A force of `(hat(i) + 2hat(j) + 3hat(k))N` is acting on a body having position vector `(3 hat(i) + hat(j) + 2hat(k))` in the same frame of reference
Considering that the two vectors mentioned in the paragraph are the two sides of a triangle, then the magnitude of area of triangle is

Answer

To solve the problem, we need to find the area of a triangle formed by two vectors: the force vector \( \mathbf{A} = \hat{i} + 2\hat{j} + 3\hat{k} \) and the position vector \( \mathbf{B} = 3\hat{i} + \hat{j} + 2\hat{k} \). The area of the triangle can be calculated using the formula:

\[ \text{Area} = \frac{1}{2} |\mathbf{A} \times \mathbf{B}| \]

### Step 1: Define the vectors
Let:
- \( \mathbf{A} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- \( \mathbf{B} = 3\hat{i} + \hat{j} + 2\hat{k} \)

### Step 2: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)
To find the cross product, we can use the determinant method:

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{vmatrix} \]

### Step 3: Expand the determinant
Calculating the determinant, we have:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

2. \( \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = (1)(2) - (3)(3) = 2 - 9 = -7 \)

3. \( \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = (1)(1) - (2)(3) = 1 - 6 = -5 \)

Putting it all together, we get:

\[ \mathbf{A} \times \mathbf{B} = \hat{i}(1) - \hat{j}(-7) + \hat{k}(-5) = \hat{i} + 7\hat{j} - 5\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now, we calculate the magnitude of \( \mathbf{A} \times \mathbf{B} \):

\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(1)^2 + (7)^2 + (-5)^2} = \sqrt{1 + 49 + 25} = \sqrt{75} \]

### Step 5: Calculate the area of the triangle
Finally, we can find the area of the triangle:

\[ \text{Area} = \frac{1}{2} |\mathbf{A} \times \mathbf{B}| = \frac{1}{2} \sqrt{75} = \frac{\sqrt{75}}{2} \]

### Final Answer
The area of the triangle is:

\[ \text{Area} = \frac{\sqrt{75}}{2} \text{ square units} \]

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