Question
Let $\overrightarrow{a} = - \widehat{i} + 2\widehat{j} + 2\widehat{k},\overrightarrow{b} = 8\widehat{i} + 7\widehat{j} - 3\widehat{k}$ and $\overrightarrow{c}$ be a vector such that $\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b}$. If $\overrightarrow{c} \cdot (\widehat{i} + \widehat{j} + \widehat{k}) = 4$, then $|\overrightarrow{a} + \overrightarrow{c}|^{2}$ is equal to :
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Solution
$\overrightarrow{a} = - \widehat{i} + 2\widehat{j} + 2\widehat{k}$
$$\begin{matrix} & \overrightarrow{b} = 8\widehat{i} + 7\widehat{j} - 3\widehat{k} \\ & \overrightarrow{c} = c_{1}\widehat{i} + c_{2}\widehat{j} + c_{3}\widehat{k} \\ & \overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b} \Rightarrow \left( 2c_{3} - 2c_{2} \right)\widehat{i} + \left( c_{3} + 2c_{1} \right)\widehat{j} - \\ & \left( c_{2} + 2c_{1} \right)\widehat{k} = 8\widehat{i} + 7\widehat{j} - 3\widehat{k} \\ & 2c_{3} - 2c_{2} = 8,c_{3} + 2c_{1} = 7,c_{2} + 2c_{1} = 3 \\ & \left( c_{1}\widehat{i} + c_{2}\widehat{j} + c_{3}\widehat{k} \right) \cdot (\widehat{i} + \widehat{j} + \widehat{k}) = 4 \\ & \ \Rightarrow c_{1} + c_{2} + c_{3} = 4,c_{1} = 2,c_{2} = - 1,c_{3} = 3 \\ & \ |\overrightarrow{a} + \overrightarrow{c}|^{2} = |\widehat{i} + \widehat{j} + 5\widehat{k}|^{2} = 27 \end{matrix}$$
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