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 two sides represented by the vectors \(\vec{a} = 3\hat{j}\) and \(\vec{b} = 2\hat{i} - \hat{k}\), we can use the formula for the area of a triangle given by two vectors:

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

### Step 1: Calculate the Cross Product \(\vec{a} \times \vec{b}\)

We will use the determinant method to find the cross product. The vectors can be arranged in a matrix as follows:

\[ \vec{a} = 0\hat{i} + 3\hat{j} + 0\hat{k} \quad \text{(which is } 0\hat{i} + 3\hat{j} + 0\hat{k}\text{)} \]
\[ \vec{b} = 2\hat{i} + 0\hat{j} - 1\hat{k} \quad \text{(which is } 2\hat{i} + 0\hat{j} - 1\hat{k}\text{)} \]

The determinant for the cross product is:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 0 \\ 2 & 0 & -1 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Calculating the determinant, we have:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 3 \\ 2 & 0 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 3 & 0 \\ 0 & -1 \end{vmatrix} = (3)(-1) - (0)(0) = -3\)
2. \(\begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = (0)(-1) - (0)(2) = 0\)
3. \(\begin{vmatrix} 0 & 3 \\ 2 & 0 \end{vmatrix} = (0)(0) - (3)(2) = -6\)

Putting these values back into the equation:

\[ \vec{a} \times \vec{b} = -3\hat{i} - 0\hat{j} - 6\hat{k} = -3\hat{i} - 6\hat{k} \]

### Step 3: Calculate the Magnitude of the Cross Product

Now, we find the magnitude of \(\vec{a} \times \vec{b}\):

\[ |\vec{a} \times \vec{b}| = \sqrt{(-3)^2 + 0^2 + (-6)^2} = \sqrt{9 + 0 + 36} = \sqrt{45} \]

### Step 4: Calculate the Area of the Triangle

Now, we can substitute this magnitude back into the area formula:

\[ \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{45} = \frac{\sqrt{45}}{2} \]

Breaking down \(\sqrt{45}\):

\[ \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \]

Thus, the area becomes:

\[ \text{Area} = \frac{3\sqrt{5}}{2} \]

### Final Answer

The area of the triangle is:

\[ \text{Area} = \frac{3\sqrt{5}}{2} \]