Vectors - Finding the midpoint

**brumby_3** · March 9th 2011, 09:32 PM

Find the midpoint of AD.

The position vectors of points A, B, D are respectively a, b, -a+2b.

I know -a+2b is equivalent to points (-1,2), but not sure what to do from here. I know how to do it when I know the points of A, but there's no numbers, so that's why I'm confused! Help is much appreciated.

**BAdhi** · March 9th 2011, 09:40 PM

$\overrightarrow{AD}=(-\underline{a}+2\underline{b})-\underline{a}=2(-\underline{a}+\underline{b})$

if mid point of $\overrightarrow{AD}$ is $E$

then $\overrightarrow{OE}=\overrightarrow{OA}+\overright arrow{AE}=\overrightarrow{OA}+\frac{\overrightarro w{AD}}{2}$

**brumby_3** · March 9th 2011, 09:50 PM

Originally Posted by BAdhi

$\overrightarrow{AD}=(-\underline{a}+2\underline{b})-\underline{a}=2(-\underline{a}+\underline{b})$

if mid point of $\overrightarrow{AD}$ is $E$

then $\overrightarrow{OE}=\overrightarrow{OA}+\overright arrow{AE}=\overrightarrow{OA}+\frac{\overrightarro w{AD}}{2}$

So is the answer [2(-a+b)/2] = -a+b or am I not reading it right?

**earboth** · March 9th 2011, 11:20 PM

Originally Posted by brumby_3

So is the answer [2(-a+b)/2] = -a+b or am I not reading it right?

Unfortunately that's wrong.

1. As BAdhi wrote you have to calculate:

$\overrightarrow{OE}= \overrightarrow{OA}+\frac12 \cdot \overrightarrow{AD}$

which simplifies to:

$\overrightarrow{OE}=\vec a + \frac12 (2(-\vec a + \vec b))=\vec b$

2. In short: The staionary vector of the midpoint of a distance is the mean of the startpoint and the endpoint vectors of the distance:

$\overrightarrow{OE}=\dfrac{\vec a + (-\vec a + 2\vec b)}2 = \vec b$

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