If the planes `vecr.(2hati-lamda hatj+3hatk)=0` and `vecr.(lamda hati+5hatj-hatk)=5` are perpendicular to each other then value of `lamda^(2)+lamda` is

Options

A. 0
B. 2
C. 3
D. 1

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Identify the normal vectors of the planes
The equations of the planes are given as:
1. \( \vec{r} \cdot (2\hat{i} - \lambda \hat{j} + 3\hat{k}) = 0 \)
2. \( \vec{r} \cdot (\lambda \hat{i} + 5\hat{j} - \hat{k}) = 5 \)

From these equations, we can identify the normal vectors:
- Normal vector of Plane 1, \( \vec{n_1} = 2\hat{i} - \lambda \hat{j} + 3\hat{k} \)
- Normal vector of Plane 2, \( \vec{n_2} = \lambda \hat{i} + 5\hat{j} - \hat{k} \)

### Step 2: Set up the condition for perpendicularity
The planes are perpendicular if their normal vectors are also perpendicular. This means that the dot product of the normal vectors must equal zero:
\[ \vec{n_1} \cdot \vec{n_2} = 0 \]

### Step 3: Calculate the dot product
Calculating the dot product:
\[ \vec{n_1} \cdot \vec{n_2} = (2\hat{i} - \lambda \hat{j} + 3\hat{k}) \cdot (\lambda \hat{i} + 5\hat{j} - \hat{k}) \]
Using the formula for the dot product:
\[ = 2\lambda + (-\lambda)(5) + 3(-1) \]
\[ = 2\lambda - 5\lambda - 3 \]
\[ = -3\lambda - 3 \]

### Step 4: Set the dot product equal to zero
Setting the dot product equal to zero gives:
\[ -3\lambda - 3 = 0 \]

### Step 5: Solve for \( \lambda \)
Solving for \( \lambda \):
\[ -3\lambda = 3 \]
\[ \lambda = -1 \]

### Step 6: Calculate \( \lambda^2 + \lambda \)
Now we need to find the value of \( \lambda^2 + \lambda \):
\[ \lambda^2 + \lambda = (-1)^2 + (-1) \]
\[ = 1 - 1 = 0 \]

### Final Answer
Thus, the value of \( \lambda^2 + \lambda \) is \( 0 \).

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