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

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(D) Given $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$.First,note that $|\vec{a}|^2

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$-5$

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(D) Given $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$.First,note that $|\vec{a}|^2 = \frac{1}{10}(3^2 + 1^2) = 1$ and $|\vec{b}|^2 = \frac{1}{49}(2^2 + 3^2 + (-6)^2) = \frac{4+9+36}{49} = 1$.Also,$\vec{a} \cdot \vec{b} = \frac{1}{7\sqrt{10}}(3(2) + 0(3) + 1(-6)) = 0$.Now,expand the expression: $(2\vec{a} - \vec{b}) \cdot [(\vec{a} 	imes \vec{b}) 	imes (\vec{a} + 2\vec{b})]$.Using the vector triple product formula $\vec{u} 	imes (\vec{v} + \vec{w}) = \vec{u} 	imes \vec{v} + \vec{u} 	imes \vec{w}$,we have:$(\vec{a} 	imes \vec{b}) 	imes (\vec{a} + 2\vec{b}) = (\vec{a} 	imes \vec{b}) 	imes \vec{a} + 2((\vec{a} 	imes \vec{b}) 	imes \vec{b})$.Using $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we get:$(\vec{a} 	imes \vec{b}) 	imes \vec{a} = -(\vec{a} 	imes (\vec{a} 	imes \vec{b})) = -((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) = - (0\vec{a} - 1\vec{b}) = \vec{b}$.$2((\vec{a} 	imes \vec{b}) 	imes \vec{b}) = 2((\vec{a} \cdot \vec{b})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a}) = 2(0\vec{b} - 1\vec{a}) = -2\vec{a}$.So,the expression becomes $(2\vec{a} - \vec{b}) \cdot (\vec{b} - 2\vec{a})$.$= -(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b}) = -|2\vec{a} - \vec{b}|^2$.$= -(4|\vec{a}|^2 + |\vec{b}|^2 - 4\vec{a} \cdot \vec{b}) = -(4(1) + 1 - 4(0)) = -5$.

If $\vec{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\vec b = \frac{1}{7}(2 \hat{i} + 3 \hat{j} - 6 \hat{k})$,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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