# Linear combination and a proof for the area.

• Apr 8th 2011, 09:03 AM
mechaniac
Linear combination and a proof for the area.
http://dl.dropbox.com/u/25711584/algebra.jpg

$|DF|=3|AF|$

$|BE|=\frac{1}{2}|BC|$

* express $\bar{AG}$ as a linear combination of the vectors u= $\bar{AB}$ and v= $\bar{AD}$.

* Show that the triangle AFG always constitutes the same fraction of the area of the Parallelogram ABCD.

i need help here. On the first question i have tryed to express the vector AG with other vectors to finally come to an linear combination of u and v. But i get problems whit any vector that ends at G. I suppose i should use the symmetry of the triangles somehow but dont really know how?

the other question i need some tips to get started.

Thanks.
• Apr 8th 2011, 10:24 AM
veileen
$\Delta AGF$ ~ $\Delta EGB$ (You know why?) $\Rightarrow \frac{AG}{EG}=\frac{AF}{EB}=\frac{GF}{BG}$
$AF=\frac{AD}{3}, EB=\frac{BC}{2}$ and $AD=BC \Rightarrow \frac{AF}{EB}=\frac{GF}{BG}=\frac{2}{3}$

Now this should be easy.
• Apr 8th 2011, 10:42 AM
earboth
Quote:

Originally Posted by mechaniac
http://dl.dropbox.com/u/25711584/algebra.jpg

$|DF|=3|AF|$

$|BE|=\frac{1}{2}|BC|$

* express $\bar{AG}$ as a linear combination of the vectors u= $\bar{AB}$ and v= $\bar{AD}$.

* Show that the triangle AFG always constitutes the same fraction of the area of the Parallelogram ABCD.

i need help here. On the first question i have tryed to express the vector AG with other vectors to finally come to an linear combination of u and v. But i get problems whit any vector that ends at G. I suppose i should use the symmetry of the triangels somehow but dont really know how?

the other question i need some tips to get started.

Thanks.

1. According to your sketch:

$\overrightarrow{AD} = \overrightarrow{BC}$

$\overrightarrow{AF}=\frac14 \overrightarrow{AD}$

2. $\overrightarrow{AG} = t \cdot \overrightarrow{AE} = t \cdot (\overrightarrow{AB} + \frac12 \overrightarrow{BC})$

$\overrightarrow{AG}= \overrightarrow{AB} + k \cdot \overrightarrow{BF} = \overrightarrow{AB} + k \cdot (-\overrightarrow{AB}+\frac14 \overrightarrow{AD})$

3. Since $\overrightarrow{BC} = \overrightarrow{AD}$ you'll get:

$t \cdot (\overrightarrow{AB} + \frac12 \overrightarrow{BC}) = \overrightarrow{AB} + k \cdot (-\overrightarrow{AB}+\frac14 \overrightarrow{BC})$

Expand both sides of the equation and compare the co-efficient of equal vectors:

$\left|\begin{array}{rcl}t&=&1-k \\ \frac12 t&=&\frac14 k\end{array}\right.$

Solve this system of equations for t and k.

4. Plug in the value for t or k into the appropriate equation to get the vector $\overrightarrow{AG}$
• Apr 8th 2011, 12:08 PM
mechaniac
Thanks for the help! i used determinants to show the area problem so dont need help on that :)