Two adjacent sides of a parallelogram ABCD are `2hati+4hatj -5 hatkand hati+2hatj+3hatk`. Then the value of `|vec(AC)xxvec(BD)|` is

Options

A. `20sqrt5`
B. `22sqrt5`
C. `24sqrt5`
D. `26sqrt5`

Explanation

To solve the problem, we need to find the value of \(|\vec{AC} \times \vec{BD}|\) given the two adjacent sides of a parallelogram \( \vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{BC} = \hat{i} + 2\hat{j} + 3\hat{k} \).

### Step-by-Step Solution:

1. **Identify the vectors:**
- Let \( \vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k} \)
- Let \( \vec{BC} = \hat{i} + 2\hat{j} + 3\hat{k} \)

2. **Use the property of the diagonals of a parallelogram:**
- The property states that \( \vec{AC} \times \vec{BD} = 2(\vec{AB} \times \vec{BC}) \).

3. **Calculate the cross product \( \vec{AB} \times \vec{BC} \):**
\[ \vec{AB} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -5 \\ 1 & 2 & 3 \end{vmatrix} \]

Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} 4 & -5 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -5 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix} \]

- Calculate each of the 2x2 determinants:
- \( \begin{vmatrix} 4 & -5 \\ 2 & 3 \end{vmatrix} = (4)(3) - (-5)(2) = 12 + 10 = 22 \)
- \( \begin{vmatrix} 2 & -5 \\ 1 & 3 \end{vmatrix} = (2)(3) - (-5)(1) = 6 + 5 = 11 \)
- \( \begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix} = (2)(2) - (4)(1) = 4 - 4 = 0 \)

- Substitute back into the equation:
\[ \vec{AB} \times \vec{BC} = 22\hat{i} - 11\hat{j} + 0\hat{k} = 22\hat{i} - 11\hat{j} \]

4. **Calculate \( \vec{AC} \times \vec{BD} \):**
\[ \vec{AC} \times \vec{BD} = 2(\vec{AB} \times \vec{BC}) = 2(22\hat{i} - 11\hat{j}) = 44\hat{i} - 22\hat{j} \]

5. **Find the magnitude:**
\[ |\vec{AC} \times \vec{BD}| = |44\hat{i} - 22\hat{j}| = \sqrt{(44)^2 + (-22)^2} = \sqrt{1936 + 484} = \sqrt{2420} \]

- Simplifying \( \sqrt{2420} \):
\[ = \sqrt{4 \times 605} = 2\sqrt{605} \]

6. **Final answer:**
\[ |\vec{AC} \times \vec{BD}| = 2\sqrt{605} \]