1. Expressing Vectors

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!

2. Hello john-1
Originally Posted by john-1
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!
That answer is not correct. It should be:
$\vec{AE} = \vec{AC}+\vec{CD}+\vec{DE}$
$=\vec{AC}-\vec{AB}-\tfrac12\vec{AC}$

$=\tfrac12\vec{AC} -\vec{AB}$

3. Hi Grandad, would you mind retyping your answer? I can't see it because of the LaTeX Error ^

4. Originally Posted by john-1
Hi Grandad, would you mind retyping your answer? I can't see it because of the LaTeX Error ^
Since the forum is being updated, LaTeX is not working yet, but you can still see what Grandad typed by quoting him. He typed:

$$\vec{AE} = \vec{AC}+\vec{CD}+\vec{DE}$$
$$=\vec{AC}-\vec{AB}-\tfrac12\vec{AC}$$

$$=\tfrac12\vec{AC} -\vec{AB}$$

I could find some way to make an image (or you could), but maybe you can see what it should look like already.

5. Using this site, I got this: (click on image for full size)

Edit: This site seems better for converting LaTeX to image. The way to get a newline is double backslash, \\.

6. Hello, john-1

$ABCDE$ is a pentagon such that $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AC} = 2\!\cdot\!\overrightarrow{ED}.$

Write $\overrightarrow{AE}$ in terms of $\overrightarrow{AB}\text{ and }\overrightarrow{AC}.$

The answer they have is: . $-\overrightarrow{AC} -\tfrac{1}{2}\overrightarrow{AC}$ . This can't be right!

We have: . $\overrightarrow{AE} \;=\;\overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DE}$

. . . . . . . . . . . $=\;\overrightarrow{AC} - \overrigjhtarrow{DC} - \overrightarrow{ED}$

. . . . . . . . . . . $=\;\overrightarrow{AC} - \overrightarrow{AB} - \frac{1}{2}\overrightarrow{AC}$

. . . . . . . . . . . $=\;\frac{1}{2}\overrightarrow{AC} - \overrightarrow{AB}$