以原点 $O$ 为基准，点 $A$、$B$ 和 $C$ 的位置向量分别为 $\mathbf{a}$、$\mathbf{b}$ 和 $\mathbf{c}$。已知 $\mathbf{a} + \mathbf{b} - \mathbf{c}$ 与 $\mathbf{a} - \mathbf{b} + \mathbf{c}$ 平行。

1. Set up the condition for parallel vectors using the vector product.

Since the two given vectors are parallel, their vector product is the zero vector:
$(\mathbf{a} + \mathbf{b} - \mathbf{c}) \times (\mathbf{a} - \mathbf{b} + \mathbf{c}) = \mathbf{0}$

2. Expand the vector product expression.

Expanding using the distributive property:
$\mathbf{a}\times\mathbf{a} - \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c} + \mathbf{b}\times\mathbf{a} - \mathbf{b}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} - \mathbf{c}\times\mathbf{a} + \mathbf{c}\times\mathbf{b} - \mathbf{c}\times\mathbf{c} = \mathbf{0}$

3. Simplify the expansion using properties of the vector product.

Applying $\mathbf{v}\times\mathbf{v} = \mathbf{0}$ and $\mathbf{u}\times\mathbf{v} = -\mathbf{v}\times\mathbf{u}$:
$\mathbf{0} - \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c} - \mathbf{a}\times\mathbf{b} - \mathbf{0} + \mathbf{b}\times\mathbf{c} + \mathbf{a}\times\mathbf{c} - \mathbf{b}\times\mathbf{c} - \mathbf{0} = \mathbf{0}$
$-2(\mathbf{a}\times\mathbf{b}) + 2(\mathbf{a}\times\mathbf{c}) = \mathbf{0}$

4. Factorise and relate back to the required displacement vectors.

Dividing by 2 and factorising:
$\mathbf{a}\times\mathbf{c} - \mathbf{a}\times\mathbf{b} = \mathbf{0}$
$\mathbf{a} \times (\mathbf{c} - \mathbf{b}) = \mathbf{0}$
Since we are given position vectors with respect to origin $O$, we know $\overrightarrow{OA} = \mathbf{a}$ and the displacement vector $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$.
Substituting these in yields:
$\overrightarrow{OA} \times \overrightarrow{BC} = \mathbf{0}$
This vector product equalling the zero vector confirms that $\overrightarrow{OA}$ and $\overrightarrow{BC}$ are parallel.

1. Set up the scalar product equation using the given angle and vector magnitudes.

The angle $OBC$ is the angle between the vectors $\overrightarrow{BO}$ and $\overrightarrow{BC}$.
By the definition of the scalar product:
$\overrightarrow{BO} \cdot \overrightarrow{BC} = |\overrightarrow{BO}| \, |\overrightarrow{BC}| \cos 60^\circ$
Given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, $|\mathbf{b}| = 1$ and $|\overrightarrow{OA}| = |\mathbf{a}| = 1$.
Evaluating the magnitudes:
$|\overrightarrow{BO}| = |-\mathbf{b}| = 1$
$|\overrightarrow{BC}| = |2\overrightarrow{OA}| = 2(1) = 2$
Substituting these values into the scalar product:
$\overrightarrow{BO} \cdot \overrightarrow{BC} = (1)(2)\cos 60^\circ = (1)(2)\left(\frac{1}{2}\right) = 1$

2. Expand the scalar product in terms of the position vectors $\mathbf{b}$ and $\mathbf{c}$.

Express $\overrightarrow{BO}$ and $\overrightarrow{BC}$ using position vectors:
$\overrightarrow{BO} = -\mathbf{b}$
$\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$
Now expand their scalar product:
$\overrightarrow{BO} \cdot \overrightarrow{BC} = (-\mathbf{b}) \cdot (\mathbf{c} - \mathbf{b})$
$= -\mathbf{b} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{b}$
$= -\mathbf{b} \cdot \mathbf{c} + |\mathbf{b}|^2$
Since $\mathbf{b}$ is a unit vector, $|\mathbf{b}|^2 = 1^2 = 1$:
$\overrightarrow{BO} \cdot \overrightarrow{BC} = 1 - \mathbf{b} \cdot \mathbf{c}$

3. Equate the expressions to deduce that the dot product is zero, confirming perpendicularity.

Equating the two expressions obtained for $\overrightarrow{BO} \cdot \overrightarrow{BC}$:
$1 - \mathbf{b} \cdot \mathbf{c} = 1$
$\mathbf{b} \cdot \mathbf{c} = 0$
Since $\mathbf{b} = \overrightarrow{OB}$ and $\mathbf{c} = \overrightarrow{OC}$:
$\overrightarrow{OB} \cdot \overrightarrow{OC} = 0$
As their scalar product is zero, it is shown that $\overrightarrow{OB}$ and $\overrightarrow{OC}$ are perpendicular.

1. Find the base area of the pyramid $HOBC$.

The area of the base $\triangle OBC$ is given by $\frac{1}{2}|\mathbf{b} \times \mathbf{c}|$.
Since $\overrightarrow{OB} \perp \overrightarrow{OC}$ (from part b), the angle between $\mathbf{b}$ and $\mathbf{c}$ is $90^\circ$.
$|\mathbf{b} \times \mathbf{c}| = |\mathbf{b}||\mathbf{c}|\sin 90^\circ$
Using $|\mathbf{b}| = 1$ (from part b) and $|\mathbf{c}| = \sqrt{3}$,
$|\mathbf{b} \times \mathbf{c}| = (1)(\sqrt{3})(1) = \sqrt{3}$
$\text{Base area} = \frac{\sqrt{3}}{2}$

2. Determine the perpendicular height of the pyramid.

The position vector of $H$ is $\mathbf{h} = \mathbf{b} \times \mathbf{c}$.
By the properties of the vector product, $\overrightarrow{OH}$ is perpendicular to both $\mathbf{b}$ and $\mathbf{c}$, meaning it is a normal vector perpendicular to the plane containing the base $OBC$.
Thus, the perpendicular height of the pyramid is simply the magnitude of $\overrightarrow{OH}$.
$\text{Height} = |\mathbf{h}| = |\mathbf{b} \times \mathbf{c}| = \sqrt{3}$

3. Calculate the exact volume of the pyramid.

$\text{Volume} = \frac{1}{3} \times \text{base area} \times \text{height}$
$\text{Volume} = \frac{1}{3} \left(\frac{\sqrt{3}}{2}\right) (\sqrt{3})$
$\text{Volume} = \frac{3}{6} = \frac{1}{2}$