If `a = 2i-3j+k, b=-i+k,c=2j-k` then the area of parallelogram whose diagonals are `a +b` and `b +c` is ........

Answer

To find the area of the parallelogram whose diagonals are given by the vectors \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{b} + \mathbf{c} \), we can follow these steps:

### Step 1: Define the vectors
Given:
- \( \mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k} \)
- \( \mathbf{b} = -\mathbf{i} + \mathbf{k} \)
- \( \mathbf{c} = 2\mathbf{j} - \mathbf{k} \)

### Step 2: Calculate \( \mathbf{a} + \mathbf{b} \)
\[ \mathbf{a} + \mathbf{b} = (2\mathbf{i} - 3\mathbf{j} + \mathbf{k}) + (-\mathbf{i} + \mathbf{k}) \]
\[ = (2 - 1)\mathbf{i} + (-3)\mathbf{j} + (1 + 1)\mathbf{k} \]
\[ = \mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \]

### Step 3: Calculate \( \mathbf{b} + \mathbf{c} \)
\[ \mathbf{b} + \mathbf{c} = (-\mathbf{i} + \mathbf{k}) + (2\mathbf{j} - \mathbf{k}) \]
\[ = (-1)\mathbf{i} + (2)\mathbf{j} + (1 - 1)\mathbf{k} \]
\[ = -\mathbf{i} + 2\mathbf{j} \]

### Step 4: Find the cross product \( (\mathbf{a} + \mathbf{b}) \times (\mathbf{b} + \mathbf{c}) \)
Let \( \mathbf{u} = \mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \) and \( \mathbf{v} = -\mathbf{i} + 2\mathbf{j} \).

Now, we compute the cross product using the determinant:
\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix} \]

### Step 5: Calculate the determinant
Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} -3 & 2 \\ 2 & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -3 \\ -1 & 2 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -3 & 2 \\ 2 & 0 \end{vmatrix} = (-3)(0) - (2)(2) = -4 \)
2. \( \begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix} = (1)(0) - (2)(-1) = 2 \)
3. \( \begin{vmatrix} 1 & -3 \\ -1 & 2 \end{vmatrix} = (1)(2) - (-3)(-1) = 2 - 3 = -1 \)

Putting it all together:
\[ \mathbf{u} \times \mathbf{v} = -4\mathbf{i} - 2\mathbf{j} - 1\mathbf{k} \]

### Step 6: Find the magnitude of the cross product
\[ |\mathbf{u} \times \mathbf{v}| = \sqrt{(-4)^2 + (-2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \]

### Step 7: Calculate the area of the parallelogram
The area of the parallelogram is given by:
\[ \text{Area} = \frac{1}{2} |\mathbf{u} \times \mathbf{v}| = \frac{1}{2} \sqrt{21} \]

### Final Answer
The area of the parallelogram is \( \frac{\sqrt{21}}{2} \) square units.

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