For two given vectors bar{a} and bar{b},if th | vedclass

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(A) Given $\overline{A} + \overline{B} = \bar{a}$ and $\overline{A} 	imes \overline{B} = \bar{b}$.Taking the cross product of $\overline{A}$ with the

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$\frac{(\bar{a} 	imes \bar{b})+\bar{a}}{\bar{a}^2}$

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(A) Given $\overline{A} + \overline{B} = \bar{a}$ and $\overline{A} 	imes \overline{B} = \bar{b}$.Taking the cross product of $\overline{A}$ with the first equation: $\overline{A} 	imes (\overline{A} + \overline{B}) = \overline{A} 	imes \bar{a}$.This simplifies to $\overline{A} 	imes \overline{A} + \overline{A} 	imes \overline{B} = \overline{A} 	imes \bar{a}$.Since $\overline{A} 	imes \overline{A} = 0$ and $\overline{A} 	imes \overline{B} = \bar{b}$,we have $\bar{b} = \overline{A} 	imes \bar{a}$,which is $\bar{b} = -(\bar{a} 	imes \overline{A})$.Now,take the cross product of $\bar{a}$ with $\bar{b} = \overline{A} 	imes \bar{a}$:$\bar{a} 	imes \bar{b} = \bar{a} 	imes (\overline{A} 	imes \bar{a})$.Using the vector triple product formula $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$:$\bar{a} 	imes \bar{b} = (\bar{a} \cdot \bar{a})\overline{A} - (\bar{a} \cdot \overline{A})\bar{a}$.Given $\bar{a} \cdot \overline{A} = 1$,we substitute this into the equation:$\bar{a} 	imes \bar{b} = \bar{a}^2 \overline{A} - 1 \cdot \bar{a}$.Rearranging for $\overline{A}$:$\bar{a}^2 \overline{A} = (\bar{a} 	imes \bar{b}) + \bar{a}$.Therefore,$\overline{A} = \frac{(\bar{a} 	imes \bar{b}) + \bar{a}}{\bar{a}^2}$.

For two given vectors $\bar{a}$ and $\bar{b}$,if the vectors $\overline{A}$ and $\overline{B}$ are such that $\overline{A}+\overline{B}=\bar{a}$,$\overline{A} \times \overline{B}=\bar{b}$,and $\overline{A} \cdot \bar{a}=1$,then $\overline{A}=$

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