If the planes `x+2y+kz=5 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)`

Answer

To find the value of \( k \) such that the planes \( x + 2y + kz = 5 \) and \( 2x + y - 2z = 0 \) are at right angles, we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The general form of a plane is given by \( Ax + By + Cz = D \). The normal vector of the plane can be represented as \( \vec{n} = (A, B, C) \).

For the first plane \( x + 2y + kz = 5 \):
- The coefficients are \( A = 1, B = 2, C = k \).
- Therefore, the normal vector \( \vec{n_1} = (1, 2, k) \).

For the second plane \( 2x + y - 2z = 0 \):
- The coefficients are \( A = 2, B = 1, C = -2 \).
- Therefore, the normal vector \( \vec{n_2} = (2, 1, -2) \).

### Step 2: Use the condition for perpendicularity
Two vectors are perpendicular if their dot product is zero. Therefore, we need to calculate the dot product of \( \vec{n_1} \) and \( \vec{n_2} \):

\[ \vec{n_1} \cdot \vec{n_2} = (1)(2) + (2)(1) + (k)(-2) \]

### Step 3: Set the dot product equal to zero
Setting the dot product equal to zero gives us the equation:

\[ 2 + 2 - 2k = 0 \]

### Step 4: Solve for \( k \)
Now, simplify the equation:

\[ 4 - 2k = 0 \]

Rearranging gives:

\[ 2k = 4 \]

Dividing both sides by 2:

\[ k = 2 \]

### Conclusion
The value of \( k \) such that the planes are perpendicular is \( k = 2 \).

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