The vectors `a =3i-2j+2k and b =-i-2k` are adjacement sides of a parallelogram. Then angle between its diagonalsc is

Options

A. `pi//4`
B. `pi//3`
C. `3pi//4`
D. `2pi//3`

Answer

To find the angle between the diagonals of a parallelogram formed by the vectors \( \mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \) and \( \mathbf{b} = -\mathbf{i} - 2\mathbf{k} \), we can follow these steps:

### Step 1: Identify the vectors
We have:
\[ \mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \]
\[ \mathbf{b} = -\mathbf{i} - 2\mathbf{k} \]

### Step 2: Find the diagonals of the parallelogram
The diagonals of the parallelogram can be represented as:
- Diagonal \( \mathbf{AC} = \mathbf{a} + \mathbf{b} \)
- Diagonal \( \mathbf{BD} = \mathbf{a} - \mathbf{b} \)

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

### Step 4: Calculate \( \mathbf{BD} \)
\[ \mathbf{BD} = \mathbf{a} - \mathbf{b} = (3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) - (-\mathbf{i} - 2\mathbf{k}) \]
\[ = (3 + 1)\mathbf{i} + (-2)\mathbf{j} + (2 + 2)\mathbf{k} = 4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} \]

### Step 5: Find the dot product of \( \mathbf{AC} \) and \( \mathbf{BD} \)
Now, we calculate the dot product \( \mathbf{AC} \cdot \mathbf{BD} \):
\[ \mathbf{AC} \cdot \mathbf{BD} = (2\mathbf{i} - 2\mathbf{j}) \cdot (4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}) \]
\[ = 2 \cdot 4 + (-2) \cdot (-2) + 0 \cdot 4 = 8 + 4 + 0 = 12 \]

### Step 6: Calculate the magnitudes of \( \mathbf{AC} \) and \( \mathbf{BD} \)
Magnitude of \( \mathbf{AC} \):
\[ |\mathbf{AC}| = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

Magnitude of \( \mathbf{BD} \):
\[ |\mathbf{BD}| = \sqrt{(4)^2 + (-2)^2 + (4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \]

### Step 7: Use the dot product to find the cosine of the angle
Using the formula for the dot product:
\[ \mathbf{AC} \cdot \mathbf{BD} = |\mathbf{AC}| |\mathbf{BD}| \cos \theta \]
Substituting the values:
\[ 12 = (2\sqrt{2})(6) \cos \theta \]
\[ 12 = 12\sqrt{2} \cos \theta \]
\[ \cos \theta = \frac{12}{12\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 8: Find the angle \( \theta \)
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \]

### Step 9: Determine the angle between the diagonals
Since the diagonals of a parallelogram intersect at an angle, the angle between the diagonals is \( 180^\circ - 45^\circ = 135^\circ \).

### Final Answer
The angle between the diagonals of the parallelogram is \( 135^\circ \).

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