If $$|\vec{a}| = 2$$, $$|\vec{b}| = 5$$ and $$|\vec{a} \times \vec{b}| = 8$$, then $$|\vec{a} \cdot \vec{b}|$$ is equal to:

We start by recalling the two standard formulas that relate the magnitudes of the dot and cross products of two vectors $$\vec a$$ and $$\vec b$$ with the angle $$\theta$$ between them.

Dot-product magnitude formula: $$|\vec a \cdot \vec b| \;=\; |\vec a|\,|\vec b|\,|\cos\theta|.$$

Cross-product magnitude formula: $$|\vec a \times \vec b| \;=\; |\vec a|\,|\vec b|\,|\sin\theta|.$$

We are given $$|\vec a| = 2,\; |\vec b| = 5,\; |\vec a \times \vec b| = 8.$$ We first use the cross-product formula to determine $$\sin\theta.$$

Substituting the known magnitudes, we have

$$8 \;=\; |\vec a \times \vec b| \;=\; |\vec a|\,|\vec b|\,|\sin\theta|$$

$$8 \;=\; 2 \times 5 \times |\sin\theta|$$

$$8 \;=\; 10\,|\sin\theta|$$

Dividing both sides by 10,

$$|\sin\theta| \;=\; \frac{8}{10} \;=\; 0.8.$$

Now we find $$|\cos\theta|$$ using the fundamental trigonometric identity $$\sin^{2}\theta + \cos^{2}\theta = 1.$$ Hence,

$$|\cos\theta| \;=\; \sqrt{1 - (\sin^{2}\theta)}$$

$$|\cos\theta| \;=\; \sqrt{1 - (0.8)^{2}}$$

$$|\cos\theta| \;=\; \sqrt{1 - 0.64}$$

$$|\cos\theta| \;=\; \sqrt{0.36}$$

$$|\cos\theta| \;=\; 0.6.$$

With $$|\cos\theta|$$ known, we substitute into the dot-product magnitude formula:

$$|\vec a \cdot \vec b| \;=\; |\vec a|\,|\vec b|\,|\cos\theta|$$

$$|\vec a \cdot \vec b| \;=\; 2 \times 5 \times 0.6$$

$$|\vec a \cdot \vec b| \;=\; 10 \times 0.6$$

$$|\vec a \cdot \vec b| \;=\; 6.$$

Hence, the correct answer is Option A.