If the diagonals of a parallelogram are `3 hati + hatj -2hatk and hati - 3 hatj + 4 hatk,` then the lengths of its sides are

Options

A. `sqrt(8), sqrt(10)`
B. `sqrt(6), sqrt(14)`
C. `sqrt(5), sqrt(12)`
D. none of these

Answer

To find the lengths of the sides of the parallelogram given its diagonals, we can follow these steps:

### Step 1: Define the diagonals
Let the diagonals of the parallelogram be represented as:
- \( \vec{AC} = 3\hat{i} + \hat{j} - 2\hat{k} \)
- \( \vec{BD} = \hat{i} - 3\hat{j} + 4\hat{k} \)

### Step 2: Find the vectors representing the sides
The diagonals of a parallelogram bisect each other. Therefore, we can express the vectors for the sides \( \vec{DA} \) and \( \vec{DC} \) in terms of the diagonals.

Using the property of diagonals:
\[ \vec{DA} + \vec{AO} = \vec{DO} \]
Where \( O \) is the midpoint of both diagonals.

### Step 3: Express the vectors in terms of the diagonals
We can express \( \vec{DA} \) as:
\[ \vec{DA} = \vec{DO} - \vec{AO} \]
Since \( \vec{DO} = \frac{1}{2} \vec{BD} \) and \( \vec{AO} = \frac{1}{2} \vec{AC} \), we can substitute:
\[ \vec{DA} = \frac{1}{2} \vec{BD} - \frac{1}{2} \vec{AC} \]

### Step 4: Calculate \( \vec{DA} \)
Substituting the values of \( \vec{AC} \) and \( \vec{BD} \):
\[ \vec{DA} = \frac{1}{2} \left( \hat{i} - 3\hat{j} + 4\hat{k} \right) - \frac{1}{2} \left( 3\hat{i} + \hat{j} - 2\hat{k} \right) \]
Calculating this:
\[ \vec{DA} = \frac{1}{2} \left( \hat{i} - 3\hat{j} + 4\hat{k} - 3\hat{i} - \hat{j} + 2\hat{k} \right) \]
\[ = \frac{1}{2} \left( (1 - 3)\hat{i} + (-3 - 1)\hat{j} + (4 + 2)\hat{k} \right) \]
\[ = \frac{1}{2} \left( -2\hat{i} - 4\hat{j} + 6\hat{k} \right) \]
\[ = -\hat{i} - 2\hat{j} + 3\hat{k} \]

### Step 5: Find the magnitude of \( \vec{DA} \)
The magnitude of \( \vec{DA} \) is calculated as follows:
\[ |\vec{DA}| = \sqrt{(-1)^2 + (-2)^2 + (3)^2} \]
\[ = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 6: Calculate the other side \( \vec{DC} \)
Using a similar approach for \( \vec{DC} \):
\[ \vec{DC} = \vec{DO} + \vec{OC} \]
Where \( \vec{OC} = \frac{1}{2} \vec{AC} \):
\[ \vec{DC} = \frac{1}{2} \vec{BD} + \frac{1}{2} \vec{AC} \]
Calculating this:
\[ \vec{DC} = \frac{1}{2} \left( \hat{i} - 3\hat{j} + 4\hat{k} + 3\hat{i} + \hat{j} - 2\hat{k} \right) \]
\[ = \frac{1}{2} \left( (1 + 3)\hat{i} + (-3 + 1)\hat{j} + (4 - 2)\hat{k} \right) \]
\[ = \frac{1}{2} \left( 4\hat{i} - 2\hat{j} + 2\hat{k} \right) \]
\[ = 2\hat{i} - \hat{j} + \hat{k} \]

### Step 7: Find the magnitude of \( \vec{DC} \)
Calculating the magnitude:
\[ |\vec{DC}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} \]
\[ = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Final Result
The lengths of the sides of the parallelogram are:
- Length of side \( DA \) = \( \sqrt{14} \)
- Length of side \( DC \) = \( \sqrt{6} \)