If the position vectors of the points A, B, C | vedclass

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(A) Given the position vectors of points $A, B, C$ are $\vec{OA} = a$,$\vec{OB} = b$,and $\vec{OC} = 3a - 2b$. First,we find the vectors $\vec{AB}$ an

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Collinear

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(A) Given the position vectors of points $A, B, C$ are $\vec{OA} = a$,$\vec{OB} = b$,and $\vec{OC} = 3a - 2b$. First,we find the vectors $\vec{AB}$ and $\vec{AC}$: $\vec{AB} = \vec{OB} - \vec{OA} = b - a$ $\vec{AC} = \vec{OC} - \vec{OA} = (3a - 2b) - a = 2a - 2b = -2(b - a)$ We observe that $\vec{AC} = -2 \vec{AB}$. Since $\vec{AC}$ is a scalar multiple of $\vec{AB}$,the vectors are parallel. Because they share a common point $A$,the points $A, B, C$ must be collinear.

If the position vectors of the points $A, B, C$ are $a, b, 3a - 2b$ respectively,then the points $A, B, C$ are

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