ABCDE is a pentagon such that AB = DC and AC=2ED. Write each vector in terms of AB and AC.
AE = ?
The answer they have is -AC -1/2AC
I can't see to figure out why this is correct. Can someone please explain it to me? Thanks!
Since the forum is being updated, LaTeX is not working yet, but you can still see what Grandad typed by quoting him. He typed:
[tex]\vec{AE} = \vec{AC}+\vec{CD}+\vec{DE}[/tex]
[tex]=\vec{AC}-\vec{AB}-\tfrac12\vec{AC}[/tex]
[tex]=\tfrac12\vec{AC} -\vec{AB}[/tex]
I could find some way to make an image (or you could), but maybe you can see what it should look like already.
Hello, john-1
$\displaystyle ABCDE$ is a pentagon such that $\displaystyle \overrightarrow{AB} = \overrightarrow{DC}$ and $\displaystyle \overrightarrow{AC} = 2\!\cdot\!\overrightarrow{ED}.$
Write $\displaystyle \overrightarrow{AE}$ in terms of $\displaystyle \overrightarrow{AB}\text{ and }\overrightarrow{AC}. $
The answer they have is: .$\displaystyle -\overrightarrow{AC} -\tfrac{1}{2}\overrightarrow{AC}$ . This can't be right!
We have: .$\displaystyle \overrightarrow{AE} \;=\;\overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DE} $
. . . . . . . . . . . $\displaystyle =\;\overrightarrow{AC} - \overrigjhtarrow{DC} - \overrightarrow{ED} $
. . . . . . . . . . . $\displaystyle =\;\overrightarrow{AC} - \overrightarrow{AB} - \frac{1}{2}\overrightarrow{AC} $
. . . . . . . . . . . $\displaystyle =\;\frac{1}{2}\overrightarrow{AC} - \overrightarrow{AB}$