The diagram above shows a toy house with horizontal rectangular base $ABCD$ where $AB = 30\text{ cm}$ and $BC = 20\text{ cm}$. There are two vertical rectangular walls, $ABFE$ and $DCGH$, where $AE = BF = CG = DH = 10\text{ cm}$. There are two vertical walls, $AELHD$ and $BFNGC$, where $LE = LH = NF = NG$ and $L$ and $N$ are $15\text{ cm}$ from the base respectively. The point $A$ is taken as the origin and vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ each of length $1\text{ cm}$, are taken along $AB$, $AD$ and $AE$ respectively.

1. Determine the coordinates and position vectors of points $P$, $H$, and $Q$.

From the given dimensions and origin $A(0,0,0)$:

• $E$ is $(0, 0, 10)$. Since $P$ lies on the line segment $AE$ and $PE = 1$, $P$ is $(0, 0, 9)$.
• $H$ is vertically above $D(0, 20, 0)$ by $10$, so $H$ is $(0, 20, 10)$.
• $Q$ lies on $BF$ with $QF = 1$, and $F$ is $(30, 0, 10)$, so $Q$ is $(30, 0, 9)$.

Position vectors:
$\vec{OP} = \begin{pmatrix} 0 \\ 0 \\ 9 \end{pmatrix}$, $\vec{OH} = \begin{pmatrix} 0 \\ 20 \\ 10 \end{pmatrix}$, $\vec{OQ} = \begin{pmatrix} 30 \\ 0 \\ 9 \end{pmatrix}$

2. Find two non-parallel direction vectors parallel to the plane $PQGH$.

$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 30 \\ 0 \\ 9 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 9 \end{pmatrix} = \begin{pmatrix} 30 \\ 0 \\ 0 \end{pmatrix} = 30\mathbf{i}$

$\vec{PH} = \vec{OH} - \vec{OP} = \begin{pmatrix} 0 \\ 20 \\ 10 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 9 \end{pmatrix} = \begin{pmatrix} 0 \\ 20 \\ 1 \end{pmatrix} = 20\mathbf{j} + \mathbf{k}$

3. Use the cross product of the direction vectors to show the given normal vector.

Let a normal vector be $\mathbf{N}$.
$\mathbf{N} = \vec{PH} \times \vec{PQ} = \begin{pmatrix} 0 \\ 20 \\ 1 \end{pmatrix} \times \begin{pmatrix} 30 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0(0) - 1(0) \\ 1(30) - 0(0) \\ 0(0) - 20(30) \end{pmatrix} = \begin{pmatrix} 0 \\ 30 \\ -600 \end{pmatrix} = 30 \begin{pmatrix} 0 \\ 1 \\ -20 \end{pmatrix}$
Since any scalar multiple is also a valid normal, dividing by 30 gives $\mathbf{n} = \begin{pmatrix} 0 \\ 1 \\ -20 \end{pmatrix} = \mathbf{j} - 20\mathbf{k}$ (shown).

4. Use the derived normal vector and a point on the plane to find the Cartesian equation.

Using $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$ with the position vector of point $P(0, 0, 9)$:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ -20 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 9 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ -20 \end{pmatrix}$
$0(x) + 1(y) - 20(z) = 0(0) + 0(1) + 9(-20)$
$y - 20z = -180$

1. Identify the normal vector to the base $ABCD$.

The plane $ABCD$ is horizontal and lies on the plane defined by $\mathbf{i}$ and $\mathbf{j}$.
Therefore, its normal vector is the vertical unit vector, $\mathbf{n}_2 = \mathbf{k}$.
From part (i), the normal vector to the plane $PQGH$ is $\mathbf{n}_1 = \mathbf{j} - 20\mathbf{k}$.

2. Evaluate the scalar product and magnitudes of the normal vectors.

Scalar product:
$\mathbf{n}_1 \cdot \mathbf{n}_2 = (\mathbf{j} - 20\mathbf{k}) \cdot \mathbf{k} = -20$

Magnitudes:
$|\mathbf{n}_1| = \sqrt{0^2 + 1^2 + (-20)^2} = \sqrt{401}$
$|\mathbf{n}_2| = \sqrt{0^2 + 0^2 + 1^2} = 1$

3. Calculate the cosine of the acute angle between the two planes.

Let $\theta$ be the acute angle between the planes. The angle is determined by the absolute value of the scalar product of their normals:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$
$\cos \theta = \frac{|-20|}{\sqrt{401} \times 1} = \frac{20}{\sqrt{401}}$

1. Determine the coordinates of $M$, the midpoint of $LN$.

From the given dimensions, the apices $L$ and $N$ are at a height of $15\text{ cm}$ from the base, so their $z$-coordinates are $15$.
Since the pentagonal walls are symmetrical and $LE = LH$, $L$ lies vertically above the midpoint of $AD$. $A$ is at $y=0$ and $D$ is at $y=20$, so the $y$-coordinate of $L$ is $10$.
Thus, $L(0, 10, 15)$.
Similarly, $N$ lies above $BC$ (where $x=30$), giving $N(30, 10, 15)$.
As $M$ is the midpoint of $LN$:
$\overrightarrow{OM} = \frac{1}{2} \left( \begin{pmatrix} 0 \\ 10 \\ 15 \end{pmatrix} + \begin{pmatrix} 30 \\ 10 \\ 15 \end{pmatrix} \right) = \begin{pmatrix} 15 \\ 10 \\ 15 \end{pmatrix}$.
So, $M(15, 10, 15)$.

2. Formulate the equation of the line $MV$.

For the length of the wire to be minimum, the line $MV$ must be perpendicular to the false ceiling plane $PQGH$.
Therefore, the direction vector of the line $MV$ is parallel to the normal of the plane, which is given in part (i) as $\mathbf{n} = \mathbf{j} - 20\mathbf{k} = \begin{pmatrix} 0 \\ 1 \\ -20 \end{pmatrix}$.
The vector equation of line $MV$ is:
$\mathbf{r} = \begin{pmatrix} 15 \\ 10 \\ 15 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -20 \end{pmatrix}$, where $\lambda \in \mathbb{R}$.

3. Find the point of intersection $V$ between the line $MV$ and the plane $PQGH$.

From part (i), the cartesian equation of the plane $PQGH$ (using normal $\mathbf{j}-20\mathbf{k}$ and point $P(0,0,9)$) is:
$y - 20z = -180$

Substitute the parametric equations of the line $MV$ ($x = 15$, $y = 10 + \lambda$, $z = 15 - 20\lambda$) into the plane equation:
$(10 + \lambda) - 20(15 - 20\lambda) = -180$
$10 + \lambda - 300 + 400\lambda = -180$
$401\lambda - 290 = -180$
$401\lambda = 110 \implies \lambda = \frac{110}{401}$

Substitute $\lambda$ back into the coordinates:
$x = 15$
$y = 10 + \frac{110}{401} = \frac{4010 + 110}{401} = \frac{4120}{401}$
$z = 15 - 20\left(\frac{110}{401}\right) = \frac{6015 - 2200}{401} = \frac{3815}{401}$

The coordinates of $V$ are $\left(15, \frac{4120}{401}, \frac{3815}{401}\right)$.