The area of a parallelogram having diagonals `a=3i+j-2k" and "b=i-3j+4k` is

Options

A. `5sqrt(3)`
B. `2sqrt(3)`
C. `4sqrt(3)`
D. none of these

Answer

To find the area of a parallelogram with given diagonals represented by vectors \( \mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \), we can use the following steps:

### Step-by-Step Solution:

1. **Identify the Diagonal Vectors:**
Let \( \mathbf{A} = \mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k} \) and \( \mathbf{B} = \mathbf{b} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \).

2. **Find the Vectors Representing the Sides of the Parallelogram:**
The sides of the parallelogram can be represented using the following formulas:
\[ \mathbf{\alpha} = \frac{\mathbf{A} + \mathbf{B}}{2} \]
\[ \mathbf{\beta} = \frac{\mathbf{A} - \mathbf{B}}{2} \]

3. **Calculate \( \mathbf{\alpha} \):**
\[ \mathbf{\alpha} = \frac{(3\mathbf{i} + \mathbf{j} - 2\mathbf{k}) + (\mathbf{i} - 3\mathbf{j} + 4\mathbf{k})}{2} \]
\[ = \frac{(3 + 1)\mathbf{i} + (1 - 3)\mathbf{j} + (-2 + 4)\mathbf{k}}{2} \]
\[ = \frac{4\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}}{2} = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \]

4. **Calculate \( \mathbf{\beta} \):**
\[ \mathbf{\beta} = \frac{(3\mathbf{i} + \mathbf{j} - 2\mathbf{k}) - (\mathbf{i} - 3\mathbf{j} + 4\mathbf{k})}{2} \]
\[ = \frac{(3 - 1)\mathbf{i} + (1 + 3)\mathbf{j} + (-2 - 4)\mathbf{k}}{2} \]
\[ = \frac{2\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}}{2} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \]

5. **Calculate the Cross Product \( \mathbf{\alpha} \times \mathbf{\beta} \):**
\[ \mathbf{\alpha} \times \mathbf{\beta} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 2 & -3 \end{vmatrix} \]

Expanding the determinant:
\[ = \mathbf{i} \begin{vmatrix} -1 & 1 \\ 2 & -3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} \]

Calculating the minors:
\[ = \mathbf{i}((-1)(-3) - (1)(2)) - \mathbf{j}((2)(-3) - (1)(1)) + \mathbf{k}((2)(2) - (-1)(1)) \]
\[ = \mathbf{i}(3 - 2) - \mathbf{j}(-6 - 1) + \mathbf{k}(4 + 1) \]
\[ = \mathbf{i}(1) + \mathbf{j}(7) + \mathbf{k}(5) \]
\[ = \mathbf{i} + 7\mathbf{j} + 5\mathbf{k} \]

6. **Find the Magnitude of the Cross Product:**
\[ |\mathbf{\alpha} \times \mathbf{\beta}| = \sqrt{1^2 + 7^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3} \]

7. **Calculate the Area of the Parallelogram:**
The area of the parallelogram is given by:
\[ \text{Area} = \frac{1}{2} |\mathbf{\alpha} \times \mathbf{\beta}| \]
\[ = \frac{1}{2} \times 5\sqrt{3} = \frac{5\sqrt{3}}{2} \]

### Final Answer:
The area of the parallelogram is \( \frac{5\sqrt{3}}{2} \).

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