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

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(B) Let the given expression be $E = (\overline{a}-2 \overline{b}) \cdot \{(\overline{a} 	imes \overline{b}) 	imes (2 \overline{a}+\overline{b})\}$.

Vedclass · Accepted Answer

$-5$

Vedclass · Answer

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

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

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