If the planes `2x-y+lamdaz-5=0` an `x+4y+2z-7=0` are perpendicular then `lamda=`

Options

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

Explanation

To solve the problem of finding the value of \(\lambda\) for which the planes \(2x - y + \lambda z - 5 = 0\) and \(x + 4y + 2z - 7 = 0\) are perpendicular, we can follow these steps:

### Step 1: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \(Ax + By + Cz + D = 0\) is \(\langle A, B, C \rangle\).

For the first plane \(2x - y + \lambda z - 5 = 0\):
- The coefficients are \(A = 2\), \(B = -1\), and \(C = \lambda\).
- Thus, the normal vector \(n_1 = \langle 2, -1, \lambda \rangle\).

For the second plane \(x + 4y + 2z - 7 = 0\):
- The coefficients are \(A = 1\), \(B = 4\), and \(C = 2\).
- Thus, the normal vector \(n_2 = \langle 1, 4, 2 \rangle\).

### 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 \(n_1\) and \(n_2\) and set it equal to zero.

The dot product \(n_1 \cdot n_2\) is given by:
\[ n_1 \cdot n_2 = (2)(1) + (-1)(4) + (\lambda)(2) \]

### Step 3: Set up the equation
Setting the dot product equal to zero gives us:
\[ 2 \cdot 1 + (-1) \cdot 4 + \lambda \cdot 2 = 0 \]
This simplifies to:
\[ 2 - 4 + 2\lambda = 0 \]

### Step 4: Solve for \(\lambda\)
Now, simplify the equation:
\[ -2 + 2\lambda = 0 \]
Adding 2 to both sides:
\[ 2\lambda = 2 \]
Dividing both sides by 2:
\[ \lambda = 1 \]

### Final Answer
Thus, the value of \(\lambda\) is:
\[ \lambda = 1 \]