In sub-parts (i) and (ii) choose the correct option and in sub-parts (iii) to (v), answer the questions as instructed.
Find the area of triangle whose two adjacent sides are given by `veca=hati+4hatj-hatk` and `vecb=hati+hatj+2hatk`

Answer

To find the area of the triangle formed by two adjacent sides represented by the vectors \(\vec{a} = \hat{i} + 4\hat{j} - \hat{k}\) and \(\vec{b} = \hat{i} + \hat{j} + 2\hat{k}\), we can follow these steps:

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

The area of the triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \|\vec{a} \times \vec{b}\| \]
First, we need to calculate the cross product \(\vec{a} \times \vec{b}\).

\[ \vec{a} = \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]

The cross product can be calculated using the determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -1 \\ 1 & 1 & 2 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Calculating the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} = (4)(2) - (-1)(1) = 8 + 1 = 9\)

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

3. \(\begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} = (1)(1) - (4)(1) = 1 - 4 = -3\)

Putting it all together:
\[ \vec{a} \times \vec{b} = 9\hat{i} - 3\hat{j} - 3\hat{k} \]

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

Now, we find the magnitude of \(\vec{a} \times \vec{b}\):
\[ \|\vec{a} \times \vec{b}\| = \sqrt{9^2 + (-3)^2 + (-3)^2} = \sqrt{81 + 9 + 9} = \sqrt{99} \]

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

Now we can calculate the area of the triangle:
\[ \text{Area} = \frac{1}{2} \|\vec{a} \times \vec{b}\| = \frac{1}{2} \sqrt{99} \]

### Final Step: Simplify the Area

We can simplify \(\sqrt{99}\) as follows:
\[ \sqrt{99} = \sqrt{9 \times 11} = 3\sqrt{11} \]
Thus, the area becomes:
\[ \text{Area} = \frac{1}{2} \times 3\sqrt{11} = \frac{3}{2} \sqrt{11} \]

### Conclusion

The area of the triangle is:
\[ \text{Area} = \frac{3}{2} \sqrt{11} \quad \text{(approximately 4.97 square units)} \]