Two adjacent sides of a parallelogram ABCD are given by $\overline{\mathrm{AB}}=2 \hat{i}+10 \hat{j}+11…

Given vectors for adjacent sides of parallelogram ABCD:

$\vec{a} = \overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$

$\vec{b} = \overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$

Magnitudes are $|\vec{a}| = \sqrt{2^2 + 10^2 + 11^2} = 15$ and $|\vec{b}| = \sqrt{(-1)^2 + 2^2 + 2^2} = 3$.

After rotation, $\vec{b'}$ remains in the parallelogram's plane and is orthogonal to $\vec{a}$, so we express it as $\vec{b'} = x\vec{a} + y\vec{b}$ with the constraints:
$729\vec{a} \cdot \vec{b'} = 0$

The dot product $\vec{a} \cdot \vec{b} = 40$ yields $225x + 40y = 0 \Rightarrow y = -\frac{45}{8}x$.

Using $|\vec{b'}|^2 = |\vec{b}|^2 = 9$:
$225x^2 + 9y^2 + 80xy = 9$
Substituting $y$ and simplifying gives $x^2 = \frac{64}{425} \Rightarrow x = \pm \frac{8}{5\sqrt{17}}$.

The cosine of the acute angle $\alpha$ between $\vec{b}$ and $\vec{b'}$ is:
$\cos \alpha = \frac{\vec{b} \cdot \vec{b'}}{9}$
$\vec{b} \cdot \vec{b'} = 40x + 9y = -\frac{85}{8}x$
Thus $\cos \alpha = -\frac{85}{72}x$.

Since $\alpha$ is acute, $\cos \alpha > 0$ implies $x < 0$, so $x = -\frac{8}{5\sqrt{17}}$.
Substituting:
$\cos \alpha = -\frac{85}{72} \left(-\frac{8}{5\sqrt{17}}\right) = \frac{\sqrt{17}}{9}$.

$\boxed{\frac{\sqrt{17}}{9}}$