{i}+{{b}_{1}}\left( {{b}_{2}}+{{b}_{3}} \right)\mathbf{j}$ …… (4) Equations (3) and (4) are identical. So, Left side of equation (1) = Right side of equation (1) Hence, $\mathbf{u}\cdot \left( \mathbf{v}+\mathbf{w} \right)=\mathbf{u}\cdot \mathbf{v}+\mathbf{u}\cdot \mathbf{w}$.