相对于原点 $O$，点 $A$ 和 $C$ 的位置向量分别为 $\mathbf{a}$ and $2\mathbf{b}$，使得
$\mathbf{a} = 5p\mathbf{i} - 2p\mathbf{j} + 4p\mathbf{k} \quad \text{and} \quad \mathbf{b} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k},$
其中 $p$ 为正常数。已知 $|\mathbf{a}| = 2|\mathbf{b}|$。

1. Calculate the magnitude of vector $\mathbf{b}$.

$|\mathbf{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

2. Find the magnitude of $\mathbf{a}$ in terms of $p$ and solve the given equation to find the exact value of $p$.

$|\mathbf{a}| = \sqrt{(5p)^2 + (-2p)^2 + (4p)^2}$
$|\mathbf{a}| = \sqrt{25p^2 + 4p^2 + 16p^2} = \sqrt{45p^2} = 3p\sqrt{5} \quad (\text{since } p > 0)$
Substituting into $|\mathbf{a}| = 2|\mathbf{b}|$:
$3p\sqrt{5} = 2(3)$
$3p\sqrt{5} = 6$
$p = \frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

1. Expand the given scalar product using distributive properties.

$(\mathbf{a} + 2\mathbf{b}) \cdot (\mathbf{a} - 2\mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - 2\mathbf{a} \cdot \mathbf{b} + 2\mathbf{b} \cdot \mathbf{a} - 4\mathbf{b} \cdot \mathbf{b}$
$= |\mathbf{a}|^2 - 4|\mathbf{b}|^2$

2. Evaluate the expression using the given condition on magnitudes.

Substitute $|\mathbf{a}| = 2|\mathbf{b}|$ into the expanded expression:
$= (2|\mathbf{b}|)^2 - 4|\mathbf{b}|^2$
$= 4|\mathbf{b}|^2 - 4|\mathbf{b}|^2 = 0$

1. Find the value of $p$ from the condition given in part (a) since it is needed for the evaluation.

$|\mathbf{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3$

$|\mathbf{a}| = \sqrt{(5p)^2 + (-2p)^2 + (4p)^2} = \sqrt{45p^2} = 3p\sqrt{5} \quad \text{(since } p > 0\text{)}$

$\text{Given } |\mathbf{a}| = 2|\mathbf{b}|: 3p\sqrt{5} = 2(3) \implies p = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

2. Evaluate the cross product $\mathbf{a} \times \mathbf{b}$ in terms of $p$, and then substitute the exact value of $p$.

$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 5p \\ -2p \\ 4p \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$
$= \begin{pmatrix} (-2p)(2) - (4p)(-2) \\ (4p)(1) - (5p)(2) \\ (5p)(-2) - (-2p)(1) \end{pmatrix}$
$= \begin{pmatrix} -4p + 8p \\ 4p - 10p \\ -10p + 2p \end{pmatrix} = \begin{pmatrix} 4p \\ -6p \\ -8p \end{pmatrix}$

$\text{Substitute } p = \frac{2\sqrt{5}}{5} \text{ to get:}$
$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} \frac{8\sqrt{5}}{5} \\ -\frac{12\sqrt{5}}{5} \\ -\frac{16\sqrt{5}}{5} \end{pmatrix} = \frac{8\sqrt{5}}{5}\mathbf{i} - \frac{12\sqrt{5}}{5}\mathbf{j} - \frac{16\sqrt{5}}{5}\mathbf{k}$

3. Set up the area of triangle $OAC$ using the evaluated cross product.

$\text{The position vectors are } \vec{OA} = \mathbf{a} \text{ and } \vec{OC} = 2\mathbf{b}.$

$\text{Area of } \triangle OAC = \frac{1}{2} |\vec{OA} \times \vec{OC}|$
$= \frac{1}{2} |\mathbf{a} \times 2\mathbf{b}|$
$= \frac{1}{2} \times 2 |\mathbf{a} \times \mathbf{b}| = |\mathbf{a} \times \mathbf{b}|$

4. Calculate the exact area of the triangle.

$\text{Area} = \sqrt{(4p)^2 + (-6p)^2 + (-8p)^2}$
$= \sqrt{16p^2 + 36p^2 + 64p^2} = \sqrt{116p^2} = 2p\sqrt{29}$

$\text{Substitute } p = \frac{2}{\sqrt{5}}:$
$\text{Area} = 2\left(\frac{2}{\sqrt{5}}\right)\sqrt{29} = \frac{4\sqrt{29}}{\sqrt{5}} = \frac{4\sqrt{145}}{5}$

1. Relate the given vectors to the base and perpendicular height of $\triangle OAC$.

In $\triangle OAC$, the base vector is $\vec{CA} = \mathbf{a} - 2\mathbf{b}$.
Let $M$ be the midpoint of $AC$. The median vector from the origin is $\vec{OM} = \frac{1}{2}(\mathbf{a} + 2\mathbf{b})$.
From part (b), we know $(\mathbf{a} + 2\mathbf{b}) \cdot (\mathbf{a} - 2\mathbf{b}) = 0$.
Therefore, $\vec{OM} \cdot \vec{CA} = \frac{1}{2}(\mathbf{a} + 2\mathbf{b}) \cdot (\mathbf{a} - 2\mathbf{b}) = 0$.
This means the median $\vec{OM}$ is perpendicular to the base $AC$ and acts as the height of the triangle.

2. Express the area of $\triangle OAC$ geometrically and equate it to the cross-product area from part (c).

Using the geometric formula for the area of a triangle ($\frac{1}{2} \times \text{base} \times \text{height}$):
$\text{Area of } \triangle OAC = \frac{1}{2} |\vec{CA}| |\vec{OM}|$
$= \frac{1}{2} |\mathbf{a} - 2\mathbf{b}| \left| \frac{1}{2}(\mathbf{a} + 2\mathbf{b}) \right|$
$= \frac{1}{4} |\mathbf{a} + 2\mathbf{b}| |\mathbf{a} - 2\mathbf{b}|$
From part (c), we also know the area of $\triangle OAC$ is given by $|\mathbf{a} \times \mathbf{b}|$.
Equating the two expressions gives the required result:
$|\mathbf{a} \times \mathbf{b}| = \frac{1}{4} (|\mathbf{a} + 2\mathbf{b}|)(|\mathbf{a} - 2\mathbf{b}|)$