Question

If $\overrightarrow{a}=2\hat{i}+5\hat{j}-7\hat{k},\overrightarrow{b}=-3\hat{i}+4\hat{j}+\hat{k}\mathrm{and}\overrightarrow{c}=\hat{i}-2\hat{j}-3\hat{k},$ compute $\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}\mathrm{and}\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)$ and verify that these are not equal.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \overrightarrow{a}=2\hat{i}+5\hat{j}-7\hat{k} \\ \overrightarrow{b}=-3\hat{i}+4\hat{j}+\hat{k} \\ \overrightarrow{c}=\hat{i}-2\hat{j}-3\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 5& -7\\ -3& 4& 1\end{array}\right| \\ =\left(5+28\right)\hat{i}-\left(2-21\right)\hat{j}+\left(8+15\right)\hat{k} \\ =33\hat{i}+19\hat{j}+23\hat{k} \\ \Rightarrow \left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 33& 19& 23\\ 1& -2& -3\end{array}\right| \\ =\left(-57+46\right)\hat{i}-\left(-99-23\right)\hat{j}+\left(-66-19\right)\hat{k} \\ \Rightarrow \left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}=-11\hat{i}+122\hat{j}-85\hat{k}...\left(1\right) \\ \therefore \overrightarrow{b}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& 4& 1\\ 1& -2& -3\end{array}\right| \\ =\left(-12+2\right)\hat{i}-\left(9-1\right)\hat{j}+\left(6-4\right)\hat{k} \\ =-10\hat{i}-8\hat{j}+2\hat{k} \\ \Rightarrow \overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 5& -7\\ -10& -8& 2\end{array}\right| \\ =\left(10-56\right)\hat{i}-\left(4-70\right)\hat{j}+\left(-16+50\right)\hat{k} \\ \Rightarrow \overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=-46\hat{i}+66\hat{j}+34\hat{k}...\left(2\right) \\ \text{From (1) and (2), we get} \\ \left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}\ne \overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right) \end{gathered}$$