Question

Sol. [Key Concept: With $a \times \ c = b$, apply $a \times (a \times \ c) = (a \cdot \ c)a - (a \cdot \ a)c = a \times \ b$ to solve for $c$, then evaluate $c \cdot (a - 2b)$.]

Using triple product identity with $\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b}$

$${\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{c}) = \overrightarrow{a} \times \overrightarrow{b} }{(\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{a} - (\overrightarrow{a} \cdot \overrightarrow{a})\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b} }$$As $\overrightarrow{a} \times \overrightarrow{b} = \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 1 & 1 \\ 0 & 1 & - 1 \end{matrix} \right| = - 2\widehat{i} + \widehat{j} + \widehat{k}$

Substituting given values

$${3\overrightarrow{a} - 3\overrightarrow{c} = - 2\widehat{i} + j + k }{3(\widehat{i} + \widehat{j} + \widehat{k}) + 2\widehat{i} + \widehat{j} + \widehat{k} = 3\overrightarrow{c} }{\overrightarrow{c} = \frac{1}{3}(5\widehat{i} + 2\widehat{j} + 2\widehat{k}) }$$Now $\overrightarrow{c} \cdot (\overrightarrow{a} - 2\overrightarrow{\text{ }b}) = \frac{1}{3}(5\widehat{i} + 2\widehat{j} + 2\widehat{k}) \cdot (\widehat{i} - \widehat{j} + 3\widehat{k})$

$$= \frac{1}{3}\lbrack 5 - 2 + 6\rbrack = 3$$