Vectors Questions, Worksheets and Revision

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} , thus 

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

We will find \overrightarrow{EF} by doing, 

\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}{8}\times 2\mathbf{a}=\dfrac{3}{4}\mathbf{a} 

Hence, 

\overrightarrow{EF}=\mathbf{b}+2\mathbf{a}+\dfrac{3}{4}\mathbf{a}=2.75\mathbf{a}+\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, 

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

Finally, 

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

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}

Therefore, 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}

Therefore, we’ve shown that 2\overrightarrow{EB}=\overrightarrow{DC}, and thus the two lines must be parallel.