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\( a\) and \(..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />
If <span class="math-tex">\(\vec a\)</span> and <span class="math-tex">\(\vec b\)</span> are vectors in space, given by <span class="math-tex">\(\vec a = \frac {\hat i - 2\hat j}{\sqrt 5}\)</span> and <span class="math-tex">\(\vec b = \frac {2\hat i + \hat j + 3\hat k}{\sqrt {14}}\)</span> then the value of <span class="math-tex">\(\left(2\vec a + \vec b\right)\left[\left(\vec a \times \vec b\right)\times \left(\vec a - 2\vec b\right)\right]\)</span> is \(|\vec{a}| = 1\), \(|\vec{b}| = 1\), and \(\vec{a} \cdot \vec{b} = \frac{2 - 2 + 0}{\sqrt{70}} = 0\). Use \((\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}\): So \((\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b}) = \vec{b} + 2\vec{a}\). \((2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b}) = 4|\vec{a}|^2 + |\vec{b}|^2 + 4\vec{a} \cdot \vec{b} = 4 + 1 + 0 = 5\) Answer: (C) 5
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