Cosine of Rotation Angle of Parallelogram Vector

To find the cosine of the angle of rotation $\alpha$, we first determine the angle between the given adjacent sides $\vec{AB}$ and $\vec{AD}$.

Step 1: Calculate the magnitudes of $\vec{AB}$ and $\vec{AD}$

Given:
$\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$
$\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$

The magnitude of $\vec{AB}$ is:
$|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$

The magnitude of $\vec{AD}$ is:
$|\vec{AD}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Step 2: Find the dot product of $\vec{AB}$ and $\vec{AD}$

$\vec{AB} \cdot \vec{AD} = (2)(-1) + (10)(2) + (11)(2) = -2 + 20 + 22 = 40$

Step 3: Determine the angle $\theta$ between $\vec{AB}$ and $\vec{AD}$

Let $\theta$ be the angle between $\vec{AB}$ and $\vec{AD}$. Then:
$\cos\theta = \frac{\vec{AB} \cdot \vec{AD}}{|\vec{AB}| |\vec{AD}|} = \frac{40}{15 \times 3} = \frac{40}{45} = \frac{8}{9}$

Since $\cos\theta > 0$, $\theta$ is an acute angle.

Step 4: Relate the rotation angle $\alpha$ to $\theta$

The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram to become $AD'$, such that $AD'$ is perpendicular to $AB$.

Since $\vec{AB}$, $\vec{AD}$, and $\vec{AD'}$ all lie in the same plane and $\vec{AD'} \perp \vec{AB}$, the angle between $\vec{AB}$ and $\vec{AD'}$ is $90^\circ$.

Since $\alpha$ is the acute angle between $\vec{AD}$ and $\vec{AD'}$, we have:
$\alpha = 90^\circ - \theta$

Step 5: Calculate $\cos\alpha$

$\cos\alpha = \cos(90^\circ - \theta) = \sin\theta$

Using the trigonometric identity $\sin\theta = \sqrt{1 - \cos^2\theta}$ for acute angle $\theta$:
$\sin\theta = \sqrt{1 - \left(\frac{8}{9}\right)^2} = \sqrt{1 - \frac{64}{81}} = \sqrt{\frac{17}{81}} = \frac{\sqrt{17}}{9}$

Thus,
$\cos\alpha = \frac{\sqrt{17}}{9}$

Correct Answer:
B