Challenging (or not) paralelogram problem - vector as a linear combination of other vectors

Question

I'm having the hardest time on this particular problem:

On the paralelogram ABCD we have

$\vec{DE} = \alpha \ \vec{DC}$

$3 \ \vec{BF} = \vec{FC}$

$G$ is the intersection between $AE$ and $DF$

What is the value of $\alpha$ so that $\vec{AG} = \ 4 \vec{GE}$? Also, write $\vec{w}$ as a linear combination of $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{AD}$.

Answer (not complete)

$$ \vec{GE} + \vec{ED} + \vec{DG} = 0 $$

$$ \frac{1}{5} (\vec{v} + \alpha \vec{u}) - \alpha \vec{u} + \vec{DG} = 0 $$

If anyone can give me a hint I appreciate.

Thank you. Best Regards.

Answer 1

We can take the position of point $A$ as $\vec{0}$ without loss of generality. Take the position vectors of the points $E$,$F$ and $G$ as $\vec{e}$,$\vec{f}$ and $\vec{g}$ respectively. Now first use the conditions given in the problem.

1. $\vec{DE}=\alpha\vec{DC}\implies e=\alpha\vec{u}+\vec{v}$
2. $3 \ \vec{BF} = \vec{FC}\implies \vec{f}=\dfrac v4+u$
3. $\vec{AG} = \ 4 \vec{GE}\implies \vec{g}=\dfrac {4e}5=\frac 45(\alpha\vec{u}+\vec{v})$

Now, use the condition that $G$ lies on the line segment $DF$. This will give you the equation $\vec{g}=\lambda \vec{v}+(1-\lambda)\vec{f}$ for some $\lambda\in\mathbb{R}$. Compraing this with the expression for $\vec{g}$ that we found in point 3, we can find the values of $\alpha$ and $\lambda$.

Answer 2

Hint: $\vec0=\vec{GE}+\vec{ED}+\vec{DG}$. Now express those three vectors in terms of $\vec u$ and $\vec v$ and exploit their linear independence.