Let bar{a} and bar{b} be the position vectors | vedclass

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(B) Given that $\bar{a}$ and $\bar{b}$ are position vectors of $A$ and $B$. Thus,$\overline{AB} = \bar{b} - \bar{a}$. Given $\overline{AC} = 3 \overli

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$4 \bar{a} - 4 \bar{b}$

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(B) Given that $\bar{a}$ and $\bar{b}$ are position vectors of $A$ and $B$. Thus,$\overline{AB} = \bar{b} - \bar{a}$. Given $\overline{AC} = 3 \overline{AB} = 3(\bar{b} - \bar{a})$. Since $\overline{AC} = \vec{c} - \vec{a}$,we have $\vec{c} - \vec{a} = 3\bar{b} - 3\bar{a}$,which implies $\vec{c} = 3\bar{b} - 2\bar{a}$. Given $\overline{BD} = 2 \overline{BA} = 2(\bar{a} - \bar{b})$. Since $\overline{BD} = \vec{d} - \vec{b}$,we have $\vec{d} - \vec{b} = 2\bar{a} - 2\bar{b}$,which implies $\vec{d} = 2\bar{a} - \bar{b}$. Now,$\overline{CD} = \vec{d} - \vec{c} = (2\bar{a} - \bar{b}) - (3\bar{b} - 2\bar{a}) = 2\bar{a} - \bar{b} - 3\bar{b} + 2\bar{a} = 4\bar{a} - 4\bar{b}$.

Let $\bar{a}$ and $\bar{b}$ be the position vectors of points $A$ and $B$ respectively. $C$ and $D$ are points on the line $AB$ such that $\overline{AC} = 3 \overline{AB}$ and $\overline{BD} = 2 \overline{BA}$. Find the vector $\overline{CD}$.

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