# linear combinations

• Mar 9th 2009, 06:14 PM
mathamatics112
linear combinations
The question is:

In a parallelogram ABCD, F is the midpoint of AD, and E divides BC internally in the ratio 3:2. If P is the point of intersection of AE and BF, show that P divides AE in the ratio 5:6.

Okay so do i make combine all these vector ratios into a single vector and then prove the 5:6 ratio? When i do this though, the vectors do not result into a 5:6 ratio. How should I attempt this question?
• Mar 10th 2009, 06:49 AM
red_dog
Let $\displaystyle \frac{AP}{PE}=k$

$\displaystyle \overrightarrow{BP}=\frac{1}{1+k}\overrightarrow{B A}+\frac{k}{1+k}\overrightarrow{BE}$

$\displaystyle \overrightarrow{BF}=\overrightarrow{BA}+\overright arrow{AF}=\overrightarrow{BA}+\frac{1}{2}\overrigh tarrow{AD}=$

$\displaystyle =\overrightarrow{BA}+\frac{1}{2}\overrightarrow{BC }=\overrightarrow{BA}+\frac{5}{6}\overrightarrow{B E}$

But the vectors BP and BF are coliniar, then $\displaystyle \overrightarrow{BP}=\alpha\overrightarrow{BF}$

$\displaystyle \frac{1}{1+k}\overrightarrow{BA}+\frac{k}{1+k}\ove rrightarrow{BE}=\alpha\overrightarrow{BA}+\frac{5\ alpha}{6}\overrightarrow{BE}$

$\displaystyle \Rightarrow\frac{1}{1+k}=\alpha, \ \frac{k}{1+k}=\frac{5\alpha}{6}\Rightarrow k=\frac{5}{6}$