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(B) Let the vertices of the triangle be $A, B, C$ with position vectors $a, b, c$ respectively. We have the vectors $AB = b - a$ and $AC = c - a$. The

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$	an A$

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(B) Let the vertices of the triangle be $A, B, C$ with position vectors $a, b, c$ respectively. We have the vectors $AB = b - a$ and $AC = c - a$. The expression is $\frac{(a-c) 	imes (b-a)}{(b-a) \cdot (c-a)}$. Note that $a - c = -(c - a) = -AC$. So,the numerator is $(-AC) 	imes (AB) = AC 	imes AB$. The denominator is $(AB) \cdot (AC) = |AB| |AC| \cos A$. The magnitude of the cross product $AC 	imes AB$ is $|AC| |AB| \sin A$. Since the cross product $AC 	imes AB$ is a vector perpendicular to the plane of the triangle,let $\hat{n}$ be the unit vector perpendicular to the plane. Then $AC 	imes AB = |AC| |AB| \sin A \hat{n}$. Thus,the expression is $\frac{|AC| |AB| \sin A \hat{n}}{|AB| |AC| \cos A} = 	an A \hat{n}$. However,in the context of scalar values for such problems,we consider the magnitude or the geometric interpretation. Given the options,the expression simplifies to $	an A$.

If $a, b$ and $c$ are position vectors of the vertices of $\triangle ABC$,then $\frac{(a-c) \times (b-a)}{(b-a) \cdot (c-a)} = $

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