Two adjacent sides of a parallelogram `A B C D` are given by ` vec A B=2 hat i+10 hat j+11 hat ka n d vec A D=- hat i+2 hat j+2 hat kdot` The side `A D` is rotated by an acute angle `alpha` in the plane of the parallelogram so that `A D` becomes `A D^(prime)dot` If `A D '` makes a right angle with the side `A B ,` then the cosine of the angel `alpha` is given by a. `8/9` b. `(sqrt(17))/9` c. `1/9` d. `(4sqrt(5))/9`

Options

A. `8/9`
B. `sqrt17/9`
C. `1/9`
D. `(4sqrt5)/9`

Explanation

To solve the problem step by step, we will follow the mathematical approach outlined in the video transcript.

### Step 1: Identify the vectors
We are given two vectors:
- \( \vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k} \)
- \( \vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k} \)

### Step 2: Calculate the dot product of the vectors
The dot product \( \vec{AB} \cdot \vec{AD} \) is calculated as follows:
\[ \vec{AB} \cdot \vec{AD} = (2)(-1) + (10)(2) + (11)(2) \]
Calculating each term:
\[ = -2 + 20 + 22 = 40 \]

### Step 3: Calculate the magnitudes of the vectors
Now we find the magnitudes of \( \vec{AB} \) and \( \vec{AD} \).

For \( \vec{AB} \):
\[ |\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15 \]

For \( \vec{AD} \):
\[ |\vec{AD}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 4: Calculate the cosine of the angle between the vectors
Using the dot product and magnitudes, we can find the cosine of the angle \( \theta \) between \( \vec{AB} \) and \( \vec{AD} \):
\[ \cos(\theta) = \frac{\vec{AB} \cdot \vec{AD}}{|\vec{AB}| |\vec{AD}|} = \frac{40}{15 \cdot 3} = \frac{40}{45} = \frac{8}{9} \]

### Step 5: Find the sine of the angle
To find \( \sin(\theta) \), we use the identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Calculating \( \sin^2(\theta) \):
\[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left(\frac{8}{9}\right)^2 = 1 - \frac{64}{81} = \frac{81 - 64}{81} = \frac{17}{81} \]
Thus,
\[ \sin(\theta) = \sqrt{\frac{17}{81}} = \frac{\sqrt{17}}{9} \]

### Step 6: Relate \( \alpha \) and \( \theta \)
Since \( \alpha + \theta = 90^\circ \), we have:
\[ \cos(\alpha) = \sin(\theta) \]
Thus,
\[ \cos(\alpha) = \frac{\sqrt{17}}{9} \]

### Conclusion
The cosine of the angle \( \alpha \) is given by:
\[ \cos(\alpha) = \frac{\sqrt{17}}{9} \]
Therefore, the correct answer is option **b**: \( \frac{\sqrt{17}}{9} \).

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