Question

Prove the property\((a + b) \times c = a \times c + b \times c\).

Short answer

The property \(({\bf{a}} + {\bf{b}}) \times {\bf{c}} = {\bf{a}} \times {\bf{c}} + {\bf{b}} \times {\bf{c}}\) is proved.

Step-by-step solution

Step 1: Formula used

Consider,\({\rm{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \),\({\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)and\({\rm{c}} = \left\langle {{c_1},{c_2},{c_3}} \right\rangle \)

Calculate the cross product of \(({\bf{a}} + b) \times c\)

Let

\(\begin{array}{l}{\bf{a}} = {a_1}{\bf{i}} + {a_2}{\bf{j}} + {a_3}{\bf{k}}\\{\bf{b}} = {b_1}{\bf{i}} + {b_2}{\bf{j}} + {b_3}{\bf{k}}\\{\bf{c}} = {c_1}{\bf{i}} + {c_2}{\bf{j}} + {c_3}{\bf{k}}\end{array}\)

\(({\bf{a}} + {\bf{b}}) \times {\bf{c}} = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1} + {b_1}}&{{a_2} + {b_2}}&{{a_3} + {b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\)

Using property of determinants, it can also written as\( = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right| = {\bf{a}} \times {\bf{c}} + {\bf{b}} \times {\bf{c}}\)

Hence, proved that the property,\(({\bf{a}} + {\bf{b}}) \times {\bf{c}} = {\bf{a}} \times {\bf{c}} + {\bf{b}} \times {\bf{c}}\).

Let,

\(\begin{array}{l}{\bf{a}} = {a_1}{\bf{i}} + {a_2}{\bf{j}} + {a_3}{\bf{k}}\\{\bf{b}} = {b_1}{\bf{i}} + {b_2}{\bf{j}} + {b_3}{\bf{k}}\\{\bf{c}} = {c_1}{\bf{i}} + {c_2}{\bf{j}} + {c_3}{\bf{k}}\end{array}\)

\({\bf{a}} \times ({\bf{b}} + {\bf{c}}) = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1} + {c_1}}&{{b_2} + {c_2}}&{{b_3} + {c_3}}\end{array}} \right|\)

Using property of determinants, we can write

\( = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right| = {\bf{a}} \times {\bf{b}} + {\bf{a}} \times {\bf{c}}\)

Hence, we proved that

\({\bf{a}} \times ({\bf{b}} + {\bf{c}}) = {\bf{a}} \times {\bf{b}} + {\bf{a}} \times {\bf{c}}\)