为了构建虚拟现实（VR）体验，需要创建一个包含几何模型的虚拟世界。以原点 $O$ 为基准，虚拟世界中的每个点都表示为坐标 $(x, y, z)$。透视投影是一种将虚拟世界中的点投影到虚拟屏幕上，同时保持物体在不同距离下的缩放特性的技术。在 VR 中正确实现这一点有助于感知深度 and 比例。此外，只有位于位于 $O$ 处的眼睛前方的区域（称为视景体，见图 1）内的点才会被投影到虚拟屏幕上。 在图 1 中，视景体中的一个三角形物体 $ABC$（其中 $A(36,36,12)$ 且 $B(24,48,72)$）被投影到虚拟屏幕上，得到图像 $A'B'C'$。已知 $A'$ 的坐标为 $(3,3,1)$，且穿过 $A'$ 和 $B'$ 的直线平行于向量 $2\mathbf{i}+\mathbf{j}-2\mathbf{k}$。

1. Form the vector equation of the line passing through the origin $O$ and point $B$.

Since the projection is from $O$, the image point $B'$ must lie on the line passing through $O(0, 0, 0)$ and $B(24, 48, 72)$.
The direction vector of this line is $\overrightarrow{OB} = \begin{pmatrix} 24 \\ 48 \\ 72 \end{pmatrix}$, which is parallel to $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$.
The equation of line $OB$ is:
$l_{OB}: \mathbf{r} = \mu \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$

2. Form the vector equation of the line passing through $A'$ and $B'$.

We are given that $A'$ has coordinates $(3, 3, 1)$ and the line $A'B'$ is parallel to $2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$.
The equation of line $A'B'$ is:
$l_{A'B'}: \mathbf{r} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$

3. Equate the two line equations to find their point of intersection, $B'$.

Since $B'$ lies on both lines, we equate their position vectors:
$\begin{pmatrix} \mu \\ 2\mu \\ 3\mu \end{pmatrix} = \begin{pmatrix} 3 + 2\lambda \\ 3 + \lambda \\ 1 - 2\lambda \end{pmatrix}$
This gives a system of linear equations:
$\mu = 3 + 2\lambda \quad \text{--- (1)}$
$2\mu = 3 + \lambda \quad \text{--- (2)}$
$3\mu = 1 - 2\lambda \quad \text{--- (3)}$

4. Solve the system of equations for the parameters $\lambda$ and $\mu$.

Substitute equation (1) into equation (2):
$2(3 + 2\lambda) = 3 + \lambda$
$6 + 4\lambda = 3 + \lambda$
$3\lambda = -3 \implies \lambda = -1$
Substitute $\lambda = -1$ into equation (1):
$\mu = 3 + 2(-1) = 1$
(Check for consistency with equation (3): $3(1) = 3$ and $1 - 2(-1) = 3$. The lines intersect consistently.)

5. Determine the position vector and coordinates of $B'$.

Using $\mu = 1$, the position vector of $B'$ is:
$\overrightarrow{OB'} = 1 \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$
Hence, the coordinates of $B'$ are $(1, 2, 3)$.

1. Find two non-parallel direction vectors that lie on the virtual screen.

From part (i), the line through $A'$ and $B'$ is parallel to $\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$.

Using the coordinates of $A'$ and $C'$, we find another direction vector on the plane:
$\overrightarrow{A'C'} = \overrightarrow{OC'} - \overrightarrow{OA'} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$

2. Determine a normal vector to the virtual screen by taking the cross product of the two direction vectors.

$\mathbf{n} = \mathbf{d}_1 \times \overrightarrow{A'C'} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} (1)(1) - (-2)(1) \\ (-2)(-2) - (2)(1) \\ (2)(1) - (1)(-2) \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$

3. Formulate the equation of the plane in scalar product form $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$.

$\mathbf{r} \cdot \mathbf{n} = \overrightarrow{OA'} \cdot \mathbf{n}$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} = 3(3) + 3(2) + 1(4)$
$\mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} = 19$

1. Find the displacement vectors $\overrightarrow{A'B'}$ and $\overrightarrow{A'C'}$.

$\overrightarrow{A'B'} = \overrightarrow{OB'} - \overrightarrow{OA'} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}$
$\overrightarrow{A'C'} = \overrightarrow{OC'} - \overrightarrow{OA'} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$

2. Evaluate the cross product of the two displacement vectors.

$\overrightarrow{A'B'} \times \overrightarrow{A'C'} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} (-1)(1) - (2)(1) \\ (2)(-2) - (-2)(1) \\ (-2)(1) - (-1)(-2) \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \\ -4 \end{pmatrix}$
(Note: This cross product may have already been calculated in part (ii) to find the normal of the plane.)

3. Calculate the area of the triangle using the magnitude of the cross product.

$\text{Area of } \triangle A'B'C' = \frac{1}{2} |\overrightarrow{A'B'} \times \overrightarrow{A'C'}|$
$\text{Area} = \frac{1}{2} \sqrt{(-3)^2 + (-2)^2 + (-4)^2}$
$\text{Area} = \frac{1}{2} \sqrt{9 + 4 + 16} = \frac{\sqrt{29}}{2} \text{ units}^2$

1. Express the position vector of $C$ in terms of an unknown scalar.

Since $C'$ is the perspective projection of $C$ from the projection centre $O$, the points $O$, $C'$, and $C$ must be collinear. Thus, $\overrightarrow{OC}$ is a scalar multiple of $\overrightarrow{OC'}$. Since $C$ lies in front of $O$, we have $\overrightarrow{OC} = k \overrightarrow{OC'}$ for some scalar $k > 0$.
$\overrightarrow{OC} = k \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} k \\ 4k \\ 2k \end{pmatrix}$

2. Formulate the vector $\overrightarrow{BC}$ in terms of $k$.

Using the position vector of $B$, $\overrightarrow{OB} = \begin{pmatrix} 24 \\ 48 \\ 72 \end{pmatrix}$:
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} k \\ 4k \\ 2k \end{pmatrix} - \begin{pmatrix} 24 \\ 48 \\ 72 \end{pmatrix} = \begin{pmatrix} k - 24 \\ 4k - 48 \\ 2k - 72 \end{pmatrix}$

3. Express the length of projection of $\overrightarrow{BC}$ onto the normal of the virtual screen.

From part (ii), the normal vector to the virtual screen is $\mathbf{n} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$. Its magnitude is $|\mathbf{n}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{29}$.
The length of projection of $\overrightarrow{BC}$ onto the normal is given by $\frac{|\overrightarrow{BC} \cdot \mathbf{n}|}{|\mathbf{n}|}$:
$\overrightarrow{BC} \cdot \mathbf{n} = 3(k - 24) + 2(4k - 48) + 4(2k - 72)$
$= 3k - 72 + 8k - 96 + 8k - 288 = 19k - 456 = 19(k - 24)$
$\text{Length of projection} = \frac{|19(k - 24)|}{\sqrt{29}}$

4. Equate the projection length to the given value and solve for $k$.

$\frac{19|k - 24|}{\sqrt{29}} = \frac{57}{29}$
$|k - 24| = \frac{57}{29} \times \frac{\sqrt{29}}{19} = \frac{3\sqrt{29}}{29}$
$k = 24 \pm \frac{3\sqrt{29}}{29}$
Since both values of $k$ are positive, $C$ stays in front of the viewing frustum for both cases. Substituting these into $\overrightarrow{OC}$ yields the possible position vectors:
$\overrightarrow{OC} = \left( 24 + \frac{3\sqrt{29}}{29} \right) \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \quad \text{and} \quad \left( 24 - \frac{3\sqrt{29}}{29} \right) \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}$