Find the area of the parallelogram whose adjacent sides are given by the vectors `veca=3hati+hatj+4hatkandvecb=hati-hatj+hatk`.

Answer

To find the area of the parallelogram formed by the vectors **a** and **b**, we will use the formula for the area of a parallelogram given by two vectors, which is the magnitude of their cross product.

### 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: Compute the cross product \(\vec{a} \times \vec{b}\)
To find the cross product, we can use the determinant of a matrix formed by the unit vectors and the components of the vectors **a** and **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
Calculating the determinant, we expand it as follows:

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

Calculating each of the 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 1 & 4 \\ -1 & 1 \end{vmatrix} = (1)(1) - (4)(-1) = 1 + 4 = 5 \]

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 3 & 4 \\ 1 & 1 \end{vmatrix} = (3)(1) - (4)(1) = 3 - 4 = -1 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (1)(1) = -3 - 1 = -4 \]

Putting it all together:
\[ \vec{a} \times \vec{b} = 5\hat{i} + 1\hat{j} - 4\hat{k} \]

### Step 4: Find the magnitude of the cross product
Now, we need to find the magnitude of the vector \(\vec{a} \times \vec{b}\):
\[ |\vec{a} \times \vec{b}| = \sqrt{(5)^2 + (1)^2 + (-4)^2} \]
\[ = \sqrt{25 + 1 + 16} = \sqrt{42} \]

### Step 5: Conclusion
The area of the parallelogram is given by the magnitude of the cross product:
\[ \text{Area} = |\vec{a} \times \vec{b}| = \sqrt{42} \text{ square units} \]