If bar{a} and bar{b} are not perpendicular to | vedclass

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(B) Given $\bar{r} 	imes \bar{a} = \bar{b} 	imes \bar{a}$,we can write $\bar{r} 	imes \bar{a} - \bar{b} 	imes \bar{a} = 0$,which implies $(\bar{r}

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$\bar{b} - \left(\frac{\bar{b} \cdot \bar{c}}{\bar{a} \cdot \bar{c}}\right) \bar{a}$

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(B) Given $\bar{r} 	imes \bar{a} = \bar{b} 	imes \bar{a}$,we can write $\bar{r} 	imes \bar{a} - \bar{b} 	imes \bar{a} = 0$,which implies $(\bar{r} - \bar{b}) 	imes \bar{a} = 0$. This means that the vector $(\bar{r} - \bar{b})$ must be parallel to $\bar{a}$. Therefore,we can write $\bar{r} - \bar{b} = k\bar{a}$ for some scalar $k$,or $\bar{r} = \bar{b} + k\bar{a}$. We are also given $\bar{r} \cdot \bar{c} = 0$. Substituting the expression for $\bar{r}$ into this equation,we get $(\bar{b} + k\bar{a}) \cdot \bar{c} = 0$. Expanding this,we have $\bar{b} \cdot \bar{c} + k(\bar{a} \cdot \bar{c}) = 0$. Solving for $k$,we get $k = -\frac{\bar{b} \cdot \bar{c}}{\bar{a} \cdot \bar{c}}$. Substituting this value of $k$ back into the expression for $\bar{r}$,we get $\bar{r} = \bar{b} - \left(\frac{\bar{b} \cdot \bar{c}}{\bar{a} \cdot \bar{c}}ight) \bar{a}$.

If $\bar{a}$ and $\bar{b}$ are not perpendicular to each other,$\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \cdot \bar{c} = 0$,then $\bar{r} =$

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