If bar{a}=frac{1}{sqrt{10}}(3 hat{i}+hat{k}) | vedclass

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(D) Let the given expression be $E = (2 \bar{a}-\bar{b}) \cdot ((\bar{a} 	imes \bar{b}) 	imes (\bar{a}+2 \bar{b}))$. Using the vector triple product

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$-5$

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(D) Let the given expression be $E = (2 \bar{a}-\bar{b}) \cdot ((\bar{a} 	imes \bar{b}) 	imes (\bar{a}+2 \bar{b}))$. Using the vector triple product identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{v} \cdot \vec{w}) \vec{u}$,we have: $(\bar{a} 	imes \bar{b}) 	imes (\bar{a}+2 \bar{b}) = (\bar{a} \cdot (\bar{a}+2 \bar{b})) \bar{b} - (\bar{b} \cdot (\bar{a}+2 \bar{b})) \bar{a}$. Since $\bar{a}$ and $\bar{b}$ are unit vectors,$\bar{a} \cdot \bar{a} = 1$ and $\bar{b} \cdot \bar{b} = 1$. Also,$\bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}} (3 	imes 2 + 0 	imes 3 + 1 	imes (-6)) = 0$. Thus,$(\bar{a} \cdot (\bar{a}+2 \bar{b})) = \bar{a} \cdot \bar{a} + 2(\bar{a} \cdot \bar{b}) = 1 + 0 = 1$. And $(\bar{b} \cdot (\bar{a}+2 \bar{b})) = \bar{b} \cdot \bar{a} + 2(\bar{b} \cdot \bar{b}) = 0 + 2(1) = 2$. So,$(\bar{a} 	imes \bar{b}) 	imes (\bar{a}+2 \bar{b}) = 1 \bar{b} - 2 \bar{a} = - (2 \bar{a} - \bar{b})$. Now,$E = (2 \bar{a}-\bar{b}) \cdot (-(2 \bar{a}-\bar{b})) = - |2 \bar{a}-\bar{b}|^2$. $|2 \bar{a}-\bar{b}|^2 = (2 \bar{a}-\bar{b}) \cdot (2 \bar{a}-\bar{b}) = 4|\bar{a}|^2 + |\bar{b}|^2 - 4(\bar{a} \cdot \bar{b}) = 4(1) + 1 - 4(0) = 5$. Therefore,$E = -5$.

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

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