A plane `P_(1)` has the equation `4x-2y+2z=3 and P_(2)` has the equation `-x+ky-2z=7`. If the angle between `P_(1)` and `P_(2)` is `(2pi)/(3)`, then the value of k can be

Options

A. 1
B. 2
C. `-17`
D. 17

Answer

To find the value of \( k \) such that the angle between the two planes \( P_1 \) and \( P_2 \) is \( \frac{2\pi}{3} \), we can follow these steps:

### Step 1: Write the equations of the planes in standard form
The equations of the planes are given as:
- Plane \( P_1: 4x - 2y + 2z = 3 \) can be rewritten as \( 4x - 2y + 2z - 3 = 0 \).
- Plane \( P_2: -x + ky - 2z = 7 \) can be rewritten as \( -x + ky - 2z - 7 = 0 \).

### Step 2: Identify the normal vectors of the planes
The normal vector of plane \( P_1 \) is \( \mathbf{n_1} = (4, -2, 2) \).
The normal vector of plane \( P_2 \) is \( \mathbf{n_2} = (-1, k, -2) \).

### Step 3: Use the formula for the angle between two planes
The cosine of the angle \( \theta \) between two planes can be calculated using the formula:
\[ \cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} \]
Given that \( \theta = \frac{2\pi}{3} \), we have:
\[ \cos \frac{2\pi}{3} = -\frac{1}{2} \]

### Step 4: Calculate the dot product \( \mathbf{n_1} \cdot \mathbf{n_2} \)
Calculating the dot product:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 4 \cdot (-1) + (-2) \cdot k + 2 \cdot (-2) = -4 - 2k - 4 = -8 - 2k \]

### Step 5: Calculate the magnitudes of the normal vectors
Calculating the magnitude of \( \mathbf{n_1} \):
\[ |\mathbf{n_1}| = \sqrt{4^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \]
Calculating the magnitude of \( \mathbf{n_2} \):
\[ |\mathbf{n_2}| = \sqrt{(-1)^2 + k^2 + (-2)^2} = \sqrt{1 + k^2 + 4} = \sqrt{k^2 + 5} \]

### Step 6: Substitute into the cosine formula
Substituting into the cosine formula:
\[ -\frac{1}{2} = \frac{-8 - 2k}{(2\sqrt{6})(\sqrt{k^2 + 5})} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ -1 \cdot (2\sqrt{6})(\sqrt{k^2 + 5}) = -2(-8 - 2k) \]
This simplifies to:
\[ 2\sqrt{6}\sqrt{k^2 + 5} = 16 + 4k \]

### Step 8: Square both sides to eliminate the square root
Squaring both sides:
\[ 4 \cdot 6 \cdot (k^2 + 5) = (16 + 4k)^2 \]
\[ 24(k^2 + 5) = 256 + 128k + 16k^2 \]
\[ 24k^2 + 120 = 16k^2 + 128k + 256 \]

### Step 9: Rearrange the equation
Rearranging gives:
\[ 24k^2 - 16k^2 - 128k + 120 - 256 = 0 \]
\[ 8k^2 - 128k - 136 = 0 \]
Dividing by 8:
\[ k^2 - 16k - 17 = 0 \]

### Step 10: Solve the quadratic equation
Using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot (-17)}}{2 \cdot 1} \]
\[ k = \frac{16 \pm \sqrt{256 + 68}}{2} = \frac{16 \pm \sqrt{324}}{2} = \frac{16 \pm 18}{2} \]
Thus, we have:
\[ k = \frac{34}{2} = 17 \quad \text{or} \quad k = \frac{-2}{2} = -1 \]

### Conclusion
The possible values of \( k \) are \( 17 \) and \( -1 \).