The area of the parallelogram represented by the vectors `vecA=2hati+3hatj` and `vecB=hati+4hatj` is

Options

A. 14 unit
B. 7.5 units
C. 10 unit
D. 5 units

Explanation

To find the area of the parallelogram represented by the vectors \(\vec{A} = 2\hat{i} + 3\hat{j}\) and \(\vec{B} = \hat{i} + 4\hat{j}\), we will use the formula for the area of a parallelogram formed by two vectors, which is given by the magnitude of their cross product:

\[ \text{Area} = |\vec{A} \times \vec{B}| \]

### Step 1: Write down the vectors
We have:
\[ \vec{A} = 2\hat{i} + 3\hat{j} \]
\[ \vec{B} = \hat{i} + 4\hat{j} \]

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

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 1 & 4 & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we can expand it as follows:

\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & 0 \\ 4 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 0 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} \]

Calculating each of these determinants:
1. \(\begin{vmatrix} 3 & 0 \\ 4 & 0 \end{vmatrix} = 3 \cdot 0 - 0 \cdot 4 = 0\)
2. \(\begin{vmatrix} 2 & 0 \\ 1 & 0 \end{vmatrix} = 2 \cdot 0 - 0 \cdot 1 = 0\)
3. \(\begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 2 \cdot 4 - 3 \cdot 1 = 8 - 3 = 5\)

Putting it all together, we have:
\[ \vec{A} \times \vec{B} = 0\hat{i} - 0\hat{j} + 5\hat{k} = 5\hat{k} \]

### Step 4: Find the magnitude
The magnitude of the vector \(5\hat{k}\) is simply:
\[ |\vec{A} \times \vec{B}| = 5 \]

### Conclusion
Thus, the area of the parallelogram represented by the vectors \(\vec{A}\) and \(\vec{B}\) is:
\[ \text{Area} = 5 \text{ square units} \]