A designer wants to construct an inclined roof that is attached to a vertical rectangular wall, positioned above a horizontal flat ground, as shown in the diagram below. Assume that the roof and wall are of negligible thickness. Points $(x, y, z)$ are defined relative to a corner of the wall at $O(0,0,0)$, where units are metres. Unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are in the directions shown in the diagram. The other corners of the wall are $A$, $B$ and $C$, such that $AB$ is 5 m and $A$ and $B$ are 3 m vertically above $O$ and $C$ respectively. The roof is attached to the wall along $AB$ and is modelled by a plane parallel to $4\mathbf{i} + 5\mathbf{j} + \mathbf{k}$.

1. Set up a point on the plane and the direction of AB

$A$ is 3 m vertically above $O$, so $A = (0,0,3)$.

Since $A$ is above $O$ and $B$ is above $C$, the top edge $AB$ is parallel to the base edge $OC$, with $|AB| = 5$ m. Taking $A=(0,0,3)$ and $B=(0,5,3)$ gives the direction of $AB$ as
$\overrightarrow{AB} = \begin{pmatrix}0\\5\\0\end{pmatrix} \parallel \begin{pmatrix}0\\1\\0\end{pmatrix}.$

2. Find the normal vector to the roof plane

The plane is parallel to both $\overrightarrow{AB}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ and $\mathbf{d}=\begin{pmatrix}4\\5\\1\end{pmatrix}$, so a normal is
$\mathbf{n} = \begin{pmatrix}0\\1\\0\end{pmatrix}\times\begin{pmatrix}4\\5\\1\end{pmatrix} = \begin{pmatrix}(1)(1)-(0)(5)\\(0)(4)-(0)(1)\\(0)(5)-(1)(4)\end{pmatrix} = \begin{pmatrix}1\\0\\-4\end{pmatrix}.$

3. Form the cartesian equation of the plane

Using $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ with $\mathbf{a}=A=(0,0,3)$:
$\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}1\\0\\-4\end{pmatrix} = \begin{pmatrix}0\\0\\3\end{pmatrix}\cdot\begin{pmatrix}1\\0\\-4\end{pmatrix}$
$x - 4z = 0 - 12 = -12.$
Hence the roof is the plane $x - 4z = -12$ (shown).

1. State the normal vectors of the wall and the roof

The wall $OABC$ lies in the plane $x = 0$ (since $O(0,0,0)$, $A(0,0,3)$, $B(0,5,3)$, $C(0,5,0)$ all have $x=0$), so its normal is
$\mathbf{n}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}.$
From part (i) the roof is $x - 4z = -12$, so its normal is
$\mathbf{n}_2 = \begin{pmatrix} 1 \\ 0 \\ -4 \end{pmatrix}.$

2. Use the dot product to find the acute angle between the planes

$\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1|\,|\mathbf{n}_2|} = \frac{|(1)(1)+(0)(0)+(0)(-4)|}{\sqrt{1}\,\sqrt{1^2+0^2+(-4)^2}} = \frac{1}{\sqrt{17}}$
$\theta = \cos^{-1}\!\left(\tfrac{1}{\sqrt{17}}\right) = 75.96^\circ \approx 76^\circ.$

3. Convert the acute angle to the required obtuse angle

The acute angle between the planes is $76^\circ$, so the obtuse angle is
$180^\circ - 75.96^\circ = 104.04^\circ \approx 104^\circ.$

1. Apply the perpendicular distance formula from a point to a plane to find the minimum length.

The minimum length of the strut is the perpendicular distance from the point $P(4, 3, 0)$ to the roof plane $x - 4z = -12$.

$\text{Length} = \frac{\left| \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ -4 \end{pmatrix} - (-12) \right|}{\sqrt{1^2 + 0^2 + (-4)^2}}$

2. Evaluate the expression to find the exact length.

$\text{Length} = \frac{|4 - (-12)|}{\sqrt{1 + 16}}$
$= \frac{16}{\sqrt{17}}$
$= \frac{16\sqrt{17}}{17}$

1. Formulate the vector equation of the wooden strut.

Since the wooden strut has a minimum length from $P(4, 3, 0)$ to the roof, it must be perpendicular to the roof plane $x - 4z = -12$.
The normal vector to the roof is $\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ -4 \end{pmatrix}$.
The vector equation of the wooden strut is:
$l_1: \mathbf{r} = \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -4 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

2. Determine the coordinates of $C$ and formulate the vector equation of the light ray.

The roof plane $x - 4z = -12$ contains the top edge of the wall $AB$. Given $A(0,0,3)$ is on this plane, and $B$ also lies on it, $B$'s $x$-coordinate must satisfy $x_B - 4(3) = -12 \implies x_B = 0$.
Thus, the rectangular wall $OABC$ lies in the $yz$-plane. With $O$ at the origin and wall length $AB = 5\text{ m}$, $C$ lies on the $y$-axis at $(0, 5, 0)$.
The light ray is emitted from $C$ in the direction $7\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}$. Its vector equation is:
$l_2: \mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 7 \\ -4 \\ 4 \end{pmatrix}, \text{ where } \mu \in \mathbb{R}$

3. Set up a system of equations to find the intersection of the two lines.

Equating the equations of $l_1$ and $l_2$:
$\begin{pmatrix} 4 + \lambda \\ 3 \\ -4\lambda \end{pmatrix} = \begin{pmatrix} 7\mu \\ 5 - 4\mu \\ 4\mu \end{pmatrix}$
Comparing the $y$ and $z$ components:
For $y$: $3 = 5 - 4\mu \implies 4\mu = 2 \implies \mu = 0.5$
For $z$: $-4\lambda = 4\mu \implies -4\lambda = 4(0.5) \implies \lambda = -0.5$

4. Verify intersection via the $x$-component and calculate the coordinates of the point.

Substitute $\lambda = -0.5$ and $\mu = 0.5$ into the $x$-components to check for consistency:
For the strut: $x = 4 + (-0.5) = 3.5$
For the light ray: $x = 7(0.5) = 3.5$
Since the $x$-components perfectly match, the two lines intersect, proving the light ray meets the wooden strut.
Substitute $\mu = 0.5$ into $l_2$ to find the exact intersection coordinates:
$\mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ 0 \end{pmatrix} + 0.5 \begin{pmatrix} 7 \\ -4 \\ 4 \end{pmatrix} = \begin{pmatrix} 3.5 \\ 3 \\ 2 \end{pmatrix}$