平面 $\pi_1$ 包含点 $(2,-2,1)$ 和直线 $\mathbf{r} = \begin{pmatrix}0 \\ 1 \\ 8\end{pmatrix} + \lambda \begin{pmatrix}0 \\ 2 \\ 3\end{pmatrix},$ 其中 $\lambda$ 为参数。 [续] 平面 $\pi_2$ 的向量方程为 $\mathbf{r} = a \begin{pmatrix} 0 \\ -1 \\ 4 \end{pmatrix} + b \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$，其中 $a$ 和 $b$ 为参数。

1. Find a second vector parallel to the plane using the given point and a point on the line

Let $A$ be the point $(2, -2, 1)$ and $B$ be the point $(0, 1, 8)$ on the line.
$\vec{AB} = \begin{pmatrix} 0 \\ 1 \\ 8 \end{pmatrix} - \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \\ 7 \end{pmatrix}$

2. Find a normal vector to the plane using the vector product

The normal vector $\mathbf{n}$ is the cross product of the line's direction vector and $\vec{AB}$.
$\mathbf{n} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} -2 \\ 3 \\ 7 \end{pmatrix} = \begin{pmatrix} (2)(7) - (3)(3) \\ (3)(-2) - (0)(7) \\ (0)(3) - (2)(-2) \end{pmatrix} = \begin{pmatrix} 5 \\ -6 \\ 4 \end{pmatrix}$

3. Determine the scalar product equation of the plane

Using the point $B(0, 1, 8)$ on the plane:
$\mathbf{r} \cdot \mathbf{n} = \mathbf{b} \cdot \mathbf{n}$
$\mathbf{a} \cdot \mathbf{n} = \begin{pmatrix} 0 \\ 1 \\ 8 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -6 \\ 4 \end{pmatrix} = (0)(5) + (1)(-6) + (8)(4) = 26$
Therefore, the equation of the plane $\pi_1$ is $\mathbf{r} \cdot \begin{pmatrix} 5 \\ -6 \\ 4 \end{pmatrix} = 26$.

1. Find the normal vector to the plane $\pi_1$ (as derived in part a).

The plane $\pi_1$ contains the point $(2,-2,1)$ and the line with point $(0,1,8)$ and direction $\begin{pmatrix}0 \\ 2 \\ 3\end{pmatrix}$.
Two direction vectors on $\pi_1$ are:
$\mathbf{d}_1 = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}$
$\mathbf{d}_2 = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ 8 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -7 \end{pmatrix}$
Normal vector $\mathbf{n}_1$ is the cross product:
$\mathbf{n}_1 = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 2 \\ -3 \\ -7 \end{pmatrix} = \begin{pmatrix} (2)(-7) - (3)(-3) \\ (3)(2) - (0)(-7) \\ (0)(-3) - (2)(2) \end{pmatrix} = \begin{pmatrix} -5 \\ 6 \\ -4 \end{pmatrix}$

2. Find the normal vector to the plane $\pi_2$.

From the vector equation of $\pi_2$, it is parallel to the vectors $\begin{pmatrix} 0 \\ -1 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$.
Normal vector $\mathbf{n}_2$ is their cross product:
$\mathbf{n}_2 = \begin{pmatrix} 0 \\ -1 \\ 4 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} (-1)(-1) - (4)(2) \\ (4)(1) - (0)(-1) \\ (0)(2) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -7 \\ 4 \\ 1 \end{pmatrix}$

3. Calculate the acute angle between the two planes using the dot product of their normal vectors.

The acute angle $\theta$ between the planes is the acute angle between $\mathbf{n}_1$ and $\mathbf{n}_2$:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1|\,|\mathbf{n}_2|}$
$\cos \theta = \frac{\left| \begin{pmatrix} -5 \\ 6 \\ -4 \end{pmatrix} \cdot \begin{pmatrix} -7 \\ 4 \\ 1 \end{pmatrix} \right|}{\sqrt{(-5)^2 + 6^2 + (-4)^2} \sqrt{(-7)^2 + 4^2 + 1^2}}$
$\cos \theta = \frac{|(-5)(-7) + (6)(4) + (-4)(1)|}{\sqrt{77} \sqrt{66}}$
$\cos \theta = \frac{|35 + 24 - 4|}{11\sqrt{42}} = \frac{55}{11\sqrt{42}} = \frac{5}{\sqrt{42}}$
$\theta = \cos^{-1}\left(\frac{5}{\sqrt{42}}\right) \approx 39.5^\circ \text{ (to 1 d.p.)}$

1. State the two plane normals

From part (a), $\pi_1: 5x - 6y + 4z = 26$, so $\mathbf{n_1} = \begin{pmatrix}5\\-6\\4\end{pmatrix}$.

For $\pi_2$, the normal is the cross product of its two direction vectors:
$\mathbf{n_2} = \begin{pmatrix}0\\-1\\4\end{pmatrix} \times \begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}-7\\4\\1\end{pmatrix},$ giving $\pi_2: -7x + 4y + z = 0$.

2. Find the direction of l

$l$ lies in both planes, so its direction is perpendicular to both normals:
$\mathbf{n_1} \times \mathbf{n_2} = \begin{pmatrix}5\\-6\\4\end{pmatrix} \times \begin{pmatrix}-7\\4\\1\end{pmatrix} = \begin{pmatrix}-22\\-33\\-22\end{pmatrix} = -11\begin{pmatrix}2\\3\\2\end{pmatrix}.$
So a direction vector for $l$ is $\begin{pmatrix}2\\3\\2\end{pmatrix}$.

3. Find a point on l by solving the planes simultaneously

Solve $5x - 6y + 4z = 26$ and $7x - 4y - z = 0$. From the second, $z = 7x - 4y$. Substituting:
$5x - 6y + 4(7x-4y) = 26 \Rightarrow 33x - 22y = 26.$
Setting $z = 0$ gives $7x - 4y = 0$, i.e. $y = \tfrac{7x}{4}$, and then $33x - 22\cdot\tfrac{7x}{4} = 26 \Rightarrow x = -\tfrac{52}{11},\; y = -\tfrac{91}{11}$.
A point on $l$ is $\left(-\tfrac{52}{11}, -\tfrac{91}{11}, 0\right)$.

4. Write the vector equation of l

$\mathbf{r} = \begin{pmatrix} -\tfrac{52}{11} \\ -\tfrac{91}{11} \\ 0 \end{pmatrix} + t\begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}, \quad t \in \mathbb{R}.$

1. Find the normal vectors for planes $\pi_1$ and $\pi_2$ (results needed from parts a and b).

For $\pi_1$, its direction vectors are $\mathbf{d}_1 = \begin{pmatrix}0 \\ 2 \\ 3\end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix}2 \\ -2 \\ 1\end{pmatrix} - \begin{pmatrix}0 \\ 1 \\ 8\end{pmatrix} = \begin{pmatrix}2 \\ -3 \\ -7\end{pmatrix}$.
Its normal vector is $\mathbf{n}_1 = \begin{pmatrix}0 \\ 2 \\ 3\end{pmatrix} \times \begin{pmatrix}2 \\ -3 \\ -7\end{pmatrix} = \begin{pmatrix}-5 \\ 6 \\ -4\end{pmatrix}$.

For $\pi_2$, its direction vectors are $\begin{pmatrix}0 \\ -1 \\ 4\end{pmatrix}$ and $\begin{pmatrix}1 \\ 2 \\ -1\end{pmatrix}$.
Its normal vector is $\mathbf{n}_2 = \begin{pmatrix}0 \\ -1 \\ 4\end{pmatrix} \times \begin{pmatrix}1 \\ 2 \\ -1\end{pmatrix} = \begin{pmatrix}-7 \\ 4 \\ 1\end{pmatrix}$.

2. Determine the direction vector of the line $l$ (result needed from part c).

Line $l$ is the intersection of $\pi_1$ and $\pi_2$, so its direction vector $\mathbf{d}_l$ is perpendicular to both $\mathbf{n}_1$ and $\mathbf{n}_2$:
$\mathbf{d}_l = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix}-5 \\ 6 \\ -4\end{pmatrix} \times \begin{pmatrix}-7 \\ 4 \\ 1\end{pmatrix} = \begin{pmatrix}22 \\ 33 \\ 22\end{pmatrix}$
We can scale this to get a simpler parallel direction vector for $l$:
$\mathbf{d}_l = \begin{pmatrix}2 \\ 3 \\ 2\end{pmatrix}$

3. Use the condition that $\pi_3$ is parallel to $l$ to solve for $m$.

The normal vector of plane $\pi_3$ is $\mathbf{n}_3 = \begin{pmatrix}3 \\ 2 \\ m\end{pmatrix}$.
Since the plane $\pi_3$ is parallel to the line $l$, the normal of $\pi_3$ is perpendicular to the direction of $l$:
$\mathbf{n}_3 \cdot \mathbf{d}_l = 0$
$\begin{pmatrix}3 \\ 2 \\ m\end{pmatrix} \cdot \begin{pmatrix}2 \\ 3 \\ 2\end{pmatrix} = 0$
$3(2) + 2(3) + m(2) = 0$
$12 + 2m = 0$
$2m = -12 \implies m = -6 \text{ (shown)}$

1. State the cartesian equations of planes $\pi_1$ and $\pi_2$ (which would have been derived in earlier parts).

For $\pi_1$, the normal vector is $\mathbf{n}_1 = \begin{pmatrix}0\\2\\3\end{pmatrix} \times \left( \begin{pmatrix}2\\-2\\1\end{pmatrix} - \begin{pmatrix}0\\1\\8\end{pmatrix} \right) = \begin{pmatrix}0\\2\\3\end{pmatrix} \times \begin{pmatrix}2\\-3\\-7\end{pmatrix} = \begin{pmatrix}-5\\6\\-4\end{pmatrix}$.
The equation of $\pi_1$ is $\mathbf{r}\cdot\begin{pmatrix}-5\\6\\-4\end{pmatrix} = \begin{pmatrix}0\\1\\8\end{pmatrix}\cdot\begin{pmatrix}-5\\6\\-4\end{pmatrix} = -26 \implies 5x - 6y + 4z = 26$.

For $\pi_2$, the normal is $\mathbf{n}_2 = \begin{pmatrix}0\\-1\\4\end{pmatrix} \times \begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}-7\\4\\1\end{pmatrix}$.
The equation of $\pi_2$ is $\mathbf{r}\cdot\begin{pmatrix}-7\\4\\1\end{pmatrix} = 0 \implies -7x + 4y + z = 0$.

2. Find the position vector of a point $A$ on the line $l$ (as established in part (c)).

The line $l$ is the intersection of $\pi_1$ and $\pi_2$. We find a point on $l$ by solving their equations simultaneously.
Let $x = 0$. The equations become:
$-6y + 4z = 26$
$4y + z = 0 \implies z = -4y$

Substituting $z = -4y$ into the first equation:
$-6y + 4(-4y) = 26 \implies -22y = 26 \implies y = -\frac{13}{11}$
Then $z = -4\left(-\frac{13}{11}\right) = \frac{52}{11}$.
Thus, the position vector of a valid point $A$ on $l$ is $\overrightarrow{OA} = \begin{pmatrix} 0 \\ -13/11 \\ 52/11 \end{pmatrix}$.

3. Evaluate the required dot product with the position vector of $A$.

We compute the dot product of $\overrightarrow{OA}$ with the given vector (which happens to be the normal to $\pi_3$ from part (d)):
$\overrightarrow{OA} \cdot (3\mathbf{i} + 2\mathbf{j} - 6\mathbf{k}) = \begin{pmatrix} 0 \\ -13/11 \\ 52/11 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ -6 \end{pmatrix}$
$= 0(3) + \left(-\frac{13}{11}\right)(2) + \left(\frac{52}{11}\right)(-6)$
$= -\frac{26}{11} - \frac{312}{11} = -\frac{338}{11}$
(Note: Because line $l$ is parallel to $\pi_3$, the dot product of the position vector of any point on $l$ with the normal vector of $\pi_3$ is a constant.)

1. Convert 'distance between line and parallel plane' to a point–plane distance

From part (d), $l$ is parallel to $\pi_3$, so $l$ never meets $\pi_3$ and every point on $l$ is at the same perpendicular distance from $\pi_3$.

Hence the distance between $l$ and $\pi_3$ equals the perpendicular distance from a point $A$ on $l$ to $\pi_3: 3x+2y-6z=0$ (which passes through $O$), given by
$\text{distance}=\frac{|\overrightarrow{OA}\cdot\mathbf{n}|}{|\mathbf{n}|},\qquad \mathbf{n}=3\mathbf{i}+2\mathbf{j}-6\mathbf{k}.$

2. Substitute the part (e) value and the magnitude of the normal

From part (e), $\overrightarrow{OA}\cdot(3\mathbf{i}+2\mathbf{j}-6\mathbf{k})=-\dfrac{338}{11}$, and
$|\mathbf{n}|=\sqrt{3^2+2^2+(-6)^2}=\sqrt{49}=7.$
Therefore
$\text{distance}=\frac{\left|-\tfrac{338}{11}\right|}{7}=\frac{338}{77}\approx 4.39\text{ units}.$

1. Recognise that l′ is parallel to l and set up the point-reflection plan

From (d), $l \parallel \pi_3$, so the reflection $l'$ has the SAME direction vector $\begin{pmatrix}2\\3\\2\end{pmatrix}$ as $l$. It is enough to reflect one point $A$ of $l$ in $\pi_3$.

$\pi_3$ passes through $O$ with normal $\mathbf{n}=\begin{pmatrix}3\\2\\-6\end{pmatrix}$, $|\mathbf{n}|^2=9+4+36=49$.

Reflection of $A$ in $\pi_3$:
$\overrightarrow{OA'}=\overrightarrow{OA}-2\,\frac{\overrightarrow{OA}\cdot\mathbf{n}}{|\mathbf{n}|^2}\,\mathbf{n}.$

2. Use OA·n from (e) to find the image point and write the equation of l′

Take the point $A=\left(0,-\tfrac{13}{11},\tfrac{52}{11}\right)$ on $l$ (from (c)). From (e), $\overrightarrow{OA}\cdot\mathbf{n}=-\dfrac{338}{11}$.

$\overrightarrow{OA'}=\overrightarrow{OA}-2\cdot\frac{-\tfrac{338}{11}}{49}\begin{pmatrix}3\\2\\-6\end{pmatrix}=\overrightarrow{OA}+\frac{676}{539}\begin{pmatrix}3\\2\\-6\end{pmatrix}.$

Componentwise: $\overrightarrow{OA'}=\begin{pmatrix}\tfrac{2028}{539}\\[2pt]\tfrac{65}{49}\\[2pt]-\tfrac{1508}{539}\end{pmatrix}.$

Hence $l':\ \mathbf{r}=\overrightarrow{OA'}+t\begin{pmatrix}2\\3\\2\end{pmatrix},\ t\in\mathbb{R}.$