The points A, B and C have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. The points A and B are fixed while C varies.

1. Apply the anti-commutative property of the vector product to equate the expression to the zero vector.

$(\mathbf{c}-\mathbf{a})\times\mathbf{b} = \mathbf{b}\times(\mathbf{c}-\mathbf{a})$
Since $\mathbf{u}\times\mathbf{v} = -\mathbf{v}\times\mathbf{u}$ for any vectors $\mathbf{u}$ and $\mathbf{v}$, we have:
$(\mathbf{c}-\mathbf{a})\times\mathbf{b} = -(\mathbf{c}-\mathbf{a})\times\mathbf{b}$
$2((\mathbf{c}-\mathbf{a})\times\mathbf{b}) = \mathbf{0}$
$(\mathbf{c}-\mathbf{a})\times\mathbf{b} = \mathbf{0}$

2. Deduce the geometrical relationship between the two vectors.

The vector product of two vectors is the zero vector if and only if the vectors are parallel (or if at least one is the zero vector).
Therefore, the vector $(\mathbf{c}-\mathbf{a})$ is parallel to $\mathbf{b}$ (which includes the possibility that $\mathbf{c}-\mathbf{a} = \mathbf{0}$), meaning $(\mathbf{c}-\mathbf{a}) = k\mathbf{b}$ for some scalar $k$.

1. Establish the relationship from part (a) and write the vector equation for C.

From part (a), using the anti-commutative property of the cross product, we have:
$(\mathbf{c}-\mathbf{a}) \times \mathbf{b} = - ((\mathbf{c}-\mathbf{a}) \times \mathbf{b})$
$2((\mathbf{c}-\mathbf{a}) \times \mathbf{b}) = \mathbf{0}$
$(\mathbf{c}-\mathbf{a}) \times \mathbf{b} = \mathbf{0}$
This implies that the vector $(\mathbf{c}-\mathbf{a})$ is parallel to $\mathbf{b}$. Thus, we can write $\mathbf{c}-\mathbf{a} = \lambda \mathbf{b}$ for some scalar parameter $\lambda \in \mathbb{R}$, which gives:
$\mathbf{c} = \mathbf{a} + \lambda \mathbf{b}$

2. Interpret the vector equation geometrically.

The equation $\mathbf{c} = \mathbf{a} + \lambda \mathbf{b}$ represents the vector equation of a straight line $\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}$.
Geometrically, the set of all possible positions of C is a straight line that passes through the fixed point A (with position vector $\mathbf{a}$) and is parallel to the vector $\mathbf{b}$ (which represents the direction of the fixed line $OB$).