How do I express vectors as linear combinations? Basically, I'm stuck at this question for some time, and it goes like this:
A, B, C, and D are consecutive points of a parallelogram. Point E divides the diagonal AC do that |AE|:|EC| = 1 : 3. Point F divides the diagonal BD so that |BD|:|BF| = 4 : 3. Let S be the point of intersection of line segments AF and ED.
All is good for this part, but here's the part I don't know how to express
Write vector AS as a linear combination of vectors c=AC and f=BD
 A: Let $A$ be the origin. Let $\vec b$, $\vec c$, $\vec d$, be the position vectors of $B, C, D$ respectively wrt A.
 We shall use the section formula for vectors, which says that if a vector with position vector $\vec r$ divides the line segment joining points M($\vec m$) and N($\vec n$) in the ratio p:q, then
$$\vec r =\frac{p \vec n+q \vec m}{p+q}. $$
Hence, the position vector of E is $$\vec e=\frac{\vec c}{4}$$ (considering the position vector of A to be $\vec 0$).
 Similarly, because $BD:BF=4:3$ implies that $BF:FD=3:1$, the position vector of F is $$\vec f=\frac{3\vec d + \vec b}{4}.$$ NOW, any point on the line segment CD, has a position vector $\vec c+\lambda\vec d$. But, AS and AF ($\vec f$) are towards the same direction, so $\vec {AS}\times \vec{AF}=\vec 0$. This means that ($\vec c+\lambda\vec d$)$\times (3\vec d+\vec b)=0$.
 Thus,$$3\vec c\times \vec d + \vec c\times \vec b +\lambda \vec d\times\vec b=\vec 0$$ Now, consider the area of $\Delta ADC$= area of $\Delta ABC = A$. Using the right hand rule to determine the direction of the vectors, you will find that $$6A\hat n - 2A\hat n +2\lambda (-\hat n)=0$$ where $\hat n$ is the unit normal vector into the plane of the screen. Hence $\lambda =2$ so $\vec {AS}= \vec c + 2\vec d$. Note that I used the formula of the area of parallelogram and triangles in vector form. If you know, fine, if you don’t, you can look it up.
Now, if the intersection of AC and BD is O, then $\vec {AO}$ + $\vec {OD}$= $\vec {DA}$, or $\frac{\vec c}{2}+\frac{\vec {BD}}{2} =\vec d.$ Now $$\vec {AS}=\vec c +2(\frac{\vec c}{2}+\frac{\vec {BD}}{2})=2\vec {AC}+\vec{BD}$$ because $\vec c= \vec {AC}$.
 P.S. I feel this (assigning one vertex as the origin and assigning position vectors to the other points wrt this reference origin) is a very useful tool for geometry problems involving a polygon as the main structure such as these.
