If$\mathbf{A}\mathbf{=}\mathbf{2}\mathbf{i}\mathbf{-}\mathbf{j}\mathbf{-}\mathbf{k}$,role="math" localid="1659148191947" $\mathbf{B}\mathbf{=}\mathbf{2}\mathbf{i}\mathbf{-}\mathbf{3}\mathbf{j}\mathbf{+}\mathbf{k}\mathbf{,}$find$\left(A\times B\right)\mathbf{C}\mathbf{,}\mathbf{}\mathbf{,}\mathbf{A}\left(B\times C\right)\mathbf{,}\left(A\times B\right)\mathbf{\times }\mathbf{C}\mathbf{,}\mathbf{A}\mathbf{\times }\left(B\times C\right)\mathbf{,}\left(A\times B\right)\mathbf{\times }\mathbf{C}\mathbf{,}\mathbf{A}\mathbf{\times }\left(B\times C\right)\mathbf{.}$

For Problems 2 to 6, given$\mathbf{A}\mathbf{=}\mathbf{i}\mathbf{+}\mathbf{j}\mathbf{-}\mathbf{2}\mathbf{k}\mathbf{,}\mathbf{B}\mathbf{=}\mathbf{2}\mathbf{i}\mathbf{-}\mathbf{j}\mathbf{+}\mathbf{3}\mathbf{k}\mathbf{,}\mathbf{C}\mathbf{=}\mathbf{j}\mathbf{-}\mathbf{5}\mathbf{k}$

The work is done by using cyclic permutations and using the identity of given parameters.

Now find the value of $\left(\mathrm{A}\times \mathrm{B}\right)\mathrm{C}$

$$\begin{gathered} \left(\mathrm{A}\times \mathrm{B}\right)\mathrm{C}=\left(\left(2\mathrm{i}-\mathrm{j}-\mathrm{k}\right)\times \left(2\mathrm{i}-3\mathrm{j}+\mathrm{k}\right)\right)\left(\mathrm{j}+\mathrm{k}\right) \\ \left(\mathrm{A}\times \mathrm{B}\right)\mathrm{C}=\left(4+3-1\right)\left(\mathrm{j}+\mathrm{k}\right) \\ \left(\mathrm{A}\times \mathrm{B}\right)\mathrm{C}=\left(6\mathrm{j}+6\mathrm{k}\right),\left(1\right) \\ \mathrm{A}\left(\mathrm{B}\times \mathrm{C}\right)=\left(2\mathrm{i}-\mathrm{j}-\mathrm{k}\right)\left(\left(2\mathrm{i}-3\mathrm{j}+\mathrm{k}\right)\times \left(\mathrm{j}+\mathrm{k}\right)\right)\mathrm{a} \\ \mathrm{A}\left(\mathrm{B}\times \mathrm{C}\right)=\left(2\mathrm{i}-\mathrm{j}-\mathrm{k}\right)\left(-3+1\right) \\ \mathrm{A}\left(\mathrm{B}\times \mathrm{C}\right)=\left(-4\mathrm{i}+2\mathrm{j}+2\mathrm{k}\right),\left(2\right) \end{gathered}$$

Solving, $\left(\mathrm{A}\times \mathrm{B}\right)\times \mathrm{C}=\mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)$using the cyclic permutations method

$$\begin{gathered} \mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=\left|\begin{array}{ccc}2& -1& -1\\ 2& 3& 1\\ 0& 1& 1\end{array}\right| \\ =2\left(-3-1\right)-\left(-1\right)\left(2-0\right)-1\left(2-0\right) \\ =-8 \end{gathered}$$

Solving $\mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)$by identity method

$$\begin{gathered} \mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=\left(\mathrm{A}\times \mathrm{C}\right)\mathrm{B}-\left(\mathrm{A}\times \mathrm{B}\right)\mathrm{C} \\ \left(\mathrm{A}\times \mathrm{B}\right)\times \mathrm{C}=-\mathrm{C}\times \left(\mathrm{A}\times \mathrm{B}\right)=-\left(\mathrm{C}\times \mathrm{B}\right)\mathrm{A}+\left(\mathrm{C}\times \mathrm{A}\right)\mathrm{B} \end{gathered}$$

Substituting the values in the equation

$$\begin{gathered} \left(\mathrm{C}\times \mathrm{A}\right)\mathrm{B}=\left(\left(\mathrm{j}+\mathrm{k}\right)\times \left(2\mathrm{i}-\mathrm{j}-\mathrm{k}\right)\right)\left(2\mathrm{i}-3\mathrm{j}+\mathrm{k}\right)) \\ \left(\mathrm{C}\times \mathrm{A}\right)\mathrm{B}=\left(-2\right)\left(2\mathrm{i}-3\mathrm{j}+\mathrm{k}\right) \\ \left(\mathrm{C}\times \mathrm{A}\right)\mathrm{B}=\left(-4\mathrm{i}+6\mathrm{j}-2\mathrm{k}\right)),\left(3\right) \end{gathered}$$

Solving localid="1659151677617" $\left(\mathrm{A}\times \mathrm{B}\right)\times \mathrm{C}$using the identity method

$$\begin{gathered} \left(\mathrm{A}\times \mathrm{B}\right)\times \mathrm{C}=-\left(-4\mathrm{i}+2\mathrm{j}+2\mathrm{k}\right)+\left(-4\mathrm{i}+6\mathrm{j}-2\mathrm{k}\right)) \\ \left(\mathrm{A}\times \mathrm{B}\right)\times \mathrm{C}=\left(4\mathrm{j}-4\mathrm{k}\right) \end{gathered}$$

Simplifying $\mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)$by identity method

$$\begin{gathered} \mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=\left(-4\mathrm{i}+6\mathrm{j}-2\mathrm{k}\right)-\left(6\mathrm{j}+6\mathrm{k}\right) \\ \mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=\left(-4\mathrm{i}+8\mathrm{k}\right) \end{gathered}$$

Therefore, the work done as:

localid="1659152677024" $$\begin{gathered} \left(\mathrm{A}\times \mathrm{B}\right)\mathrm{C}=\left(6\mathrm{j}+6\mathrm{k}\right), \\ \mathrm{A}\left(\mathrm{B}\times \mathrm{C}\right)=\left(-4\mathrm{i}+2\mathrm{j}+2\mathrm{k}\right), \\ \left(\mathrm{A}\times \mathrm{B}\right)\times \mathrm{C}=\mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=-8 \\ \left(\mathrm{A}\times \mathrm{B}\right)\times \mathrm{C}=\left(4\mathrm{j}-4\mathrm{k}\right) \\ \mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=\left(-4\mathrm{j}-8\mathrm{k}\right) \end{gathered}$$