In triangle $PQR$, the vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ represent the vectors $\vec{PQ}$, $\vec{QR}$ and $\vec{PR}$ respectively.

1. State the geometrical interpretation of $|\mathbf{c} \times \mathbf{\hat{a}}|$.

$|\mathbf{c} \times \mathbf{\hat{a}}|$ represents the perpendicular distance from point $R$ to the line passing through $P$ and $Q$ (i.e. the height of $\triangle PQR$ when $PQ$ is taken as the base).

2. Express the area of triangle $PQR$ in terms of the base and this height.

Taking $PQ$ as the base, the length of the base is $|\mathbf{a}|$.

$\text{Area of } \triangle PQR = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} |\mathbf{a}| |\mathbf{c} \times \mathbf{\hat{a}}|$

3. Substitute the definition of the unit vector $\mathbf{\hat{a}}$ to show the required area formula.

Since $\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$, we can substitute this into the expression:

$\text{Area} = \frac{1}{2} |\mathbf{a}| \left| \mathbf{c} \times \frac{\mathbf{a}}{|\mathbf{a}|} \right|$

Because $\frac{1}{|\mathbf{a}|}$ is a positive scalar, it can be factored out of the magnitude:

$\text{Area} = \frac{1}{2} |\mathbf{a}| \left( \frac{1}{|\mathbf{a}|} \right) |\mathbf{c} \times \mathbf{a}|$
$\text{Area} = \frac{1}{2} |\mathbf{c} \times \mathbf{a}| \text{ units}^2 \text{ (shown)}$

1. Use vector addition and properties of the cross product to show that $|\mathbf{c} \times \mathbf{a}| = |\mathbf{b} \times \mathbf{a}|$.

By the triangle law of vector addition, $\vec{PR} = \vec{PQ} + \vec{QR}$, so $\mathbf{c} = \mathbf{a} + \mathbf{b}$.

Taking the cross product with $\mathbf{a}$ on both sides:
$\mathbf{c} \times \mathbf{a} = (\mathbf{a} + \mathbf{b}) \times \mathbf{a}$
$\mathbf{c} \times \mathbf{a} = \mathbf{a} \times \mathbf{a} + \mathbf{b} \times \mathbf{a}$

Since the cross product of any vector with itself is the zero vector ($\mathbf{a} \times \mathbf{a} = \mathbf{0}$):
$\mathbf{c} \times \mathbf{a} = \mathbf{b} \times \mathbf{a}$

Taking the magnitude of both sides yields:
$|\mathbf{c} \times \mathbf{a}| = |\mathbf{b} \times \mathbf{a}|$

2. Express the magnitude of both cross products using the definition involving angles.

The magnitude of a cross product is given by $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin \theta$, where $\theta$ is the angle between the vectors when placed tail-to-tail.

For $\mathbf{c}$ and $\mathbf{a}$, both originate from point $P$, so the angle between them is exactly the interior angle $\angle QPR$. Thus:
$|\mathbf{c} \times \mathbf{a}| = |\mathbf{c}| |\mathbf{a}| \sin \angle QPR$

For $\mathbf{b}$ ($\vec{QR}$) and $\mathbf{a}$ ($\vec{PQ}$), placing them tail-to-tail at $Q$ means extending $\vec{PQ}$. The angle between them is the exterior angle of the triangle at $Q$, which is $180^\circ - \angle PQR$. Thus:
$|\mathbf{b} \times \mathbf{a}| = |\mathbf{b}| |\mathbf{a}| \sin(180^\circ - \angle PQR)$

Since $\sin(180^\circ - \theta) = \sin \theta$, this simplifies to:
$|\mathbf{b} \times \mathbf{a}| = |\mathbf{b}| |\mathbf{a}| \sin \angle PQR$

3. Equate the two expressions and rearrange to deduce the Sine Rule.

Substitute the expressions from Step 2 into the equality from Step 1:
$|\mathbf{c}| |\mathbf{a}| \sin \angle QPR = |\mathbf{b}| |\mathbf{a}| \sin \angle PQR$

Since $\mathbf{a}$ forms a side of the triangle, its magnitude is non-zero ($|\mathbf{a}| \neq 0$). Dividing both sides by $|\mathbf{a}|$ gives:
$|\mathbf{c}| \sin \angle QPR = |\mathbf{b}| \sin \angle PQR$

Dividing both sides by $\sin \angle PQR \sin \angle QPR$, we obtain:
$\frac{|\mathbf{c}|}{\sin \angle PQR} = \frac{|\mathbf{b}|}{\sin \angle QPR} \quad \text{(Shown)}$