以原点 O 为基准，点 A, B, C, D 的位置向量分别为 $2i - j + k,\quad i - 2k,\quad -i + 3j + 2k,\quad -i + j + 4k,$ 。求：

1. Set up the vector product of two non-parallel vectors in the plane OAB

The points A and B have position vectors $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}$ and $\mathbf{b} = \mathbf{i} - 2\mathbf{k}$.
Since O is the origin, $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$ are two vectors lying in the plane OAB.
A vector perpendicular to the plane OAB is given by the cross product $\mathbf{a} \times \mathbf{b}$.
$\mathbf{n} = (2\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (\mathbf{i} - 2\mathbf{k})$

2. Evaluate the vector product

$\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 0 & -2 \end{vmatrix}$
$\mathbf{n} = \mathbf{i}( (-1)(-2) - (1)(0) ) - \mathbf{j}( (2)(-2) - (1)(1) ) + \mathbf{k}( (2)(0) - (-1)(1) )$
$\mathbf{n} = 2\mathbf{i} + 5\mathbf{j} + \mathbf{k}$

1. Find the normal vector to the plane OCD and recall the normal vector to OAB.

From part (i), the normal vector to plane OAB is $\mathbf{n}_1 = \mathbf{a} \times \mathbf{b} = \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix}$.

The normal vector to plane OCD is given by the cross product of position vectors $\mathbf{c}$ and $\mathbf{d}$:
$\mathbf{n}_2 = \mathbf{c} \times \mathbf{d} = \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix} \times \begin{pmatrix} -1 \\ 1 \\ 4 \end{pmatrix}$
$\mathbf{n}_2 = \begin{pmatrix} (3)(4) - (2)(1) \\ (2)(-1) - (-1)(4) \\ (-1)(1) - (3)(-1) \end{pmatrix} = \begin{pmatrix} 10 \\ 2 \\ 2 \end{pmatrix}$

2. Apply the scalar product formula to find the acute angle between the two planes.

The acute angle $\theta$ between the planes is the acute angle between their normal vectors:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\mathbf{n}_1 \cdot \mathbf{n}_2 = \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 2 \\ 2 \end{pmatrix} = 2(10) + 5(2) + 1(2) = 32$
$|\mathbf{n}_1| = \sqrt{2^2 + 5^2 + 1^2} = \sqrt{30}$
$|\mathbf{n}_2| = \sqrt{10^2 + 2^2 + 2^2} = \sqrt{108}$

3. Calculate the value of the angle to the required precision.

$\cos \theta = \frac{32}{\sqrt{30} \sqrt{108}} = \frac{32}{\sqrt{3240}}$
$\theta = \cos^{-1}\left( \frac{32}{\sqrt{3240}} \right) \approx 55.795^\circ$
$\theta \approx 55.8^\circ \text{ (to nearest } 0.1^\circ)$