Find the vector and Cartesian equation of the plane that passes through the point (1,4,6) and the normal vector to the plane is `hati-2hatj+hatk`.

Answer

To find the vector and Cartesian equation of the plane that passes through the point (1, 4, 6) with a normal vector \( \hat{i} - 2\hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Identify the point and normal vector
We have the point \( A(1, 4, 6) \) and the normal vector \( \mathbf{n} = \hat{i} - 2\hat{j} + \hat{k} \).

### Step 2: Write the position vector of point A
The position vector \( \mathbf{A} \) corresponding to the point \( A(1, 4, 6) \) can be expressed as:
\[ \mathbf{A} = 1\hat{i} + 4\hat{j} + 6\hat{k} = \hat{i} + 4\hat{j} + 6\hat{k} \]

### Step 3: Use the plane equation formula
The general equation of a plane can be given by:
\[ \mathbf{r} - \mathbf{A} \cdot \mathbf{n} = 0 \]
Where \( \mathbf{r} \) is the position vector of any point on the plane.

### Step 4: Substitute the known values
Let \( \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k} \). The equation becomes:
\[ (x\hat{i} + y\hat{j} + z\hat{k}) - (\hat{i} + 4\hat{j} + 6\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 0 \]
This simplifies to:
\[ \mathbf{r} \cdot \mathbf{n} - \mathbf{A} \cdot \mathbf{n} = 0 \]

### Step 5: Calculate \( \mathbf{A} \cdot \mathbf{n} \)
Now, we need to calculate \( \mathbf{A} \cdot \mathbf{n} \):
\[ \mathbf{A} \cdot \mathbf{n} = (1\hat{i} + 4\hat{j} + 6\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) \]
Calculating the dot product:
\[ = 1 \cdot 1 + 4 \cdot (-2) + 6 \cdot 1 = 1 - 8 + 6 = -1 \]

### Step 6: Write the vector equation of the plane
Now we can substitute back into the equation:
\[ \mathbf{r} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = -1 \]
This is the vector equation of the plane.

### Step 7: Write the Cartesian equation
Now, we can express \( \mathbf{r} \) in terms of its components:
\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = -1 \]
Calculating the dot product:
\[ x - 2y + z = -1 \]
Rearranging gives us the Cartesian equation of the plane:
\[ x - 2y + z + 1 = 0 \]

### Final Result
- **Vector Equation of the Plane**: \( \mathbf{r} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = -1 \)
- **Cartesian Equation of the Plane**: \( x - 2y + z + 1 = 0 \)