The vertices of a triangle are A(1, 3, 2), B(2, -1, 3) and C(-1, 4, 2). Find the angle A.

Options

A. `cos^(-1)(-sqrt(2/5))`
B. `cos^(-1)(-1/sqrt(5))`
C. `sin^(-1)(-sqrt(2/5))`
D. None of these

Answer

To find the angle A in triangle ABC with vertices A(1, 3, 2), B(2, -1, 3), and C(-1, 4, 2), we will use the cosine rule. The steps are as follows:

### Step 1: Find the vectors AB and AC
The vectors AB and AC can be calculated as follows:

- **Vector AB**:
\[ \text{AB} = B - A = (2 - 1, -1 - 3, 3 - 2) = (1, -4, 1) \]

- **Vector AC**:
\[ \text{AC} = C - A = (-1 - 1, 4 - 3, 2 - 2) = (-2, 1, 0) \]

### Step 2: Calculate the dot product of AB and AC
The dot product of vectors AB and AC is given by:
\[ \text{AB} \cdot \text{AC} = (1)(-2) + (-4)(1) + (1)(0) = -2 - 4 + 0 = -6 \]

### Step 3: Calculate the magnitudes of vectors AB and AC
The magnitudes of vectors AB and AC are calculated as follows:

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

- **Magnitude of AC**:
\[ |\text{AC}| = \sqrt{(-2)^2 + 1^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5} \]

### Step 4: Use the cosine formula to find cos(A)
Using the formula for the cosine of angle A:
\[ \cos A = \frac{\text{AB} \cdot \text{AC}}{|\text{AB}| \cdot |\text{AC}|} \]
Substituting the values we found:
\[ \cos A = \frac{-6}{(3\sqrt{2})(\sqrt{5})} = \frac{-6}{3\sqrt{10}} = \frac{-2}{\sqrt{10}} \]

### Step 5: Find angle A
To find angle A, we take the inverse cosine:
\[ A = \cos^{-1}\left(\frac{-2}{\sqrt{10}}\right) \]

### Final Answer
Thus, the angle A is:
\[ A = \cos^{-1}\left(\frac{-2}{\sqrt{10}}\right) \]