The vertices of a triangle ABC are A (2,3,1) , B(-2, 2,0), and C(0,1,-1).
What is the cosine of angle ABC ?

Options

A. `(1)/(sqrt(3))`
B. `(1)/(sqrt(2)) `
C. `(2)/(sqrt(6))`
D. None of these

Explanation

To find the cosine of angle ABC in triangle ABC with vertices A(2, 3, 1), B(-2, 2, 0), and C(0, 1, -1), we will follow these steps:

### Step 1: Define the vectors BA and BC
First, we need to find the vectors BA and BC.

- **Vector BA**: This is obtained by subtracting the coordinates of point A from point B.
\[ \text{BA} = A - B = (2 - (-2), 3 - 2, 1 - 0) = (4, 1, 1) \]

- **Vector BC**: This is obtained by subtracting the coordinates of point B from point C.
\[ \text{BC} = C - B = (0 - (-2), 1 - 2, -1 - 0) = (2, -1, -1) \]

### Step 2: Calculate the dot product of vectors BA and BC
Next, we calculate the dot product of vectors BA and BC.
\[ \text{BA} \cdot \text{BC} = (4)(2) + (1)(-1) + (1)(-1) = 8 - 1 - 1 = 6 \]

### Step 3: Calculate the magnitudes of vectors BA and BC
Now, we need to find the magnitudes of both vectors.

- **Magnitude of BA**:
\[ |\text{BA}| = \sqrt{4^2 + 1^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2} \]

- **Magnitude of BC**:
\[ |\text{BC}| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 4: Use the cosine formula
The cosine of the angle θ between two vectors can be found using the formula:
\[ \cos \theta = \frac{\text{BA} \cdot \text{BC}}{|\text{BA}| \cdot |\text{BC}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{6}{(3\sqrt{2})(\sqrt{6})} = \frac{6}{3\sqrt{12}} = \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the cosine of angle ABC is:
\[ \cos \angle ABC = \frac{1}{\sqrt{3}} \]