Two adjacent sides of a triangle are represented by the vectors `vec(a)=3hat(i)+4hat(j) and vec(b)=-5hat(i)+7hat(j)`. The area of the triangle is

Options

A. 41 sq units
B. 37 sq units
C. `(41)/(2)` sq units
D. none of these

Explanation

To find the area of the triangle formed by the vectors \(\vec{a}\) and \(\vec{b}\), we can use the formula:

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

### Step 1: Identify the vectors
We are given:
\[ \vec{a} = 3\hat{i} + 4\hat{j} \]
\[ \vec{b} = -5\hat{i} + 7\hat{j} \]

### Step 2: Calculate the cross product \(\vec{a} \times \vec{b}\)
For two vectors in the plane, the cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ -5 & 7 & 0 \end{vmatrix} \]

### Step 3: Compute the determinant
Calculating the determinant, we have:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 4 & 0 \\ 7 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 0 \\ -5 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 4 \\ -5 & 7 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 4 & 0 \\ 7 & 0 \end{vmatrix} = 4 \cdot 0 - 7 \cdot 0 = 0\)

2. \(\begin{vmatrix} 3 & 0 \\ -5 & 0 \end{vmatrix} = 3 \cdot 0 - (-5) \cdot 0 = 0\)

3. \(\begin{vmatrix} 3 & 4 \\ -5 & 7 \end{vmatrix} = (3 \cdot 7) - (4 \cdot -5) = 21 + 20 = 41\)

So, we have:

\[ \vec{a} \times \vec{b} = 0\hat{i} - 0\hat{j} + 41\hat{k} = 41\hat{k} \]

### Step 4: Find the magnitude of the cross product
The magnitude of the cross product is:

\[ \|\vec{a} \times \vec{b}\| = |41| = 41 \]

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

\[ \text{Area} = \frac{1}{2} \|\vec{a} \times \vec{b}\| = \frac{1}{2} \times 41 = 20.5 \]

### Final Answer
The area of the triangle is \(20.5\) square units.

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