Let the angle between $\vec A$ and $\vec B$ be $\theta$.

Use the given condition: $\vec A \cdot \vec B = |\vec A \times \vec B|$

Now, $\vec A \cdot \vec B = AB\cos\theta$ and $|\vec A \times \vec B| = AB\sin\theta$

So, $AB\cos\theta = AB\sin\theta$

Assuming $A,B \neq 0$, divide by $AB$: $\cos\theta = \sin\theta$ $\tan\theta = 1$ $\theta = 45^\circ$

Since $|\vec A \times \vec B| \ge 0$, we also need $\vec A\cdot\vec B \ge 0$, so $\cos\theta \ge 0$, which is consistent with $\theta=45^\circ$.

Now find $|\vec A-\vec B|$ using $|\vec A-\vec B|^2 = A^2 + B^2 - 2AB\cos\theta$

Substitute $\cos 45^\circ = \frac{1}{\sqrt2}$: $|\vec A-\vec B|^2 = A^2 + B^2 - 2AB\left(\frac{1}{\sqrt2}\right)$ $|\vec A-\vec B|^2 = A^2 + B^2 - \sqrt2\,AB$

Therefore, $|\vec A-\vec B| = \sqrt{A^2+B^2-\sqrt2\,AB}$

Compare with the options:

• A: $\sqrt{A^2+B^2+\sqrt2 AB}$
• B: $\sqrt{A^2+B^2}$
• C: $\sqrt{A^2+B^2-\sqrt2 AB}$
• D: $\sqrt{A^2+B^2+2AB}$

Hence, the correct option is C.