点 A, B 和 C 的位置向量分别为 $2i,\;3j$ 和 $4k$。求一个垂直于包含 A, B 和 C 的平面 $\Pi_1$ 的向量。 平面 $\Pi_2$ 的方程为 $\mathbf{r} = i + 4j + 2k + \lambda(i - j) + \mu(j - k).$ 求平面 $\Pi_1$ 与 $\Pi_2$ 之间的锐角。

1. Find two non-parallel vectors on the plane.

$\vec{AB} = \mathbf{b} - \mathbf{a} = 3\mathbf{j} - 2\mathbf{i} = \begin{pmatrix} -2 \\ 3 \\ 0 \end{pmatrix}$
$\vec{AC} = \mathbf{c} - \mathbf{a} = 4\mathbf{k} - 2\mathbf{i} = \begin{pmatrix} -2 \\ 0 \\ 4 \end{pmatrix}$

2. Calculate the cross product of these two vectors to find a normal vector.

$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} -2 \\ 3 \\ 0 \end{pmatrix} \times \begin{pmatrix} -2 \\ 0 \\ 4 \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} (3)(4) - (0)(0) \\ (0)(-2) - (-2)(4) \\ (-2)(0) - (3)(-2) \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} 12 \\ 8 \\ 6 \end{pmatrix}$

3. State a simpler vector parallel to the cross product.

$\begin{pmatrix} 12 \\ 8 \\ 6 \end{pmatrix} = 2 \begin{pmatrix} 6 \\ 4 \\ 3 \end{pmatrix}$
A suitable perpendicular vector is $6\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$.
(Note: Any non-zero scalar multiple of this vector is a correct answer.)

1. Recall or compute the normal vector to plane $\Pi_1$ from the earlier part.

From the earlier part, the normal to $\Pi_1$ can be found using the cross product of two vectors on the plane. Let $A$, $B$, and $C$ have position vectors $\mathbf{a} = 2\mathbf{i}$, $\mathbf{b} = 3\mathbf{j}$, and $\mathbf{c} = 4\mathbf{k}$.
$\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} -2 \\ 3 \\ 0 \end{pmatrix}, \quad \vec{AC} = \mathbf{c} - \mathbf{a} = \begin{pmatrix} -2 \\ 0 \\ 4 \end{pmatrix}$
$\mathbf{n}_1 = \vec{AB} \times \vec{AC} = \begin{pmatrix} -2 \\ 3 \\ 0 \end{pmatrix} \times \begin{pmatrix} -2 \\ 0 \\ 4 \end{pmatrix} = \begin{pmatrix} 12 \\ 8 \\ 6 \end{pmatrix} = 2\begin{pmatrix} 6 \\ 4 \\ 3 \end{pmatrix}$
We take $\mathbf{n}_1 = \begin{pmatrix} 6 \\ 4 \\ 3 \end{pmatrix}$.

2. Determine a normal vector to plane $\Pi_2$.

The plane $\Pi_2$ is given in parametric form:
$\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$
The direction vectors parallel to the plane are $\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$.
A normal vector to $\Pi_2$ is found using the cross product $\mathbf{n}_2 = \mathbf{d}_1 \times \mathbf{d}_2$:
$\mathbf{n}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (-1)(-1) - (0)(1) \\ (0)(0) - (1)(-1) \\ (1)(1) - (-1)(0) \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

3. Set up the dot product to find the acute angle between the two planes.

The acute angle $\theta$ between the planes is equal to the acute angle between their respective normal vectors:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$

4. Evaluate the dot product and the magnitudes of the normal vectors.

$\mathbf{n}_1 \cdot \mathbf{n}_2 = \begin{pmatrix} 6 \\ 4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 6(1) + 4(1) + 3(1) = 13$
$|\mathbf{n}_1| = \sqrt{6^2 + 4^2 + 3^2} = \sqrt{36 + 16 + 9} = \sqrt{61}$
$|\mathbf{n}_2| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

5. Calculate the acute angle.

Substitute the values into the angle formula:
$\cos \theta = \frac{13}{\sqrt{61}\sqrt{3}} = \frac{13}{\sqrt{183}}$
$\theta = \cos^{-1}\left(\frac{13}{\sqrt{183}}\right) \approx 16.102^\circ \approx 16.1^\circ \text{ (to 1 d.p.)}$