Chapter 13 - Vector Geometry - 13.4 The Cross Product - Exercises - Page 678: 51

Using the formula for the cross product, we obtain $\left( {{\bf{u}} + {\bf{v}}} \right) \times {\bf{w}} = {\bf{u}} \times {\bf{w}} + {\bf{v}} \times {\bf{w}}$

Let the components of ${\bf{u}}$, ${\bf{v}}$ and ${\bf{w}}$ be given by ${\bf{u}} = \left( {{u_1},{u_2},{u_3}} \right)$, ${\bf{v}} = \left( {{v_1},{v_2},{v_3}} \right)$ and ${\bf{w}} = \left( {{w_1},{w_2},{w_3}} \right)$, respectively. So, ${\bf{u}} + {\bf{v}} = \left( {{u_1} + {v_1},{u_2} + {v_2},{u_3} + {v_3}} \right)$. Using the formula for the cross product, we have $\left( {{\bf{u}} + {\bf{v}}} \right) \times {\bf{w}} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ {{u_1} + {v_1}}&{{u_2} + {v_2}}&{{u_3} + {v_3}}\\ {{w_1}}&{{w_2}}&{{w_3}} \end{array}} \right|$ $\left( {{\bf{u}} + {\bf{v}}} \right) \times {\bf{w}} = \left| {\begin{array}{*{20}{c}} {{u_2} + {v_2}}&{{u_3} + {v_3}}\\ {{w_2}}&{{w_3}} \end{array}} \right|{\bf{i}}$ $ - \left| {\begin{array}{*{20}{c}} {{u_1} + {v_1}}&{{u_3} + {v_3}}\\ {{w_1}}&{{w_3}} \end{array}} \right|{\bf{j}}$ $ + \left| {\begin{array}{*{20}{c}} {{u_1} + {v_1}}&{{u_2} + {v_2}}\\ {{w_1}}&{{w_2}} \end{array}} \right|{\bf{k}}$ $\left( {{\bf{u}} + {\bf{v}}} \right) \times {\bf{w}} = \left( {\left( {{u_2} + {v_2}} \right){w_3} - \left( {{u_3} + {v_3}} \right){w_2}} \right){\bf{i}} - \left( {\left( {{u_1} + {v_1}} \right){w_3} - \left( {{u_3} + {v_3}} \right){w_1}} \right){\bf{j}} + \left( {\left( {{u_1} + {v_1}} \right){w_2} - \left( {{u_2} + {v_2}} \right){w_1}} \right){\bf{k}}$ $\left( {{\bf{u}} + {\bf{v}}} \right) \times {\bf{w}} = \left( {{u_2}{w_3} - {u_3}{w_2} + {v_2}{w_3} - {v_3}{w_2}} \right){\bf{i}}$ $ - \left( {{u_1}{w_3} - {u_3}{w_1} + {v_1}{w_3} - {v_3}{w_1}} \right){\bf{j}}$ $ + \left( {{u_1}{w_2} - {u_2}{w_1} + {v_1}{w_2} - {v_2}{w_1}} \right){\bf{k}}$ $\left( {{\bf{u}} + {\bf{v}}} \right) \times {\bf{w}} = \left( {{u_2}{w_3} - {u_3}{w_2}} \right){\bf{i}} + \left( {{v_2}{w_3} - {v_3}{w_2}} \right){\bf{i}}$ $ - \left( {{u_1}{w_3} - {u_3}{w_1}} \right){\bf{j}} - \left( {{v_1}{w_3} - {v_3}{w_1}} \right){\bf{j}}$ $ + \left( {{u_1}{w_2} - {u_2}{w_1}} \right){\bf{k}} + \left( {{v_1}{w_2} - {v_2}{w_1}} \right){\bf{k}}$ Since ${\bf{u}} \times {\bf{w}} = \left( {{u_2}{w_3} - {u_3}{w_2}} \right){\bf{i}} - \left( {{u_1}{w_3} - {u_3}{w_1}} \right){\bf{j}} + \left( {{u_1}{w_2} - {u_2}{w_1}} \right){\bf{k}}$ ${\bf{v}} \times {\bf{w}} = \left( {{v_2}{w_3} - {v_3}{w_2}} \right){\bf{i}} - \left( {{v_1}{w_3} - {v_3}{w_1}} \right){\bf{j}} + \left( {{v_1}{w_2} - {v_2}{w_1}} \right){\bf{k}}$ Therefore, $\left( {{\bf{u}} + {\bf{v}}} \right) \times {\bf{w}} = {\bf{u}} \times {\bf{w}} + {\bf{v}} \times {\bf{w}}$