In a computer game, a slope can be modelled as a plane $p$ containing three points, $A(1, 0, -3)$, $B(1, 4, -15)$ and $C(2, 3, -14)$.

1. Find two non-parallel vectors lying on the plane, such as $\vec{AB}$ and $\vec{AC}$.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 1 \\ 4 \\ -15 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ -3 \end{pmatrix} = \begin{pmatrix} 0 \\ 4 \\ -12 \end{pmatrix}$

$\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 2 \\ 3 \\ -14 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ -11 \end{pmatrix}$

2. Calculate a normal vector to the plane using the cross product of $\vec{AB}$ and $\vec{AC}$.

$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 0 \\ 4 \\ -12 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ -11 \end{pmatrix} = \begin{pmatrix} 4(-11) - (-12)(3) \\ -12(1) - 0(-11) \\ 0(3) - 4(1) \end{pmatrix}$

$\mathbf{n} = \begin{pmatrix} -44 + 36 \\ -12 - 0 \\ 0 - 4 \end{pmatrix} = \begin{pmatrix} -8 \\ -12 \\ -4 \end{pmatrix}$

3. Use the normal vector and point $A$ to formulate the scalar product equation of the plane.

A simpler normal vector parallel to $\mathbf{n}$ is obtained by dividing by $-4$:
$\mathbf{n}' = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$

Equation of plane $p$ is $\mathbf{r} \cdot \mathbf{n}' = \vec{OA} \cdot \mathbf{n}'$:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = 1(2) + 0(3) + (-3)(1) = 2 + 0 - 3$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = -1 \text{ (shown)}$

1. State the geometric condition for the path to lie entirely on the slope (plane $p$).

Since Griffles travels on the slope, the path lies on plane $p$. Therefore, the direction vector of the line must be perpendicular to the normal vector of the plane, which means their scalar product is zero.

2. Evaluate the scalar product to form the equation.

$\begin{pmatrix} 1 \\ m \\ n \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = 0$
$1(2) + m(3) + n(1) = 0$
$3m + n + 2 = 0$

1. Find the vector equation of the line passing through $B$ and $C$.

Let the line passing through $B$ and $C$ be $l_{BC}$.
The direction vector $\vec{BC} = \vec{OC} - \vec{OB} = \begin{pmatrix} 2 \\ 3 \\ -14 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ -15 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.
The vector equation of the line is $\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ -15 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.
For the point to lie on the line segment $BC$, the parameter $\mu$ is restricted to $0 \le \mu \le 1$.

2. Set up a system of equations to find the intersection between Griffles' path and the line $BC$.

Equating the position vectors of Griffles' path and $l_{BC}$:
$\begin{pmatrix} 1 \\ 0 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1/3 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -15 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$
This yields the system of linear equations:
$1 + \lambda = 1 + \mu \quad \text{--- (1)}$
$\frac{1}{3}\lambda = 4 - \mu \quad \text{--- (2)}$
$-3 - 3\lambda = -15 + \mu \quad \text{--- (3)}$

3. Solve the system to find the intersection point and show it lies outside the segment $BC$.

From (1), we have $\lambda = \mu$.
Substituting into (2):
$\frac{1}{3}\mu = 4 - \mu$
$\frac{4}{3}\mu = 4 \implies \mu = 3$
(Checking consistency with equation (3): $-3 - 3(3) = -12$ and $-15 + 3 = -12$. The lines indeed intersect).
Since the intersection occurs at $\mu = 3$, which falls outside the required range of $0 \le \mu \le 1$ for the line segment $BC$, the intersection point does not lie on the obstacle's base.
Alternatively, the point of intersection is $(4, 1, -12)$, which has an $x$-coordinate of $4$, falling outside the $x$-coordinates of $B$ and $C$ ($1$ and $2$ respectively).
Thus, Griffles successfully navigates the obstacle without colliding into it (shown).

1. Find the vector $\vec{AE}$ and state the direction vector of the path.

Given the starting point $A(1, 0, -3)$ and the sensor at $E(5, 2, 1)$, the vector from $A$ to $E$ is:
$\vec{AE} = \vec{OE} - \vec{OA} = \begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix}$
From part (b) and (c), the path has direction vector $\mathbf{u} = \begin{pmatrix} 1 \\ 1/3 \\ -3 \end{pmatrix}$. To avoid fractions, we can use a parallel direction vector $\mathbf{d} = 3\mathbf{u} = \begin{pmatrix} 3 \\ 1 \\ -9 \end{pmatrix}$.

2. Calculate the squared magnitude of $\vec{AE}$, $\mathbf{d}$, and their dot product.

To find the shortest distance from $E$ to the path, we can use the dot product to find the length of the projection of $\vec{AE}$ onto the path's direction.
$|\vec{AE}|^2 = 4^2 + 2^2 + 4^2 = 36$
$|\mathbf{d}|^2 = 3^2 + 1^2 + (-9)^2 = 91$
$\vec{AE} \cdot \mathbf{d} = (4)(3) + (2)(1) + (4)(-9) = 12 + 2 - 36 = -22$

3. Use Pythagoras' theorem to find the shortest perpendicular distance.

The square of the shortest distance $h$ from $E$ to the line is given by $h^2 = |\vec{AE}|^2 - \left( \frac{\vec{AE} \cdot \mathbf{d}}{|\mathbf{d}|} \right)^2$.
$h^2 = 36 - \frac{(-22)^2}{91} = 36 - \frac{484}{91} = \frac{3276 - 484}{91} = \frac{2792}{91}$
$h = \sqrt{\frac{2792}{91}} \approx 5.539 \text{ units}$

4. Compare the shortest distance with the sensor's range and conclude.

Since the minimum possible distance from the sensor $E$ to any point on Griffles' path is approximately $5.54$ units, which is strictly greater than the sensor's trigger range of $5$ units, Griffles will never enter the zone. Thus, he does not activate the sensor.

1. State the general equation of plane $q$ using the fact that it is parallel to $p$.

Since plane $q$ is parallel to plane $p$, it shares the same normal vector $\mathbf{n} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$.
Let the cartesian equation of $q$ be $2x + 3y + z = d$.

2. Calculate the perpendicular distance from point $E$ to plane $p$.

The cartesian equation of plane $p$ is $2x + 3y + z = -1$, which can be written as $2x + 3y + z + 1 = 0$.
Using the perpendicular distance formula for point $E(5, 2, 1)$ to plane $p$:
$\text{Distance} = \frac{|2(5) + 3(2) + 1(1) + 1|}{\sqrt{2^2 + 3^2 + 1^2}}$
$\text{Distance} = \frac{|10 + 6 + 1 + 1|}{\sqrt{14}} = \frac{18}{\sqrt{14}}$

3. Use the given equidistant condition to find $d$ and state the final cartesian equation of $q$.

Since $E$ is equidistant from planes $p$ and $q$, the distance from $E$ to $q$ is also $\frac{18}{\sqrt{14}}$.
$\frac{|2(5) + 3(2) + 1(1) - d|}{\sqrt{14}} = \frac{18}{\sqrt{14}}$
$|17 - d| = 18$
$17 - d = 18 \quad \text{or} \quad 17 - d = -18$
$d = -1 \quad \text{or} \quad d = 35$
Since $q$ is a different plane from $p$ ($d = -1$ corresponds to plane $p$), we have $d = 35$.
The cartesian equation of $q$ is $2x + 3y + z = 35$.