The vector equation of the plane through the point `(2, 1, -1)` and passing through the line of intersection of the plane `rcdot(hat(i)+3hat(j)-hat(k))=0 and rcdot(hat(j)+2hat(k))=0,` is

Options

A. `rcdot(hat(i)+9hat(j)+11hat(k))=0`
B. `rcdot(hat(i)+9hat(j)+11hat(k))=6`
C. `hat(r)cdot(hat(i)-3hat(k)-13hat(k))=0`
D. None of these

Answer

To find the vector equation of the plane through the point \( (2, 1, -1) \) and passing through the line of intersection of the given planes, we can follow these steps:

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

From these equations, we can identify the normal vectors:
- For the first plane, \( \mathbf{n_1} = \hat{i} + 3\hat{j} - \hat{k} \)
- For the second plane, \( \mathbf{n_2} = \hat{j} + 2\hat{k} \)

### Step 2: Write the general equation of the plane
The general equation of a plane that is a linear combination of the two planes can be expressed as:
\[ \mathbf{r} \cdot \mathbf{n_1} + \lambda \mathbf{n_2} = 0 \]
where \( \lambda \) is a scalar.

### Step 3: Substitute the normal vectors into the equation
Substituting the normal vectors into the equation gives:
\[ \mathbf{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) + \lambda (\hat{j} + 2\hat{k}) = 0 \]

### Step 4: Substitute the point into the equation
The plane passes through the point \( (2, 1, -1) \), which we can express as the position vector:
\[ \mathbf{p} = 2\hat{i} + 1\hat{j} - 1\hat{k} \]
Substituting this point into the equation results in:
\[ (2\hat{i} + 1\hat{j} - 1\hat{k}) \cdot (\hat{i} + 3\hat{j} - \hat{k}) + \lambda (1\hat{j} + 2(-1)\hat{k}) = 0 \]

### Step 5: Calculate the dot products
Calculating the dot product:
1. \( (2\hat{i}) \cdot (\hat{i}) = 2 \)
2. \( (1\hat{j}) \cdot (3\hat{j}) = 3 \)
3. \( (-1\hat{k}) \cdot (-\hat{k}) = 1 \)

Thus, we have:
\[ 2 + 3 + 1 + \lambda(1 - 4) = 0 \]
This simplifies to:
\[ 6 - 3\lambda = 0 \]

### Step 6: Solve for \( \lambda \)
Solving for \( \lambda \):
\[ 3\lambda = 6 \implies \lambda = 2 \]

### Step 7: Substitute \( \lambda \) back into the plane equation
Now substitute \( \lambda \) back into the plane equation:
\[ \mathbf{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) + 2(\hat{j} + 2\hat{k}) = 0 \]
This gives:
\[ \mathbf{r} \cdot (\hat{i} + 3\hat{j} - \hat{k} + 2\hat{j} + 4\hat{k}) = 0 \]
Simplifying the normal vector:
\[ \mathbf{n} = \hat{i} + (3 + 2)\hat{j} + (-1 + 4)\hat{k} = \hat{i} + 5\hat{j} + 3\hat{k} \]

### Step 8: Write the final vector equation of the plane
Thus, the vector equation of the plane is:
\[ \mathbf{r} \cdot (\hat{i} + 5\hat{j} + 3\hat{k}) = 0 \]