Vectors

First we can find the vector \overrightarrow{AC}

 

\overrightarrow{AC} =\overrightarrow{AB} +\overrightarrow{BC} =3\mathbf{a} +2\mathbf{b}

 

Since M is the midpoint of AC we know that \overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AC}  

 

\overrightarrow{AM}=\dfrac{1}{2}(3\mathbf{a} +2\mathbf{b})  = \dfrac{3}{2}\mathbf{a} +\mathbf{b}

The vector \overrightarrow{EF} is the same as:

 

\overrightarrow{EF}=\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CF}

 

First, we can find the vector \overrightarrow{ED} as \dfrac{1}{2}\mathbf{b} .

 

Then we also know \overrightarrow{DC} = \overrightarrow{AB}, hence \overrightarrow{DC}=2\mathbf{a} .

 

The final section \overrightarrow{CF} = \dfrac{3}{5}\times 2\mathbf{a}=\dfrac{6}{5}\mathbf{a}

 

Hence,

 

\overrightarrow{EF}= \dfrac{1}{2} \mathbf{b}+2\mathbf{a}+\dfrac{6}{5}\mathbf{a}=3.2\mathbf{a}+ \dfrac{1}{2} \mathbf{b}

First we can find the vector \overrightarrow{BC}

 

\overrightarrow{BC} =-\overrightarrow{AB} +\overrightarrow{AC} =-\mathbf{a} +2\mathbf{b}

 

Then we can find:

 

\overrightarrow{BD} =\dfrac{3}{5}\overrightarrow{BC}  =-\dfrac{3}{5}\mathbf{a} +\dfrac{6}{5}\mathbf{b}

 

Next:

 

\begin{aligned} \overrightarrow{AD} &=\overrightarrow{AB}+\overrightarrow{BD}  \\ &=\mathbf{a} +\bigg(-\dfrac{3}{5}\mathbf{a} +\dfrac{6}{5}\mathbf{b}\bigg) \\ & =\dfrac{2}{5}\mathbf{a}+\dfrac{6}{5}\mathbf{b} \end{aligned}

 

Finally:

 

\begin{aligned} \overrightarrow{AE} &=\dfrac{1}{3}\overrightarrow{AD} \\ &=\dfrac{1}{3}\bigg(\dfrac{2}{5}\mathbf{a}+\dfrac{6}{5}\mathbf{b}\bigg) \\ &= \dfrac{2}{15}\mathbf{a}+\dfrac{2}{5}\mathbf{b} \end{aligned}

We will find \overrightarrow{AC} by doing,

 

\overrightarrow{AC}=\overrightarrow{AB}-\overrightarrow{OB}+\overrightarrow{OC}

 

Thus

 

\overrightarrow{AC} =\mathbf{b}-\mathbf{a}+\mathbf{b}= 2\mathbf{b}-\mathbf{a}

We will find \overrightarrow{EB} by doing

 

\overrightarrow{EB}=\overrightarrow{EA}+\overrightarrow{AB}

 

The first vector is straightforward, because we know \overrightarrow{AE}, and that is just the same vector in the opposite direction. So, we get:

 

\overrightarrow{EA}=-\overrightarrow{AE}=-(3\mathbf{a}-2\mathbf{b})=-3\mathbf{a}+2\mathbf{b}

 

Now we need \overrightarrow{AB}. Since B is the midpoint of AC (given in the question), we must have that \overrightarrow{AB}=\frac{1}{2} \overrightarrow{AC}. Therefore, looking at the diagram, we get that:

 

\overrightarrow{AB}=\dfrac{1}{2}\overrightarrow{AC}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{DC}\right)

 

We’re given the second part of this, \overrightarrow{DC}=2\mathbf{a}+4\mathbf{b}, and since E is the midpoint of AD, we can also work out the first part:

 

\overrightarrow{AD}=2\overrightarrow{AE}=2(3\mathbf{a}-2\mathbf{b})=6\mathbf{a}-4\mathbf{b}

 

Now, we have everything we need and can go back through our work, filling in the gaps we find:

 

\overrightarrow{AB}=\dfrac{1}{2}\left(6\mathbf{a}-4\mathbf{b}+2\mathbf{a} +4\mathbf{b}\right)=\dfrac{1}{2}\left(8\mathbf{a}\right)=4\mathbf{a}

 

Finally we have that:

 

\overrightarrow{EB}=\overrightarrow{EA}+\overrightarrow{AB}=-3\mathbf{a}+2\mathbf{b}+4\mathbf{a}=\mathbf{a}+2\mathbf{b}

 

If \overrightarrow{EB} and \overrightarrow{DC} are parallel, then one must be a multiple of the other. Well, if we multiply \overrightarrow{EB} by 2 then,

 

2\times\overrightarrow{EB}=2(\mathbf{a}+2\mathbf{b})=2\mathbf{a}+4\mathbf{b}=\overrightarrow{DC}

 

We have shown that 2\overrightarrow{EB}=\overrightarrow{DC}, therefore, the two lines must be parallel.