Vertices of a triangle are given by `hat(i)+3hat(j)+2hat(k), 2hat(i)-hat(j)+hat(k)` and `-hat(i)+2hat(j)+3hat(k)`, then area of triangle is (in units)

Options

A. `(sqrt(107))/(2)`
B. `sqrt((107)/(6))`
C. `sqrt((107)/(2))`
D. `(sqrt(207))/(3)`

Answer

To find the area of the triangle formed by the vertices given in vector form, we will follow these steps:

### Step 1: Identify the vertices
The vertices of the triangle are given as:
- Vertex A: \( \mathbf{A} = \hat{i} + 3\hat{j} + 2\hat{k} \)
- Vertex B: \( \mathbf{B} = 2\hat{i} - \hat{j} + \hat{k} \)
- Vertex C: \( \mathbf{C} = -\hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 2: Find the vectors representing two sides of the triangle
We can find two vectors that represent the sides of the triangle using the vertices:
- Vector \( \mathbf{AB} = \mathbf{B} - \mathbf{A} \)
- Vector \( \mathbf{AC} = \mathbf{C} - \mathbf{A} \)

Calculating these:
1. \( \mathbf{AB} = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + 3\hat{j} + 2\hat{k}) \)
\[ = (2 - 1)\hat{i} + (-1 - 3)\hat{j} + (1 - 2)\hat{k} = \hat{i} - 4\hat{j} - \hat{k} \]

2. \( \mathbf{AC} = (-\hat{i} + 2\hat{j} + 3\hat{k}) - (\hat{i} + 3\hat{j} + 2\hat{k}) \)
\[ = (-1 - 1)\hat{i} + (2 - 3)\hat{j} + (3 - 2)\hat{k} = -2\hat{i} - \hat{j} + \hat{k} \]

### Step 3: Calculate the cross product of the two vectors
The area of the triangle can be found using the formula:
\[ \text{Area} = \frac{1}{2} |\mathbf{AB} \times \mathbf{AC}| \]
To find \( \mathbf{AB} \times \mathbf{AC} \), we set up the determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & -1 \\ -2 & -1 & 1 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} -4 & -1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ -2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -4 \\ -2 & -1 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} -4 & -1 \\ -1 & 1 \end{vmatrix} = (-4)(1) - (-1)(-1) = -4 - 1 = -5 \)

2. \( \begin{vmatrix} 1 & -1 \\ -2 & 1 \end{vmatrix} = (1)(1) - (-1)(-2) = 1 - 2 = -1 \)

3. \( \begin{vmatrix} 1 & -4 \\ -2 & -1 \end{vmatrix} = (1)(-1) - (-4)(-2) = -1 - 8 = -9 \)

Now substituting these back:
\[ \mathbf{AB} \times \mathbf{AC} = -5\hat{i} + 1\hat{j} - 9\hat{k} \]

### Step 4: Find the magnitude of the cross product
The magnitude of the cross product is:
\[ |\mathbf{AB} \times \mathbf{AC}| = \sqrt{(-5)^2 + 1^2 + (-9)^2} = \sqrt{25 + 1 + 81} = \sqrt{107} \]

### Step 5: Calculate the area of the triangle
Finally, the area of the triangle is:
\[ \text{Area} = \frac{1}{2} |\mathbf{AB} \times \mathbf{AC}| = \frac{1}{2} \sqrt{107} \]

Thus, the area of the triangle is \( \frac{\sqrt{107}}{2} \) square units.

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