Suppose the diagonals of a parallelogram are represented by vectors `i+3j-2k" and "3i+j-4k`. If A is the area of this parallelogram, then A =

Answer

To find the area \( A \) of the parallelogram given its diagonals represented by the vectors \( \mathbf{d_1} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} \) and \( \mathbf{d_2} = 3\mathbf{i} + \mathbf{j} - 4\mathbf{k} \), we can use the formula:

\[ A = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| \]

### Step 1: Calculate the Cross Product \( \mathbf{d_1} \times \mathbf{d_2} \)

The cross product of two vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \) is given by the determinant:

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]

For our vectors:

\[ \mathbf{d_1} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}, \quad \mathbf{d_2} = \begin{pmatrix} 3 \\ 1 \\ -4 \end{pmatrix} \]

The determinant becomes:

\[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -2 \\ 3 & 1 & -4 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Calculating the determinant:

\[ = \mathbf{i} \begin{vmatrix} 3 & -2 \\ 1 & -4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -2 \\ 3 & -4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 3 & -2 \\ 1 & -4 \end{vmatrix} = (3)(-4) - (-2)(1) = -12 + 2 = -10 \)
2. \( \begin{vmatrix} 1 & -2 \\ 3 & -4 \end{vmatrix} = (1)(-4) - (-2)(3) = -4 + 6 = 2 \)
3. \( \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} = (1)(1) - (3)(3) = 1 - 9 = -8 \)

Putting it all together:

\[ \mathbf{d_1} \times \mathbf{d_2} = -10\mathbf{i} - 2\mathbf{j} - 8\mathbf{k} \]

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

The magnitude of the vector \( \mathbf{d_1} \times \mathbf{d_2} \) is given by:

\[ \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \sqrt{(-10)^2 + (-2)^2 + (-8)^2} \]

Calculating:

\[ = \sqrt{100 + 4 + 64} = \sqrt{168} \]

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

Now, substituting back into the area formula:

\[ A = \frac{1}{2} \left| \mathbf{d_1} \times \mathbf{d_2} \right| = \frac{1}{2} \sqrt{168} \]

Simplifying \( \sqrt{168} \):

\[ \sqrt{168} = \sqrt{4 \times 42} = 2\sqrt{42} \]

Thus, the area becomes:

\[ A = \frac{1}{2} \times 2\sqrt{42} = \sqrt{42} \]

### Final Answer

The area \( A \) of the parallelogram is:

\[ \boxed{\sqrt{42}} \]