If the vertices A,B,C of a triangle ABC are `(1,2,3), (-1,0,0) , (0,1,2)` , respectively, then find `angleABC`.

Options

A. `cos^(-1)(10/sqrt52)`
B. `cos^(-1)(10/sqrt102)`
C. `cos^(-1)(10/sqrt103)`
D. `cos^(-1)(5/sqrt102)`

Explanation

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

### Step 1: Find the vectors \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \)

1. **Calculate the vector \( \overrightarrow{BA} \)**:
\[ \overrightarrow{BA} = A - B = (1, 2, 3) - (-1, 0, 0) = (1 + 1, 2 - 0, 3 - 0) = (2, 2, 3) \]

2. **Calculate the vector \( \overrightarrow{BC} \)**:
\[ \overrightarrow{BC} = C - B = (0, 1, 2) - (-1, 0, 0) = (0 + 1, 1 - 0, 2 - 0) = (1, 1, 2) \]

### Step 2: Compute the dot product \( \overrightarrow{BA} \cdot \overrightarrow{BC} \)

\[ \overrightarrow{BA} \cdot \overrightarrow{BC} = (2, 2, 3) \cdot (1, 1, 2) = 2 \cdot 1 + 2 \cdot 1 + 3 \cdot 2 = 2 + 2 + 6 = 10 \]

### Step 3: Find the magnitudes of \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \)

1. **Magnitude of \( \overrightarrow{BA} \)**:
\[ |\overrightarrow{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \]

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

### Step 4: Use the cosine formula to find \( \angle ABC \)

The cosine of the angle \( \theta \) between the two vectors is given by:
\[ \cos \theta = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| \cdot |\overrightarrow{BC}|} \]

Substituting the values we found:
\[ \cos \theta = \frac{10}{\sqrt{17} \cdot \sqrt{6}} = \frac{10}{\sqrt{102}} \]

### Step 5: Calculate \( \theta \)

Now, we can find the angle \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right) \]

### Final Answer

Thus, the angle \( \angle ABC \) is:
\[ \angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right) \]