下图显示了一个以 $A$、$B$、$C$ 和 $D$ 为顶点的三棱锥。以原点 $O$ 为基准，点 $A$、$B$、$C$ 和 $D$ 的位置向量分别为 $\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$、$2\mathbf{i} - \mathbf{j} + 4\mathbf{k}$、$4\mathbf{i} + 5\mathbf{j} - \mathbf{k}$ 和 $-2\mathbf{i} - 5\mathbf{k}$。设包含点 $A$、$B$ 和 $C$ 的平面为 $\Pi$。

1. Find two distinct direction vectors on the plane $\Pi$.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}$

$\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 4 \\ 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix}$

2. Find a normal vector to the plane using the vector product of the two direction vectors.

$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} (-3)(-4) - (1)(3) \\ (1)(3) - (1)(-4) \\ (1)(3) - (-3)(3) \end{pmatrix} = \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}$

This matches the normal vector given in the required form.

3. Determine the constant $\alpha$ using the scalar product equation of a plane.

The equation of the plane is $\mathbf{r} \cdot \mathbf{n} = \vec{OA} \cdot \mathbf{n}$

$\alpha = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}$
$\alpha = (1)(9) + (2)(7) + (3)(12) = 9 + 14 + 36 = 59$

$\therefore \mathbf{r} \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} = 59$ (Shown)

1. Find the direction vector of the line $BD$.

$\vec{BD} = \vec{OD} - \vec{OB} = \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \\ -9 \end{pmatrix}$

2. Calculate the acute angle between the line and the plane using the normal vector.

From part (a), the normal vector to $\Pi$ is $\mathbf{n} = \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}$.
Let $\theta$ be the acute angle between the line $BD$ and $\Pi$.

$\sin \theta = \frac{|\vec{BD} \cdot \mathbf{n}|}{|\vec{BD}| |\mathbf{n}|}$
$\sin \theta = \frac{\left| \begin{pmatrix} -4 \\ 1 \\ -9 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} \right|}{\sqrt{(-4)^2 + 1^2 + (-9)^2} \sqrt{9^2 + 7^2 + 12^2}}$
$\sin \theta = \frac{|-36 + 7 - 108|}{\sqrt{98} \sqrt{274}}$
$\sin \theta = \frac{|-137|}{7\sqrt{2} \times \sqrt{2}\sqrt{137}} = \frac{137}{14\sqrt{137}} = \frac{\sqrt{137}}{14}$

$\theta = \sin^{-1}\left(\frac{\sqrt{137}}{14}\right) \approx 56.726^\circ \approx 56.7^\circ$ (to 1 d.p.)

1. State the Cartesian equation of plane $\Pi$, using the value of $\alpha$ found in part (a).

Using point $A(1, 2, 3)$ to find $\alpha$:
$\alpha = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} = 9(1) + 7(2) + 12(3) = 59$
The equation of plane $\Pi$ is $\mathbf{r} \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} = 59$.

2. Form the vector equation of the line perpendicular to $\Pi$ passing through $D$.

The normal vector to the plane, $\mathbf{n} = \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}$, serves as the direction vector for the perpendicular line.
The point $D$ has position vector $\mathbf{d} = -2\mathbf{i} - 5\mathbf{k} = \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix}$.
The equation of the line is:
$\mathbf{r} = \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

3. Substitute the line equation into the plane equation to find the value of $\lambda$ at the intersection point.

$\left[ \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} \right] \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} = 59$
$\left( -18 + 0 - 60 \right) + \lambda \left( 9^2 + 7^2 + 12^2 \right) = 59$
$-78 + \lambda (81 + 49 + 144) = 59$
$-78 + 274\lambda = 59$
$274\lambda = 137 \implies \lambda = \frac{1}{2}$

4. Calculate the position vector of the foot of the perpendicular.

Substitute $\lambda = \frac{1}{2}$ back into the equation of the line:
$\vec{OF} = \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}$
$\vec{OF} = \begin{pmatrix} -2 + 4.5 \\ 0 + 3.5 \\ -5 + 6 \end{pmatrix} = \begin{pmatrix} 2.5 \\ 3.5 \\ 1 \end{pmatrix}$
Writing in $\mathbf{i}, \mathbf{j}, \mathbf{k}$ format, the position vector is $\frac{5}{2}\mathbf{i} + \frac{7}{2}\mathbf{j} + \mathbf{k}$.

1. Find the cross product of vectors $\vec{AB}$ and $\vec{AC}$.

We first determine vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}$
$\vec{AC} = \begin{pmatrix} 4 \\ 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix}$

Their cross product gives the normal vector $\mathbf{n}$:
$\vec{AB} \times \vec{AC} = \begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} (-3)(-4)-(1)(3) \\ (1)(3)-(1)(-4) \\ (1)(3)-(-3)(3) \end{pmatrix} = \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}$

2. Calculate the area of the triangular base $ABC$.

The area of a triangle is half the magnitude of the cross product of its edge vectors:
$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$
$\text{Area} = \frac{1}{2} \sqrt{9^2 + 7^2 + 12^2}$
$\text{Area} = \frac{\sqrt{81 + 49 + 144}}{2} = \frac{\sqrt{274}}{2} \text{ units}^2$

3. Determine the perpendicular height of the pyramid from apex $D$ to the base plane $\Pi$.

To find the height, we evaluate the perpendicular distance from $D(-2, 0, -5)$ to plane $\Pi$.
First, compute the constant $\alpha$ for the plane equation $\mathbf{r} \cdot \mathbf{n} = \alpha$ using point $A$:
$\alpha = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} = 9 + 14 + 36 = 59$

The perpendicular height $h$ is given by:
$h = \frac{| \mathbf{r}_D \cdot \mathbf{n} - \alpha |}{|\mathbf{n}|}$
$h = \frac{\left| \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix} - 59 \right|}{\sqrt{274}}$
$h = \frac{| -18 + 0 - 60 - 59 |}{\sqrt{274}} = \frac{137}{\sqrt{274}}$

(Note: This height is also equivalent to $|\vec{DF}|$, where $F$ is the foot of the perpendicular found earlier in part (c)).

4. Calculate the volume of the pyramid.

Using the given formula for the volume of a pyramid:
$\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}$
$\text{Volume} = \frac{1}{3} \left( \frac{\sqrt{274}}{2} \right) \left( \frac{137}{\sqrt{274}} \right)$
$\text{Volume} = \frac{137}{6} \text{ units}^3$