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 solve the problem, we need to find the value of \( k \) such that the angle between the two planes \( P_1 \) and \( P_2 \) is \( \frac{2\pi}{3} \).

### Step-by-Step Solution:

1. **Identify the Normal Vectors**:
The equations of the planes are given as:
- Plane \( P_1: 4x - 2y + 2z = 3 \)
The normal vector \( \vec{n_1} = (4, -2, 2) \).
- Plane \( P_2: -x + ky - 2z = 7 \)
The normal vector \( \vec{n_2} = (-1, k, -2) \).

2. **Use the Formula for the Angle Between Two Planes**:
The cosine of the angle \( \theta \) between two planes can be calculated using the formula:
\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]
where \( \vec{n_1} \cdot \vec{n_2} \) is the dot product of the normal vectors and \( |\vec{n_1}| \) and \( |\vec{n_2}| \) are the magnitudes of the normal vectors.

3. **Calculate the Dot Product \( \vec{n_1} \cdot \vec{n_2} \)**:
\[ \vec{n_1} \cdot \vec{n_2} = (4)(-1) + (-2)(k) + (2)(-2) = -4 - 2k - 4 = -8 - 2k \]

4. **Calculate the Magnitudes**:
\[ |\vec{n_1}| = \sqrt{4^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \]
\[ |\vec{n_2}| = \sqrt{(-1)^2 + k^2 + (-2)^2} = \sqrt{1 + k^2 + 4} = \sqrt{k^2 + 5} \]

5. **Set Up the Equation Using the Given Angle**:
Since the angle \( \theta = \frac{2\pi}{3} \), we know:
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]
Therefore, we can set up the equation:
\[ -\frac{1}{2} = \frac{-8 - 2k}{(2\sqrt{6})(\sqrt{k^2 + 5})} \]

6. **Cross Multiply to Eliminate the Fraction**:
\[ -1 \cdot (2\sqrt{6})(\sqrt{k^2 + 5}) = -2(-8 - 2k) \]
Simplifying gives:
\[ 2\sqrt{6}\sqrt{k^2 + 5} = 16 + 4k \]

7. **Square Both Sides**:
\[ 4 \cdot 6 \cdot (k^2 + 5) = (16 + 4k)^2 \]
\[ 24(k^2 + 5) = 256 + 128k + 16k^2 \]
Expanding and rearranging gives:
\[ 24k^2 + 120 = 16k^2 + 128k + 256 \]
\[ 8k^2 - 128k - 136 = 0 \]

8. **Divide the Equation by 8**:
\[ k^2 - 16k - 17 = 0 \]

9. **Factor the Quadratic Equation**:
\[ (k - 17)(k + 1) = 0 \]
Therefore, the solutions are:
\[ k = 17 \quad \text{or} \quad k = -1 \]

### Final Answer:
The values of \( k \) can be \( 17 \) or \( -1 \).