关于原点 $O$，定点 $A$ 和 $B$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$，其中 $\mathbf{a}$ 和 $\mathbf{b}$ 是非零且不平行的向量。

1. Rewrite the position vector equation to identify the geometric locus

$\mathbf{r} = \lambda \mathbf{a} + (1 - \lambda)\mathbf{b}$
$\mathbf{r} = \mathbf{b} + \lambda(\mathbf{a} - \mathbf{b})$
This matches the standard vector equation of a line, $\mathbf{r} = \mathbf{p} + t\mathbf{d}$, where the line passes through a point with position vector $\mathbf{b}$ (point $B$) and is parallel to the direction vector $\mathbf{a} - \mathbf{b}$ (which is the vector $\vec{BA}$). As $\lambda$ varies over all real numbers, $R$ traces out the entire line passing through both $A$ and $B$.

1. Use the definition of the scalar product to show that it evaluates to zero.

The scalar product of the position vectors is given by:
$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\angle AOB)$
Since $\angle AOB = 90^\circ$ and $\cos(90^\circ) = 0$, we have:
$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|(0) = 0$

1. Formulate the dot product equation for the shortest distance from the origin to the line.

For $R^*$ to be the closest point to the origin, the position vector $\vec{OR^*}$ must be perpendicular to the line passing through $A$ and $B$. Therefore, it must be orthogonal to the direction vector $\vec{BA} = \mathbf{a} - \mathbf{b}$.
$\vec{OR^*} \cdot \vec{BA} = 0$
$\left( \lambda \mathbf{a} + (1 - \lambda)\mathbf{b} \right) \cdot (\mathbf{a} - \mathbf{b}) = 0$

2. Expand the dot product and solve for $\lambda$ using the orthogonality of $\mathbf{a}$ and $\mathbf{b}$.

Expanding the dot product gives:
$\lambda (\mathbf{a} \cdot \mathbf{a}) - \lambda (\mathbf{a} \cdot \mathbf{b}) + (1 - \lambda)(\mathbf{b} \cdot \mathbf{a}) - (1 - \lambda)(\mathbf{b} \cdot \mathbf{b}) = 0$
Since $\angle AOB = 90^\circ$, from part (b) we know $\mathbf{a} \cdot \mathbf{b} = 0$. Using the property $\mathbf{x} \cdot \mathbf{x} = |\mathbf{x}|^2$:
$\lambda |\mathbf{a}|^2 - (1 - \lambda)|\mathbf{b}|^2 = 0$
$\lambda |\mathbf{a}|^2 = |\mathbf{b}|^2 - \lambda |\mathbf{b}|^2$
$\lambda (|\mathbf{a}|^2 + |\mathbf{b}|^2) = |\mathbf{b}|^2 \implies \lambda = \frac{|\mathbf{b}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2}$

3. Substitute $\lambda$ back to find the position vector of $R^*$.

Substitute $\lambda$ into the expression for $\mathbf{r}^*$:
$1 - \lambda = 1 - \frac{|\mathbf{b}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2} = \frac{|\mathbf{a}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2}$
$\mathbf{r}^* = \frac{|\mathbf{b}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2} \mathbf{a} + \frac{|\mathbf{a}|^2}{|\mathbf{a}|^2 + |\mathbf{b}|^2} \mathbf{b} = \frac{|\mathbf{b}|^2 \mathbf{a} + |\mathbf{a}|^2 \mathbf{b}}{|\mathbf{a}|^2 + |\mathbf{b}|^2}$

4. Use the ratio theorem to state the ratio $AR^* : BR^*$.

By the ratio theorem, if a point $R^*$ divides the line segment $AB$ in the ratio $AR^* : BR^* = \mu : \nu$, its position vector is:
$\mathbf{r}^* = \frac{\nu \mathbf{a} + \mu \mathbf{b}}{\mu + \nu}$
Comparing this standard formula to our result for $\mathbf{r}^*$, we can equate coefficients to see that $\nu = |\mathbf{b}|^2$ and $\mu = |\mathbf{a}|^2$.
Therefore, the ratio is:
$AR^* : BR^* = \mu : \nu = |\mathbf{a}|^2 : |\mathbf{b}|^2$