The area of the triangle whose adjacent sides are : `vec(a)=3hat(i)+hat(j)+4hat(k)` and `vec(b)=hat(i)-hat(j)+hat(k)` is :

Options

A. `(1)/(2)sqrt(42)`
B. 42
C. `sqrt(42)`
D. `sqrt(21)`

Answer

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

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

where \(\|\vec{a} \times \vec{b}\|\) is the magnitude of the cross product of the vectors \(\vec{a}\) and \(\vec{b}\).

### Step 1: Define the vectors
Given:
\[ \vec{a} = 3\hat{i} + \hat{j} + 4\hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]

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

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 4 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 4 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 4 \\ -1 & 1 \end{vmatrix} = (1)(1) - (4)(-1) = 1 + 4 = 5\)
2. \(\begin{vmatrix} 3 & 4 \\ 1 & 1 \end{vmatrix} = (3)(1) - (4)(1) = 3 - 4 = -1\)
3. \(\begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (1)(1) = -3 - 1 = -4\)

Putting it all together:

\[ \vec{a} \times \vec{b} = 5\hat{i} - (-1)\hat{j} - 4\hat{k} = 5\hat{i} + 1\hat{j} - 4\hat{k} \]

### Step 4: Find the magnitude of \(\vec{a} \times \vec{b}\)
The magnitude is given by:

\[ \|\vec{a} \times \vec{b}\| = \sqrt{(5)^2 + (1)^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42} \]

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

\[ \text{Area} = \frac{1}{2} \|\vec{a} \times \vec{b}\| = \frac{1}{2} \sqrt{42} \]

### Final Answer
The area of the triangle is:

\[ \text{Area} = \frac{\sqrt{42}}{2} \]