Question

Show the \(a \times (b \times c) \ne (a \times b) \times c\).

Short answer

\(a \times (b \times c) \ne (a \times b) \times c\) is shown.

Step-by-step solution

Formula used

Consider the general expression to find the cross product of\({\bf{a}}\)and\({\bf{b}}\).

\({\rm{a}} \times {\rm{b}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right| \ldots \ldots \ldots (1)\)

Modify equation (1)

\({\rm{b}} \times {\rm{c}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\)

Find the \(a \times (b \times c)\)

As given, \({\rm{a}} = \langle 1,0,1\rangle ,{\rm{b}} = \langle 2,1, - 1\rangle \) and \({\rm{c}} = \langle 0,1,3\rangle \)

Substitute 2 for \({b_1},2\) for \({b_2}, - 1\) for \({b_3},0\) for \({c_1},1\) for \({c_2}\) and 3 for \({c_3}\),

\(\begin{array}{l}{\rm{b}} \times {\rm{c}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\2&1&{ - 1}\\0&1&3\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&{ - 1}\\1&3\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}2&{ - 1}\\0&3\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}2&1\\0&1\end{array}} \right|{\rm{k}}\\{\rm{b}} \times {\rm{c}} = (3 + 1){\rm{i}} - (6 + 0){\rm{j}} + (2 - 0){\rm{k}}\\{\rm{b}} \times {\rm{c}} = 4{\rm{i}} - 6{\rm{j}} + 2{\rm{k}}\end{array}\)

Find \(a \times (b \times c)\)

\(\begin{array}{l}a \times (b \times c) = \left| {\begin{array}{*{20}{c}}i&j&k\\1&0&1\\4&{ - 6}&2\end{array}} \right|\\a \times (b \times c) = \left| {\begin{array}{*{20}{c}}0&1\\{ - 6}&2\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}1&1\\4&2\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}1&0\\4&{ - 6}\end{array}} \right|k\\a \times (b \times c) = (0 + 6)i - (2 - 4)j + ( - 6 - 0)k\\a \times (b \times c) = 6i + 2j - 6k \ldots \ldots \ldots (2)\end{array}\)

Substitute 1 for \({a_1},0\) for \({a_2},1\) for \({a_3},2\) for \({b_1},2\) for \({b_2}\) and \( - 1\) for \({b_3}\) in equation (1),

\(\begin{array}{l}a \times b = \left| {\begin{array}{*{20}{c}}i&j&k\\1&0&1\\2&1&{ - 1}\end{array}} \right|\\a \times b = \left| {\begin{array}{*{20}{c}}0&1\\1&{ - 1}\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}1&1\\2&{ - 1}\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}1&0\\2&1\end{array}} \right|k\\a \times b = (0 - 1)i - ( - 1 - 2)j + (1 - 0)k\\a \times b = - i + 3j + k\end{array}\)

Find the \((a \times b) \times c\)

Find \((a \times b) \times c\)

\(\begin{array}{l}(a \times b) \times c = \left| {\begin{array}{*{20}{c}}i&j&k\\{ - 1}&3&1\\0&1&3\end{array}} \right|\\(a \times b) \times c = \left| {\begin{array}{*{20}{l}}3&1\\1&3\end{array}} \right|i - \left| {\begin{array}{*{20}{c}}{ - 1}&1\\0&3\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}{ - 1}&3\\0&1\end{array}} \right|k\\(a \times b) \times c = (9 - 1)i - ( - 3 - 0)j + ( - 1 - 0)k\\(a \times b) \times c = 8i + 3j - k \ldots \ldots \ldots (3)\end{array}\)

Since equation (2) and equation (3) are not equal, then the given condition \(a \times (b \times c) \ne (a \times b) \times c\)is obtained.

Thus, \(a \times (b \times c) \ne (a \times b) \times c\) is shown.