Two adjacent sides of a parallelogram ABCD are given by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side AD is rotated by an acute angle $\alpha$ in the plane of parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then $\cos \alpha =$
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Angle between two vectors $\vec{u}, \vec{v}$ is given by $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$.
Step 1: Understanding the Question:
We are given the vectors for two sides of a parallelogram. One side is rotated in the plane to become perpendicular to the other. We need to find the cosine of that rotation angle. Step 3: Detailed Explanation:
1. Let $\theta$ be the original angle between $\vec{AB}$ and $\vec{AD}$.
$\vec{AB} \cdot \vec{AD} = (2)(-1) + (10)(2) + (11)(2) = -2 + 20 + 22 = 40$.
$|\vec{AB}| = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.
$|\vec{AD}| = \sqrt{1 + 4 + 4} = 3$.
$\cos \theta = \frac{40}{15 \cdot 3} = \frac{40}{45} = \frac{8}{9}$.
2. The side AD is rotated in the plane by $\alpha$ to become AD' such that AD' $\perp$ AB.
This means the new angle between the sides is $90^\circ$.
The rotation angle $\alpha$ is the difference between $90^\circ$ and the original angle $\theta$.
So, $\alpha = 90^\circ - \theta$.
3. Calculate $\cos \alpha$:
$\cos \alpha = \cos(90^\circ - \theta) = \sin \theta$.
Using $\sin \theta = \sqrt{1 - \cos^2 \theta}$:
$\sin \theta = \sqrt{1 - (\frac{8}{9})^2} = \sqrt{1 - \frac{64}{81}} = \sqrt{\frac{17}{81}} = \frac{\sqrt{17}}{9}$. Step 4: Final Answer:
The value of $\cos \alpha$ is $\frac{\sqrt{17}}{9}$.