The scalar product of two vectors is defined as $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$. Since the range of the cosine function is $[-1, 1]$, it follows that $\cos \theta \le 1$. Therefore, $\vec{a} \cdot \vec{b} \le |\vec{a}||\vec{b}|$. This statement is always true.
The triangle inequality for vectors states that $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$. Option (B) suggests the reverse inequality, which is generally false.
The expression $|\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}|$ is only true under specific conditions (e.g., when vectors are collinear and in the same direction), but it is not true for any two arbitrary vectors.
The magnitude of the cross product is $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta$. Since $|\sin \theta| \le 1$, it follows that $|\vec{a} \times \vec{b}| \le |\vec{a}||\vec{b}|$. Option (D) suggests the reverse, which is false.
Final Answer: (A)
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