If $$\overrightarrow {\text{a}} $$ and $$\overrightarrow {\text{b}} $$ are two arbitrary vectors with magnitudes a and b, respectively, $${\left| {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right|^2}$$ will be equal to A. $${{\text{a}}^2}{{\text{b}}^2} – {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$ B. $${\text{ab}} – \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$ C. $${{\text{a}}^2}{{\text{b}}^2} + {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$ D. $${\text{ab}} + \overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} $$

The correct answer is $\boxed{{\text{a}}^2}{{\text{b}}^2} – {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$.

The cross product of two vectors $\overrightarrow {\text{a}}$ and $\overrightarrow {\text{b}}$ is denoted by $\overrightarrow {\text{a}} \times \overrightarrow {\text{b}}$. The magnitude of the cross product is given by the formula

$$\left| {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right| = \sqrt{{\text{a}}^2{{\text{b}}^2} – {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}}$$

Therefore, the square of the magnitude of the cross product is given by

$${\left| {\overrightarrow {\text{a}} \times \overrightarrow {\text{b}} } \right|^2} = {\text{a}}^2{{\text{b}}^2} – {\left( {\overrightarrow {\text{a}} \cdot \overrightarrow {\text{b}} } \right)^2}$$

Option A is incorrect because it does not include the factor of $-1$ in front of the dot product. Option B is incorrect because it does not include the square of the magnitudes of the vectors. Option C is incorrect because it does not include the negative sign in front of the dot product. Option D is incorrect because it includes the dot product instead of the cross product.