If the vectors `(hati+ hatj + hatk)` and `3 hati` from two sides of a triangle, then area of triangle is :

Options

A. `sqrt3` unit
B. `2sqrt3` unit
C. `(3)/(sqrt2)` unit
D. `3sqrt2` unit

Explanation

To find the area of the triangle formed by the vectors \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \) and \( \mathbf{B} = 3\hat{i} \), we will follow these steps:

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

### Step 2: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)
To find the area of the triangle, we first need to compute the cross product \( \mathbf{A} \times \mathbf{B} \).

Using the determinant form:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 3 & 0 & 0 \end{vmatrix} \]

### Step 3: Compute the determinant
Calculating the determinant, we have:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} = (1)(0) - (1)(0) = 0 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = (1)(0) - (1)(3) = -3 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = (1)(0) - (1)(3) = -3 \]

Thus, we have:
\[ \mathbf{A} \times \mathbf{B} = 0\hat{i} + 3\hat{j} - 3\hat{k} = 3\hat{j} - 3\hat{k} \]

### Step 4: Calculate the magnitude of the cross product
The magnitude of \( \mathbf{A} \times \mathbf{B} \) is given by:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(0)^2 + (3)^2 + (-3)^2} = \sqrt{0 + 9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Step 5: Calculate the area of the triangle
The area \( A \) of the triangle formed by the two vectors is given by:
\[ A = \frac{1}{2} |\mathbf{A} \times \mathbf{B}| = \frac{1}{2} (3\sqrt{2}) = \frac{3\sqrt{2}}{2} \]

### Final Answer
Thus, the area of the triangle is \( \frac{3\sqrt{2}}{2} \).

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