The points `A(1,1,2), B (3,4,2) and C(5,6,4)` . The exterior angle of the triangle at the vertex B is

Options

A. `cos^(-1) [-5//sqrt((39))]`
B. `cos^(-1) [5//sqrt((39))]`
C. `cos^(-1)(5//9)`
D. none of these

Answer

To find the exterior angle of triangle ABC at vertex B, we will follow these steps:

### Step 1: Find the vectors BA and BC
The coordinates of the points are:
- A(1, 1, 2)
- B(3, 4, 2)
- C(5, 6, 4)

**Vector BA** is calculated as:
\[ \text{BA} = A - B = (1 - 3, 1 - 4, 2 - 2) = (-2, -3, 0) \]

**Vector BC** is calculated as:
\[ \text{BC} = C - B = (5 - 3, 6 - 4, 4 - 2) = (2, 2, 2) \]

### Step 2: Calculate the dot product of vectors BA and BC
The dot product \( \text{BA} \cdot \text{BC} \) is calculated as follows:
\[ \text{BA} \cdot \text{BC} = (-2)(2) + (-3)(2) + (0)(2) = -4 - 6 + 0 = -10 \]

### Step 3: Calculate the magnitudes of vectors BA and BC
The magnitude of vector BA is:
\[ |\text{BA}| = \sqrt{(-2)^2 + (-3)^2 + 0^2} = \sqrt{4 + 9 + 0} = \sqrt{13} \]

The magnitude of vector BC is:
\[ |\text{BC}| = \sqrt{(2)^2 + (2)^2 + (2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

### Step 4: Calculate the cosine of the angle θ between vectors BA and BC
Using the formula for the cosine of the angle between two vectors:
\[ \cos \theta = \frac{\text{BA} \cdot \text{BC}}{|\text{BA}| \cdot |\text{BC}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{-10}{\sqrt{13} \cdot 2\sqrt{3}} = \frac{-10}{2\sqrt{39}} = \frac{-5}{\sqrt{39}} \]

### Step 5: Find the angle θ
To find θ, we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{-5}{\sqrt{39}}\right) \]

### Step 6: Calculate the exterior angle at vertex B
The exterior angle at vertex B is given by:
\[ \text{Exterior Angle} = \pi - \theta \]
Substituting the value of θ:
\[ \text{Exterior Angle} = \pi - \cos^{-1}\left(\frac{-5}{\sqrt{39}}\right) \]

### Final Answer
Thus, the exterior angle at vertex B is:
\[ \text{Exterior Angle} = \cos^{-1}\left(\frac{5}{\sqrt{39}}\right) \]