The vertices of a `Delta 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)/(sqrt3)`
B. `(1)/(sqrt2)`
C. `(2)/(sqrt6)`
D. None of these

Answer

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
We need to define the vectors BA and BC:
- Vector BA = A - B
- Vector BC = C - B

### Step 2: Calculate vector BA
Using the coordinates of points A and B:
- A = (2, 3, 1)
- B = (-2, 2, 0)

The vector BA can be calculated as:
\[ BA = A - B = (2 - (-2), 3 - 2, 1 - 0) = (2 + 2, 3 - 2, 1 - 0) = (4, 1, 1) \]

### Step 3: Calculate vector BC
Using the coordinates of points B and C:
- B = (-2, 2, 0)
- C = (0, 1, -1)

The vector BC can be calculated as:
\[ BC = C - B = (0 - (-2), 1 - 2, -1 - 0) = (0 + 2, 1 - 2, -1 - 0) = (2, -1, -1) \]

### Step 4: Calculate the magnitudes of vectors BA and BC
The magnitude of vector BA is:
\[ |BA| = \sqrt{(4^2 + 1^2 + 1^2)} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2} \]

The magnitude of vector BC is:
\[ |BC| = \sqrt{(2^2 + (-1)^2 + (-1)^2)} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 5: Calculate the dot product of vectors BA and BC
The dot product \( BA \cdot BC \) is calculated as follows:
\[ BA \cdot BC = (4)(2) + (1)(-1) + (1)(-1) = 8 - 1 - 1 = 6 \]

### Step 6: Use the cosine formula
The cosine of angle ABC can be calculated using the formula:
\[ \cos(\angle ABC) = \frac{BA \cdot BC}{|BA| \cdot |BC|} \]
Substituting the values we have:
\[ \cos(\angle ABC) = \frac{6}{(3\sqrt{2})(\sqrt{6})} \]

### Step 7: Simplify the expression
Calculating the denominator:
\[ |BA| \cdot |BC| = 3\sqrt{2} \cdot \sqrt{6} = 3\sqrt{12} = 3 \cdot 2\sqrt{3} = 6\sqrt{3} \]
Thus:
\[ \cos(\angle ABC) = \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer
The cosine of angle ABC is:
\[ \cos(\angle ABC) = \frac{1}{\sqrt{3}} \]

---