If the planes `x+2y+kz=0` and `2x+y-2z=0`, are at right angles, then the value of k is

Options

A. 2
B. `-2`
C. `(1)/(2)`
D. `-(1)/(2)`

Explanation

To find the value of \( k \) such that the planes \( x + 2y + kz = 0 \) and \( 2x + y - 2z = 0 \) are at right angles, we can use the condition for the perpendicularity of two planes.

### Step-by-Step Solution:

1. **Identify the normal vectors of the planes**:
The normal vector of the first plane \( x + 2y + kz = 0 \) is given by the coefficients of \( x, y, z \):
\[ \mathbf{n_1} = (1, 2, k) \]
The normal vector of the second plane \( 2x + y - 2z = 0 \) is:
\[ \mathbf{n_2} = (2, 1, -2) \]

2. **Use the condition for perpendicularity**:
Two planes are perpendicular if the dot product of their normal vectors is zero:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 0 \]
This gives us:
\[ (1)(2) + (2)(1) + (k)(-2) = 0 \]

3. **Expand the dot product**:
Simplifying the equation:
\[ 2 + 2 - 2k = 0 \]
Combine the constants:
\[ 4 - 2k = 0 \]

4. **Solve for \( k \)**:
Rearranging the equation gives:
\[ 2k = 4 \]
Dividing both sides by 2:
\[ k = 2 \]

### Final Answer:
The value of \( k \) is \( 2 \).

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