If the vectors `(hat(i) + hat(j) + hat(k))` and `3hat(i)` form two sides of a triangle, then area of the triangle is :

Options

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

Answer

To find the area of the triangle formed by the vectors \(\hat{i} + \hat{j} + \hat{k}\) and \(3\hat{i}\), we will use the formula for the area of a triangle given two vectors \(\mathbf{a}\) and \(\mathbf{b}\):

\[ \text{Area} = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| \]

### Step 1: Identify the vectors
Let:
\[ \mathbf{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{b} = 3\hat{i} \]

### Step 2: Set up the cross product
The cross product \(\mathbf{a} \times \mathbf{b}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\mathbf{a}\) and \(\mathbf{b}\):

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 3 & 0 & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
We can calculate the determinant as follows:

\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} = 0\)

2. \(\begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = (1)(0) - (1)(3) = -3\)

3. \(\begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = (1)(0) - (1)(3) = -3\)

Putting it all together:

\[ \mathbf{a} \times \mathbf{b} = 0\hat{i} - (-3)\hat{j} + (-3)\hat{k} = 3\hat{j} - 3\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now we find the magnitude of the vector \(3\hat{j} - 3\hat{k}\):

\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(0)^2 + (3)^2 + (-3)^2} = \sqrt{0 + 9 + 9} = \sqrt{18} = 3\sqrt{2} \]

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

\[ \text{Area} = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| = \frac{1}{2} (3\sqrt{2}) = \frac{3\sqrt{2}}{2} \]

### Final Answer
Thus, the area of the triangle is:

\[ \frac{3\sqrt{2}}{2} \text{ unit}^2 \]