Find the planes passing through the intersection of plane `rcdot(2hat(i)-3hat(j)+4hat(k))=1` and `rcdot(hat(i)-hat(j))+4=0` and perpendicular to planes `rcdot(2hat(i)-hat(j)+hat(k))=-8`

Answer

To solve the problem, we need to find the planes that pass through the intersection of two given planes and are perpendicular to a third plane. Let's break down the solution step by step.

### Step 1: Identify the given planes

The two planes provided in the question are:

1. \( P_1: \mathbf{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 1 \)
2. \( P_2: \mathbf{r} \cdot (\hat{i} - \hat{j}) + 4 = 0 \)

We can rewrite these equations in a more standard form:

- For \( P_1 \): \( \mathbf{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) - 1 = 0 \)
- For \( P_2 \): \( \mathbf{r} \cdot (\hat{i} - \hat{j}) + 4 = 0 \) can be rewritten as \( \mathbf{r} \cdot (\hat{i} - \hat{j}) = -4 \)

### Step 2: Find the equation of the plane through the intersection

The equation of a plane passing through the intersection of two planes \( P_1 \) and \( P_2 \) can be expressed as:

\[ P_1 + \lambda P_2 = 0 \]

Where \( \lambda \) is a constant. Substituting the equations of \( P_1 \) and \( P_2 \):

\[ \mathbf{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) - 1 + \lambda (\mathbf{r} \cdot (\hat{i} - \hat{j}) + 4) = 0 \]

This simplifies to:

\[ \mathbf{r} \cdot (2 + \lambda)\hat{i} + (-3 - \lambda)\hat{j} + 4\hat{k} = 1 - 4\lambda \]

### Step 3: Identify the normal vector of the new plane

The normal vector of the new plane is:

\[ \mathbf{n} = (2 + \lambda)\hat{i} + (-3 - \lambda)\hat{j} + 4\hat{k} \]

### Step 4: Use the condition of perpendicularity

The new plane must be perpendicular to the plane given by:

\[ P_3: \mathbf{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = -8 \]

The normal vector of this plane is:

\[ \mathbf{n_3} = 2\hat{i} - \hat{j} + \hat{k} \]

For the planes to be perpendicular, the dot product of their normal vectors must equal zero:

\[ \mathbf{n} \cdot \mathbf{n_3} = 0 \]

Substituting the normal vector \( \mathbf{n} \):

\[ (2 + \lambda) \cdot 2 + (-3 - \lambda)(-1) + 4 \cdot 1 = 0 \]

### Step 5: Solve for \( \lambda \)

Expanding the equation:

\[ 2(2 + \lambda) + (3 + \lambda) + 4 = 0 \]

This simplifies to:

\[ 4 + 2\lambda + 3 + \lambda + 4 = 0 \]

Combining like terms:

\[ 2\lambda + \lambda + 11 = 0 \]

Thus:

\[ 3\lambda + 11 = 0 \implies \lambda = -\frac{11}{3} \]

### Step 6: Substitute \( \lambda \) back into the plane equation

Now substitute \( \lambda \) back into the equation of the plane:

\[ \mathbf{r} \cdot \left(-5\hat{i} + 2\hat{j} + 12\hat{k}\right) = 47 \]

### Final Answer

The desired plane is:

\[ \mathbf{r} \cdot \left(-5\hat{i} + 2\hat{j} + 12\hat{k}\right) = 47 \]

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