If `A = (1,2, 5), B = (5, 7, 9) and `C = (3, 2, -1)` then a unit vector normal to the plane of triangle ABC is .........

Answer

To find a unit vector normal to the plane of triangle ABC with vertices A = (1, 2, 5), B = (5, 7, 9), and C = (3, 2, -1), we will follow these steps:

### Step 1: Find the vectors AB and AC
The vectors AB and AC can be calculated using the coordinates of points A, B, and C.

- **Vector AB**:
\[ \text{AB} = B - A = (5 - 1, 7 - 2, 9 - 5) = (4, 5, 4) \]

- **Vector AC**:
\[ \text{AC} = C - A = (3 - 1, 2 - 2, -1 - 5) = (2, 0, -6) \]

### Step 2: Calculate the cross product of vectors AB and AC
The normal vector to the plane formed by points A, B, and C can be found by taking the cross product of vectors AB and AC.

Using the determinant method for the cross product:
\[ \text{AB} \times \text{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 4 \\ 2 & 0 & -6 \end{vmatrix} \]

Calculating the determinant:
- For \(\hat{i}\):
\[ \hat{i}(5 \cdot (-6) - 4 \cdot 0) = -30 \hat{i} \]

- For \(\hat{j}\):
\[ -\hat{j}(4 \cdot (-6) - 4 \cdot 2) = -(-24 - 8) = 32 \hat{j} \]

- For \(\hat{k}\):
\[ \hat{k}(4 \cdot 0 - 5 \cdot 2) = -10 \hat{k} \]

Thus, the cross product is:
\[ \text{AB} \times \text{AC} = (-30, 32, -10) \]

### Step 3: Calculate the magnitude of the normal vector
The magnitude of the vector \((-30, 32, -10)\) is calculated as follows:
\[ \text{Magnitude} = \sqrt{(-30)^2 + 32^2 + (-10)^2} = \sqrt{900 + 1024 + 100} = \sqrt{2024} \]

### Step 4: Find the unit vector
To find the unit vector, we divide the normal vector by its magnitude:
\[ \text{Unit Vector} = \frac{1}{\sqrt{2024}} \cdot (-30, 32, -10) = \left(-\frac{30}{\sqrt{2024}}, \frac{32}{\sqrt{2024}}, -\frac{10}{\sqrt{2024}}\right) \]

### Final Answer
The unit vector normal to the plane of triangle ABC is:
\[ \left(-\frac{30}{\sqrt{2024}}, \frac{32}{\sqrt{2024}}, -\frac{10}{\sqrt{2024}}\right) \]

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