Problem 107

Analysis

This question examines the square of a vector being equal to the square of its magnitude and involves the operation of the dot product of vectors, making it a medium-level question.

From $|\vec{AO} + \vec{BO}| = |\vec{AO} - \vec{BO}|$, we can deduce that $\vec{AO} \cdot \vec{BO} = 0$, which means $AO \perpendicular BO$. Therefore, $\triangle AOB$ is an isosceles right triangle with side length $2$, from which we can derive the result.

Solution

Given $|\vec{AO} + \vec{BO}| = |\vec{AO} - \vec{BO}| \Rightarrow (\vec{AO} + \vec{BO})^2 = (\vec{AO} - \vec{BO})^2 \Rightarrow \vec{AO} \cdot \vec{BO} = 0$,

$\therefore AO \perpendicular BO$,
$\therefore \triangle AOB$ is an isosceles right triangle with side length $2$,

Thus, $\vec{AO} \cdot \vec{AB} = |\vec{AO}||\vec{AB}|\cos 45^{\circ} = 2 \times 2\sqrt{2} \times \frac{\sqrt{2}}{2} = 4$.
Therefore, the answer is $\boxed{4}$.