Let's analyze each option:

(A) \(\vec{a}.\vec{b}\ge 0\)

The dot product \(\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta\), where \(\theta\) is the angle between the vectors. If \(\theta\) is obtuse (greater than 90 degrees), then \(\cos\theta\) is negative, making the dot product negative. So, this statement is not always true.

(B) \(\vec{a}.\vec{b} = \vec{b}.\vec{a}\)

The dot product is commutative, meaning the order of the vectors does not matter. \(\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta\) and \(\vec{b}.\vec{a} = |\vec{b}||\vec{a}|\cos\theta\). Since multiplication is commutative, \(|\vec{a}||\vec{b}| = |\vec{b}||\vec{a}|\), so \(\vec{a}.\vec{b} = \vec{b}.\vec{a}\). This statement is always true.

(C) \(|\vec{a}.\vec{b}| = |\vec{a}||\vec{b}|\)

We know that \(\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta\). Therefore, \(|\vec{a}.\vec{b}| = ||\vec{a}||\vec{b}|\cos\theta| = |\vec{a}||\vec{b}||\cos\theta|\). This is only equal to \(|\vec{a}||\vec{b}|\) when \(|\cos\theta| = 1\), which means \(\theta = 0\) or \(\theta = \pi\). This is not always true.

(D) \(|\vec{a}+\vec{b}| = |\vec{a}| + |\vec{b}|\)

This is only true when \(\vec{a}\) and \(\vec{b}\) are in the same direction (i.e., they are parallel and point in the same direction). In general, \(|\vec{a}+\vec{b}|^2 = (\vec{a}+\vec{b}).(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}.\vec{b}\). So, \(|\vec{a}+\vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta}\). This is only equal to \(|\vec{a}| + |\vec{b}|\) when \(\cos\theta = 1\), which means \(\theta = 0\). This is not always true.