If \(\overrightarrow{{a}}\) and \(\overrightarrow{{b}}\) are vectors in space, given by \(\overrightarrow{{a}}=\frac{\hat{i}-2\hat{j}}{\sqrt[]{5}}\) and \(\overrightarrow{{b}}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt[]{14}}\), then the value of\((2\vec{a} + \vec{b}).[(\vec{a} × \vec{b}) × (\vec{a} – 2\vec{b})]\) is
\(|\vec{a}| = |\vec{b}| = 1\) and \(\vec{a} \cdot \vec{b} = \frac{2 - 2}{\sqrt{70}} = 0\).
\((\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a} = \vec{b}\) and \((\vec{a} \times \vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a} = -\vec{a}\).
So \((\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b}) = \vec{b} + 2\vec{a}\).
\((2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b}) = 4 + 1 + 0 = 5\)
Answer: (C) 5
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