Let a and b be the vectors along the diagonals of a parallelogram having area 2 √ 2. Let the angle between a and b be acute, | a |=1, and | a · b |=|…

Let $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ be the vectors along the diagonals of a parallelogram having area $\displaystyle 2 \sqrt{2}$. Let the angle between $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ be acute, $\displaystyle |\vec{a}|=1$, and $\displaystyle |\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$. If $\displaystyle \vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$, then an angle between $\displaystyle \vec{b}$ and $\displaystyle \vec{c}$ is