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(A) Let the three points be $A$,$B$,and $C$ with position vectors $\vec{OA} = \vec{a}$,$\vec{OB} = \vec{b}$,and $\vec{OC} = 3\vec{a} - 2\vec{b}$.To ch

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Collinear

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(A) Let the three points be $A$,$B$,and $C$ with position vectors $\vec{OA} = \vec{a}$,$\vec{OB} = \vec{b}$,and $\vec{OC} = 3\vec{a} - 2\vec{b}$.To check if the points are collinear,we examine the vectors $\vec{AB}$ and $\vec{BC}$.$\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$.$\vec{BC} = \vec{OC} - \vec{OB} = (3\vec{a} - 2\vec{b}) - \vec{b} = 3\vec{a} - 3\vec{b} = 3(\vec{a} - \vec{b})$.Since $\vec{BC} = -3(\vec{b} - \vec{a}) = -3\vec{AB}$,the vectors $\vec{AB}$ and $\vec{BC}$ are parallel.Because they share a common point $B$,the points $A$,$B$,and $C$ must be collinear.

If the position vectors of three points are $a$,$b$,and $(3a - 2b)$,then these points are .....

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