If a→=−110(3i^+k^),b→=17(2i^+3j^−6k→), then the value of (2a→−b→)⋅{(a→×b→)×(a→+2b→)}, is

Clearly, a→and b→ are perpendicular unit vectors. Now,(2a→−b→)⋅{(a→×b→)×(a→+2b→)}=[2a→−b→,a→×b→,a→+2b→]=−[a→×b→,2a→−b→,a→+2b→]=−(a→×b→)⋅{(2a→−b→)×(a→+2b→)}=−(a→×b→)⋅5(a→×b→)=−5|a→×b→|=−5|a→|2|b→|2 [∵a→⊥b→]=−5 [∵|a→|=|b→|=1]

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If $\vec{a} = - \frac{1}{\sqrt{10}} (3 \hat{i} + \hat{k}), \vec{b} = \frac{1}{7} (2 \hat{i} + 3 \hat{j} - 6 \vec{k})$ , then the value of
$(2 \vec{a} - \vec{b}) \cdot {(\vec{a} \times \vec{b}) \times (\vec{a} + 2 \vec{b})}$ , is

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answer is -5.

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Detailed Solution

Clearly, $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors.

Now,

$\begin{array}{l} (2 \vec{a} - \vec{b}) \cdot {(\vec{a} \times \vec{b}) \times (\vec{a} + 2 \vec{b})} \\ = [2 \vec{a} - \vec{b}, \vec{a} \times \vec{b}, \vec{a} + 2 \vec{b}] \\ = - [\vec{a} \times \vec{b}, 2 \vec{a} - \vec{b}, \vec{a} + 2 \vec{b}] \\ = - (\vec{a} \times \vec{b}) \cdot {(2 \vec{a} - \vec{b}) \times (\vec{a} + 2 \vec{b})} \end{array}$