It is given that $\overrightarrow{PQ}=\begin{pmatrix}-5\\6\end{pmatrix}$ and the position vector of $P$ is $\begin{pmatrix}2\\-4\end{pmatrix}$. Find the magnitude $\left|\overrightarrow{PQ}\right|$. Answer ... units the coordinates of $Q$. Answer (... , ...) ABCDEF is a regular hexagon. G is a point on DC extended such that $2DG = 3DC$. $\overrightarrow{AD}=\mathbf{x}$ and $\overrightarrow{AE}=\mathbf{y}$. Express, as simply as possible, in terms of $\mathbf{x}$ and/or $\mathbf{y}$, Express, as simply as possible, in terms of $\mathbf{x}$ and/or $\mathbf{y}$, $\overrightarrow{AB}$ $\overrightarrow{AF}$ Answer ... $\overrightarrow{BG}$ Answer ... M is the mid-point of AF.
H is a point on BE where 4HE = 3BE. Find $\overrightarrow{HG}$. Answer ... Using vectors, determine whether H lies on MG.

1. Find the magnitude of the given vector

Given $\overrightarrow{PQ}=\begin{pmatrix}-5\\6\end{pmatrix}$,

$|\overrightarrow{PQ}|=\sqrt{(-5)^2+6^2}=\sqrt{25+36}=\sqrt{61}$

1. Write the position vector of $Q$ using the position vector of $P$ and $\overrightarrow{PQ}$.

$\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}$

2. Add the corresponding components and convert the position vector to coordinates.

$\overrightarrow{OQ}=\begin{pmatrix}2\\-4\end{pmatrix}+\begin{pmatrix}-5\\6\end{pmatrix}=\begin{pmatrix}-3\\2\end{pmatrix}$
Hence the coordinates of $Q$ are $(-3,2)$.

1. Express \(\overrightarrow{AB}\) using the given vectors \(\overrightarrow{AD}\) and \(\overrightarrow{AE}\).

Since
$\overrightarrow{AE}=\overrightarrow{AD}+\overrightarrow{DE}$,

$\overrightarrow{DE}=\mathbf{y}-\mathbf{x}$.

In a regular hexagon, $\overrightarrow{AB}=\overrightarrow{ED}$, so

$\overrightarrow{AB}=-(\overrightarrow{DE})=-(\mathbf{y}-\mathbf{x})=\mathbf{x}-\mathbf{y}$.

1. Express $\overrightarrow{AF}$ using the midpoint of $AD$ and the given vector $\overrightarrow{AE}$.

Let $O$ be the centre of the regular hexagon. Since $AD$ is a diagonal joining opposite vertices,

$\overrightarrow{AO}=\frac{1}{2}\overrightarrow{AD}=\frac{1}{2}\mathbf{x}.$

In a regular hexagon, $\overrightarrow{AF}=\overrightarrow{OE}$.

Therefore

$\overrightarrow{AF}=\overrightarrow{OE}=\overrightarrow{AE}-\overrightarrow{AO}=\mathbf{y}-\frac{1}{2}\mathbf{x}.$

1. Express $\overrightarrow{DC}$ using the known side vectors of the regular hexagon.

From earlier parts, $\overrightarrow{AB}=\mathbf{x}-\mathbf{y}$ and $\overrightarrow{BC}=\dfrac{1}{2}\mathbf{x}$.

Since
$\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}$,
$\mathbf{x}=(\mathbf{x}-\mathbf{y})+\dfrac{1}{2}\mathbf{x}+\overrightarrow{CD}$
$\overrightarrow{CD}=\mathbf{y}-\dfrac{1}{2}\mathbf{x}$
Hence
$\overrightarrow{DC}=\dfrac{1}{2}\mathbf{x}-\mathbf{y}.$

2. Use $2DG=3DC$ to find $\overrightarrow{DG}$, then add vectors along $B \to D \to G$.

Given $2DG=3DC$ and $G$ is on $DC$ produced,
$\overrightarrow{DG}=\dfrac{3}{2}\overrightarrow{DC}$
$=\dfrac{3}{2}\left(\dfrac{1}{2}\mathbf{x}-\mathbf{y}\right)$
$=\dfrac{3}{4}\mathbf{x}-\dfrac{3}{2}\mathbf{y}.$

Now
$\overrightarrow{BG}=\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DG}.$
Since $\overrightarrow{BA}=-(\mathbf{x}-\mathbf{y})=\mathbf{y}-\mathbf{x}$,
$\overrightarrow{BG}=(\mathbf{y}-\mathbf{x})+\mathbf{x}+\left(\dfrac{3}{4}\mathbf{x}-\dfrac{3}{2}\mathbf{y}\right)$
$=\dfrac{3}{4}\mathbf{x}-\dfrac{1}{2}\mathbf{y}.$

1. Express $\overrightarrow{AH}$ using the ratio on $BE$.

From earlier parts,
$\overrightarrow{AB}=\mathbf{x}-\mathbf{y}$ and $\overrightarrow{AE}=\mathbf{y}$.

$\overrightarrow{BE}=\overrightarrow{AE}-\overrightarrow{AB}=\mathbf{y}-(\mathbf{x}-\mathbf{y})=2\mathbf{y}-\mathbf{x}.$

Since $4HE=3BE$, $\overrightarrow{HE}=\dfrac{3}{4}\overrightarrow{BE}$, so
$\overrightarrow{BH}=\dfrac{1}{4}\overrightarrow{BE}=\dfrac{1}{4}(2\mathbf{y}-\mathbf{x}).$

Hence
$\overrightarrow{AH}=\overrightarrow{AB}+\overrightarrow{BH}=(\mathbf{x}-\mathbf{y})+\dfrac{1}{4}(2\mathbf{y}-\mathbf{x})=\dfrac{3}{4}\mathbf{x}-\dfrac{1}{2}\mathbf{y}.$

2. Find $\overrightarrow{HG}$ from the position vectors of $H$ and $G$.

From earlier part,
$\overrightarrow{BG}=\dfrac{3}{4}\mathbf{x}-\dfrac{1}{2}\mathbf{y}.$

So
$\overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{BG}=(\mathbf{x}-\mathbf{y})+\left(\dfrac{3}{4}\mathbf{x}-\dfrac{1}{2}\mathbf{y}\right)=\dfrac{7}{4}\mathbf{x}-\dfrac{3}{2}\mathbf{y}.$

Therefore
$\overrightarrow{HG}=\overrightarrow{AG}-\overrightarrow{AH}=\left(\dfrac{7}{4}\mathbf{x}-\dfrac{3}{2}\mathbf{y}\right)-\left(\dfrac{3}{4}\mathbf{x}-\dfrac{1}{2}\mathbf{y}\right)=\mathbf{x}-\mathbf{y}.$

1. Find the vector from $A$ to $M$ using the fact that $M$ is the midpoint of $AF$.

Since $M$ is the midpoint of $AF$,
$\overrightarrow{AM}=\frac12\overrightarrow{AF}.$
From earlier, $\overrightarrow{AF}=\mathbf{y}-\frac12\mathbf{x}$.

Hence
$\overrightarrow{AM}=\frac12\left(\mathbf{y}-\frac12\mathbf{x}\right)=\frac12\mathbf{y}-\frac14\mathbf{x}.$

2. Compare $\overrightarrow{HM}$ with $\overrightarrow{HG}$ to decide whether $H$, $M$ and $G$ are collinear.

From the earlier part,
$\overrightarrow{HG}=\mathbf{x}-\mathbf{y}.$
Also, from earlier results,
$\overrightarrow{AH}=\frac34\mathbf{x}-\frac12\mathbf{y}.$
So
$\overrightarrow{HM}=\overrightarrow{AM}-\overrightarrow{AH}$
$=\left(\frac12\mathbf{y}-\frac14\mathbf{x}\right)-\left(\frac34\mathbf{x}-\frac12\mathbf{y}\right)$
$=\mathbf{y}-\mathbf{x}$
$=-(\mathbf{x}-\mathbf{y})$
$=-\overrightarrow{HG}.$
Therefore $\overrightarrow{HM}$ is a scalar multiple of $\overrightarrow{HG}$, so $M$, $H$ and $G$ are collinear. Since the scalar is negative, $M$ and $G$ are on opposite sides of $H$.

Hence $H$ lies on $MG$.