The diagram shows a parallelogram $ABCD$. $E$ is such a point on $CD$ such that $BD:EB=1:3$. Write expressions for these vectors. (a) DC (b) CD (c) AC (d) AE (e) DE
I solved the first three: $DC = b, \,CD = -b, \,AC = b + a$. I am confused with the next one that requires finding $BE$. I know that $BD$ is $1/4$ of something and $EB$ is $3/4$ of something, but I'm not sure what is that something. Please, help me understand this part. The answer for $AE$ is supposed to be $b+$3/4$a$ I can do the rest of it once I understand it. Thanks!
Edit: added a new picture link of the question
EDIT: Your question and answer don't match. If AE = $b +3a/4$, then the condition on E should be that CE:EB=1:3.
A red flag to me is that the question is using ":" notation when the points involved are not collinear.
As you can see, the picture might have confused you, because in general, the point E might lie outside the line segment CD.
Note the absolute value on $\lambda$, so there's always a positive and a negative solution, corresponding to two points either side of D.