Question

Find $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$, when
(i) $\overrightarrow{a}=2\hat{i}-3\hat{j},\overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}\mathrm{and}\overrightarrow{c}=3\hat{i}-\hat{k}$
(ii) $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k},\overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k}\mathrm{and}\overrightarrow{c}=\hat{j}+\hat{k}$
(iii) $\overrightarrow{a}=2\hat{i}+3\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k}\mathrm{and}\overrightarrow{c}=-3\hat{i}+\hat{j}+2\hat{k}$

Answer 1

(iii)
$$\begin{gathered} \mathrm{given}\overrightarrow{a}=2\hat{i}+3\hat{j}+\hat{k} \\ \overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k} \\ \overrightarrow{c}=-3\hat{i}+\hat{j}+2\hat{k} \\ \mathrm{then}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\left|\begin{array}{ccc}2& 3& 1\\ 1& -2& 1\\ -3& 1& 2\end{array}\right| \end{gathered}$$

= 2(−4 − 1) −3(2 + 3) + (1 + (−6))

= 2(−5) − 3(5) + 1(−5)

= −10 − 15 − 5

i.e $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=-30$