If `2hat(i)+hat(j)-3hat(k)` and `m hat(i)+3hat(j)-hat(k)` are perpendicular to each other, then find 'm'. Also find the area of the rectangle having these two vectors as sides.

Answer

To solve the problem, we need to find the value of \( m \) such that the vectors \( \mathbf{A} = 2\hat{i} + \hat{j} - 3\hat{k} \) and \( \mathbf{B} = m\hat{i} + 3\hat{j} - \hat{k} \) are perpendicular. We will also find the area of the rectangle formed by these two vectors.

### Step 1: Set up the dot product equation
Since the vectors are perpendicular, their dot product must equal zero:
\[ \mathbf{A} \cdot \mathbf{B} = 0 \]
Calculating the dot product:
\[ (2\hat{i} + \hat{j} - 3\hat{k}) \cdot (m\hat{i} + 3\hat{j} - \hat{k}) = 0 \]
This expands to:
\[ 2m + 1 \cdot 3 + (-3)(-1) = 0 \]
Simplifying this gives:
\[ 2m + 3 + 3 = 0 \]
\[ 2m + 6 = 0 \]

### Step 2: Solve for \( m \)
Now, we solve for \( m \):
\[ 2m = -6 \implies m = -3 \]

### Step 3: Find the area of the rectangle
The area of the rectangle formed by the two vectors is given by the magnitude of the cross product of the vectors \( \mathbf{A} \) and \( \mathbf{B} \):
\[ \text{Area} = |\mathbf{A} \times \mathbf{B}| \]

### Step 4: Calculate the cross product
The cross product \( \mathbf{A} \times \mathbf{B} \) can be calculated using the determinant of a matrix:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ -3 & 3 & -1 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -3 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ -3 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ -3 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[ 1 \cdot (-1) - (-3) \cdot 3 = -1 + 9 = 8 \]
2. For \( \hat{j} \):
\[ 2 \cdot (-1) - (-3) \cdot (-3) = -2 - 9 = -11 \]
3. For \( \hat{k} \):
\[ 2 \cdot 3 - 1 \cdot (-3) = 6 + 3 = 9 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 8\hat{i} + 11\hat{j} + 9\hat{k} \]

### Step 5: Find the magnitude of the cross product
Now, we find the magnitude:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{8^2 + 11^2 + 9^2} = \sqrt{64 + 121 + 81} \]
Calculating the sum:
\[ = \sqrt{266} \]

### Final Result
Thus, the value of \( m \) is \( -3 \) and the area of the rectangle is \( \sqrt{266} \).