The cosine of the angle of the triangle with vertices `A(1,-1,2),B(6,11,2)` and `C(1,2,6)` is

Options

A. `63/65`
B. `36/65`
C. `16/65`
D. `13/64`

Explanation

To find the cosine of the angle at vertex A of the triangle with vertices A(1, -1, 2), B(6, 11, 2), and C(1, 2, 6), we will follow these steps:

### Step 1: Determine the position vectors of points A, B, and C.
The position vectors are:
- \( \vec{A} = (1, -1, 2) \)
- \( \vec{B} = (6, 11, 2) \)
- \( \vec{C} = (1, 2, 6) \)

### Step 2: Calculate the vectors AB and AC.
The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} = (6 - 1, 11 - (-1), 2 - 2) = (5, 12, 0) \]

The vector \( \vec{AC} \) is given by:
\[ \vec{AC} = \vec{C} - \vec{A} = (1 - 1, 2 - (-1), 6 - 2) = (0, 3, 4) \]

### Step 3: Calculate the dot product \( \vec{AB} \cdot \vec{AC} \).
The dot product is calculated as follows:
\[ \vec{AB} \cdot \vec{AC} = (5, 12, 0) \cdot (0, 3, 4) = (5 \cdot 0) + (12 \cdot 3) + (0 \cdot 4) = 0 + 36 + 0 = 36 \]

### Step 4: Calculate the magnitudes of vectors AB and AC.
The magnitude of \( \vec{AB} \) is:
\[ |\vec{AB}| = \sqrt{5^2 + 12^2 + 0^2} = \sqrt{25 + 144 + 0} = \sqrt{169} = 13 \]

The magnitude of \( \vec{AC} \) is:
\[ |\vec{AC}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5 \]

### Step 5: Use the dot product and magnitudes to find \( \cos A \).
Using the formula for the cosine of the angle:
\[ \cos A = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| \cdot |\vec{AC}|} \]
Substituting the values we found:
\[ \cos A = \frac{36}{13 \cdot 5} = \frac{36}{65} \]

### Final Answer:
The cosine of the angle at vertex A is \( \frac{36}{65} \).

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