6 equilateral triangles each with sides 5 units are arranged to form a regular hexagon as shown in the diagram below. By expressing $\overrightarrow{OQ}$ in terms of $\overrightarrow{OP}$ and $\overrightarrow{OR}$, or otherwise, find $\overrightarrow{OP}\cdot(\overrightarrow{OP}+\overrightarrow{OQ}+\overrightarrow{OR})$. $\overrightarrow{OE}$, $\overrightarrow{OF}$ and $\overrightarrow{OG}$ are non-zero position vectors relative to a fixed origin O for points E, F and G respectively. It is given that $2\overrightarrow{OE}-5\overrightarrow{OF}+3\overrightarrow{OG}=0$. Show that E, F and G are collinear. State the ratio of EF : EG and find $\overrightarrow{EF}\times\overrightarrow{EG}$.

1. Express $\overrightarrow{OQ}$ in terms of $\overrightarrow{OP}$ and $\overrightarrow{OR}$.

From the regular hexagon, we can use vector addition:
$\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ}$
Since the side $PQ$ is parallel to the major diagonal $OR$ and has half its length, we have $\overrightarrow{PQ} = \frac{1}{2}\overrightarrow{OR}$.
Therefore,
$\overrightarrow{OQ} = \overrightarrow{OP} + \frac{1}{2}\overrightarrow{OR}$

2. Substitute $\overrightarrow{OQ}$ into the given expression and expand.

$\overrightarrow{OP}\cdot(\overrightarrow{OP}+\overrightarrow{OQ}+\overrightarrow{OR})$
$= \overrightarrow{OP}\cdot\left(\overrightarrow{OP} + \overrightarrow{OP} + \frac{1}{2}\overrightarrow{OR} + \overrightarrow{OR}\right)$
$= \overrightarrow{OP}\cdot\left(2\overrightarrow{OP} + \frac{3}{2}\overrightarrow{OR}\right)$
$= 2(\overrightarrow{OP}\cdot\overrightarrow{OP}) + \frac{3}{2}(\overrightarrow{OP}\cdot\overrightarrow{OR})$
$= 2|\overrightarrow{OP}|^2 + \frac{3}{2}(\overrightarrow{OP}\cdot\overrightarrow{OR})$

3. Evaluate the magnitudes and the scalar product $\overrightarrow{OP}\cdot\overrightarrow{OR}$.

The side length of the hexagon is given as 5, so $|\overrightarrow{OP}| = 5$.
The major diagonal $OR$ spans two equilateral triangles, so $|\overrightarrow{OR}| = 2 \times 5 = 10$.
The angle between $\overrightarrow{OP}$ and the major diagonal $\overrightarrow{OR}$ is half of the regular hexagon's interior angle ($120^\circ / 2 = 60^\circ$).
$\overrightarrow{OP}\cdot\overrightarrow{OR} = |\overrightarrow{OP}| |\overrightarrow{OR}| \cos(60^\circ) = (5)(10)\left(\frac{1}{2}\right) = 25$

4. Calculate the final numerical value.

Substitute the evaluated components back into the expanded expression:
$2|\overrightarrow{OP}|^2 + \frac{3}{2}(\overrightarrow{OP}\cdot\overrightarrow{OR}) = 2(5)^2 + \frac{3}{2}(25)$
$= 2(25) + 37.5 = 50 + 37.5 = 87.5$

1. Rearrange the given vector equation to express $\overrightarrow{EF}$ in terms of $\overrightarrow{EG}$.

$2\overrightarrow{OE} - 5\overrightarrow{OF} + 3\overrightarrow{OG} = \mathbf{0}$
$5\overrightarrow{OF} = 2\overrightarrow{OE} + 3\overrightarrow{OG}$
Subtract $5\overrightarrow{OE}$ from both sides to introduce the displacement vectors from $E$:
$5\overrightarrow{OF} - 5\overrightarrow{OE} = 3\overrightarrow{OG} - 3\overrightarrow{OE}$
$5(\overrightarrow{OF} - \overrightarrow{OE}) = 3(\overrightarrow{OG} - \overrightarrow{OE})$
$5\overrightarrow{EF} = 3\overrightarrow{EG}$
$\overrightarrow{EF} = \frac{3}{5}\overrightarrow{EG}$

2. Use the scalar multiple relationship to show collinearity.

Since $\overrightarrow{EF} = \frac{3}{5}\overrightarrow{EG}$, $\overrightarrow{EF}$ is a scalar multiple of $\overrightarrow{EG}$. This establishes that the line segments $EF$ and $EG$ are parallel.
Because they share a common point $E$, the points $E$, $F$, and $G$ must be collinear.

3. State the ratio of the lengths $EF : EG$.

From $\overrightarrow{EF} = \frac{3}{5}\overrightarrow{EG}$, taking the magnitude on both sides gives:
$|\overrightarrow{EF}| = \frac{3}{5}|\overrightarrow{EG}|$
$EF = \frac{3}{5}EG$
Thus, the ratio $EF : EG = 3 : 5$.

4. Evaluate the cross product $\overrightarrow{EF} \times \overrightarrow{EG}$.

Since $E$, $F$, and $G$ are collinear, the vectors $\overrightarrow{EF}$ and $\overrightarrow{EG}$ are parallel.
Alternatively, substituting the earlier result:
$\overrightarrow{EF} \times \overrightarrow{EG} = \left(\frac{3}{5}\overrightarrow{EG}\right) \times \overrightarrow{EG} = \frac{3}{5} (\overrightarrow{EG} \times \overrightarrow{EG})$
Since the cross product of any vector with itself (or a parallel vector) is the zero vector:
$\overrightarrow{EF} \times \overrightarrow{EG} = \mathbf{0}$