The points with position vectors bar{a}+bar{b | vedclass

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(D) Let the position vectors of points $A, B$,and $C$ be $\vec{p} = \bar{a}+\bar{b}$,$\vec{q} = \bar{a}-\bar{b}$,and $\vec{r} = \bar{a}+k\bar{b}$ resp

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for all real values of $k$

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(D) Let the position vectors of points $A, B$,and $C$ be $\vec{p} = \bar{a}+\bar{b}$,$\vec{q} = \bar{a}-\bar{b}$,and $\vec{r} = \bar{a}+k\bar{b}$ respectively.For the points to be collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.$\vec{AB} = \vec{q} - \vec{p} = (\bar{a}-\bar{b}) - (\bar{a}+\bar{b}) = -2\bar{b}$.$\vec{BC} = \vec{r} - \vec{q} = (\bar{a}+k\bar{b}) - (\bar{a}-\bar{b}) = (k+1)\bar{b}$.Since $\vec{AB}$ and $\vec{BC}$ are both scalar multiples of the vector $\bar{b}$,they are parallel for any real value of $k$.Specifically,$\vec{AB} = \left( \frac{-2}{k+1} ight) \vec{BC}$ for $k 
eq -1$.If $k = -1$,then $\vec{BC} = 0$,which means point $C$ coincides with point $B$,and the points are still considered collinear.Thus,the points are collinear for all real values of $k$.

The points with position vectors $\bar{a}+\bar{b}$,$\bar{a}-\bar{b}$,and $\bar{a}+k\bar{b}$ are collinear:

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