Two sides of a triangle are given by `hati+hatj+hatk and -hati+2hatj+3hatk` then area of triangle is

Options

A. `sqrt(26)`
B. `sqrt(26)//2`
C. `sqrt(46)`
D. `26`

Answer

To find the area of the triangle given two sides represented as vectors, we can follow these steps:

### Step 1: Define the Vectors
Let the two sides of the triangle be represented as vectors:
- Vector **A** = \( \hat{i} + \hat{j} + \hat{k} \)
- Vector **B** = \( -\hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 2: Calculate the Cross Product
The area of the triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} |\mathbf{A} \times \mathbf{B}| \]
To find the cross product \( \mathbf{A} \times \mathbf{B} \), we can use the determinant method:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 2 & 3 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Now, we will calculate the determinant:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} = (1)(3) - (1)(-1) = 3 + 1 = 4 \)
3. \( \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} = (1)(2) - (1)(-1) = 2 + 1 = 3 \)

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = \hat{i}(1) - \hat{j}(4) + \hat{k}(3) = \hat{i} - 4\hat{j} + 3\hat{k} \]

### Step 4: Find the Magnitude of the Cross Product
Now, we need to find the magnitude of the cross product:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(1)^2 + (-4)^2 + (3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \]

### Step 5: Calculate the Area of the Triangle
Finally, we can calculate the area of the triangle:
\[ \text{Area} = \frac{1}{2} |\mathbf{A} \times \mathbf{B}| = \frac{1}{2} \sqrt{26} \]

### Final Answer
Thus, the area of the triangle is \( \frac{1}{2} \sqrt{26} \).

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