Two sides of a triangle is represented by `vec(a) = 3hat(j)` and `vec(b) = 2hat(i) - hat(k)`. The area of triangle is :

Options

A. 5
B. `3sqrt(5)`
C. `(3)/(2)sqrt(5)`
D. `sqrt(5)`

Answer

To find the area of the triangle formed by the vectors \(\vec{a} = 3\hat{j}\) and \(\vec{b} = 2\hat{i} - \hat{k}\), we can follow these steps:

### Step 1: Understand the formula for the area of a triangle
The area \(A\) of a triangle formed by two vectors \(\vec{a}\) and \(\vec{b}\) can be calculated using the formula:
\[ A = \frac{1}{2} |\vec{a} \times \vec{b}| \]
where \(\vec{a} \times \vec{b}\) is the cross product of the vectors.

### Step 2: Set up the vectors
Given:
\[ \vec{a} = 3\hat{j} = 0\hat{i} + 3\hat{j} + 0\hat{k} \]
\[ \vec{b} = 2\hat{i} - \hat{k} = 2\hat{i} + 0\hat{j} - 1\hat{k} \]

### Step 3: Write the cross product in determinant form
The cross product \(\vec{a} \times \vec{b}\) can be calculated using the determinant of a matrix:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 0 \\ 2 & 0 & -1 \end{vmatrix} \]

### Step 4: Calculate the determinant
Expanding the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 3 \\ 2 & 0 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 3 & 0 \\ 0 & -1 \end{vmatrix} = 3 \cdot (-1) - 0 \cdot 0 = -3\)
2. \(\begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = 0 \cdot (-1) - 0 \cdot 2 = 0\)
3. \(\begin{vmatrix} 0 & 3 \\ 2 & 0 \end{vmatrix} = 0 \cdot 0 - 3 \cdot 2 = -6\)

Putting it all together:
\[ \vec{a} \times \vec{b} = -3\hat{i} - 0\hat{j} - 6\hat{k} = -3\hat{i} - 6\hat{k} \]

### Step 5: Find the magnitude of the cross product
The magnitude of \(\vec{a} \times \vec{b}\) is:
\[ |\vec{a} \times \vec{b}| = \sqrt{(-3)^2 + 0^2 + (-6)^2} = \sqrt{9 + 0 + 36} = \sqrt{45} \]

### Step 6: Calculate the area of the triangle
Now, substituting back into the area formula:
\[ A = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{45} = \frac{\sqrt{9 \cdot 5}}{2} = \frac{3\sqrt{5}}{2} \]

### Final Answer
The area of the triangle is:
\[ \boxed{\frac{3\sqrt{5}}{2}} \]