Question

Sol. [Key Concept: $2(a \times \ b) + 3(b \times \ c) = 0$ makes $bparallel(2a - 3c)$, so $c = \frac{2a - \lambda\ b}{3}$; use $a \cdot \ c = 15$ to find $\lambda$, then $c \cdot (i + j - 3k) = - 5$.]

$$\ 2(\overrightarrow{a} \times \overrightarrow{b}) - 3(\overrightarrow{c} \times \overrightarrow{b}) = \overrightarrow{0}$$

$$(2\overrightarrow{a} - 3\overrightarrow{c}) \times \overrightarrow{b} = \overrightarrow{0}$$

$$\overrightarrow{b}\|(2\overrightarrow{a} - 3\overrightarrow{c})$$

$$2\overrightarrow{a} - 3\overrightarrow{c} = \lambda\overrightarrow{b}$$

$$\begin{matrix} & & \ \\ & & \\ & & \\ & \overrightarrow{c} = \frac{2\overrightarrow{a} - \lambda\overrightarrow{b}}{3} & \text{(1)} \\ & \overrightarrow{a} \cdot \overrightarrow{c} = 15 & \text{(1)} \\ & \left( \frac{2\overrightarrow{a} - \lambda\overrightarrow{b}}{3} \right) \cdot \overrightarrow{a} = 15 & \text{(1)} \\ & \frac{2|a|^{2} - \lambda\overrightarrow{b} \cdot \overrightarrow{a}}{3} = 15 & \text{(1)} \\ & 2(26) - \lambda(40 - 2 - 3) = 45 & \text{(1)} \\ & \lambda = \frac{1}{5} & \text{(1)} \\ & \ \Rightarrow \overrightarrow{c} \cdot (\widehat{i} + \widehat{j} - 3\widehat{k}) = \frac{\left( 2(4\widehat{i} - \widehat{j} + 3k) - \frac{1}{5}(10\widehat{i} + 2\widehat{j} - k) \cdot (\widehat{i} + \widehat{j} - 3\widehat{k}) \right)}{3} & \text{(1)} \\ & \ = \frac{2(4 - 1 - 9) - \frac{1}{5}(10 + 2 + 3)}{3} = - 5 & \text{(1)} \end{matrix}$$