Relative to the origin O, the position vectors of two points A and B are a and b respectively. The length of a is k units and b is a unit vector. The angle between a and b is $\frac{\pi}{6}$ radians. It is given that the point M lies on the line segment AB such that AM:AB is 3:4, find exact area of $\Delta OAM$, giving your answer in terms of k.

1. Determine the position vector of point M using the ratio theorem.

Since $M$ lies on the line segment $AB$ such that $AM:AB = 3:4$, the ratio of the lengths $AM:MB$ is $3:1$.

By the ratio theorem:
$\mathbf{m} = \frac{1\mathbf{a} + 3\mathbf{b}}{1 + 3} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}$

2. Express the area of $\Delta OAM$ in terms of the vector product of $\mathbf{a}$ and $\mathbf{m}$.

$\text{Area of } \Delta OAM = \frac{1}{2} |\mathbf{a} \times \mathbf{m}|$

3. Substitute the expression for $\mathbf{m}$ and simplify the vector product using algebraic properties.

$\mathbf{a} \times \mathbf{m} = \mathbf{a} \times \left( \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b} \right)$
$= \frac{1}{4}(\mathbf{a} \times \mathbf{a}) + \frac{3}{4}(\mathbf{a} \times \mathbf{b})$

Since the cross product of a vector with itself is zero ($\mathbf{a} \times \mathbf{a} = \mathbf{0}$):
$\mathbf{a} \times \mathbf{m} = \frac{3}{4}(\mathbf{a} \times \mathbf{b})$

Substituting this back into the area formula:
$\text{Area of } \Delta OAM = \frac{1}{2} \left| \frac{3}{4}(\mathbf{a} \times \mathbf{b}) \right| = \frac{3}{8} |\mathbf{a} \times \mathbf{b}|$

4. Evaluate the magnitude of the vector product $\mathbf{a} \times \mathbf{b}$ and calculate the final area.

Using the magnitude formula for a cross product, where $|\mathbf{a}| = k$, $|\mathbf{b}| = 1$, and the angle is $\frac{\pi}{6}$:
$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\left(\frac{\pi}{6}\right)$
$|\mathbf{a} \times \mathbf{b}| = (k)(1)\left(\frac{1}{2}\right) = \frac{k}{2}$

Therefore, the exact area is:
$\text{Area} = \frac{3}{8} \left( \frac{k}{2} \right) = \frac{3}{16}k$