The vector equation of a plane passing through a point having position vector `2hati+3hatj-4hatk` and perpendicular to the vector `2hati-hatj+2hatk`, is

Options

A. `vecr.(2hati-hatj+2hatk)=7`
B. `vecr.(2hati-hatj+2hatk)=-7`
C. `vecr.(2hati-hatj+2hatk=4`
D. none of these

Explanation

To find the vector equation of a plane that passes through a specific point and is perpendicular to a given vector, we can follow these steps:

### Step 1: Identify the given information
- The position vector of the point through which the plane passes is given as \( \mathbf{a} = 2\hat{i} + 3\hat{j} - 4\hat{k} \).
- The normal vector \( \mathbf{n} \) to the plane is given as \( \mathbf{n} = 2\hat{i} - \hat{j} + 2\hat{k} \).

### Step 2: Write the general form of the vector equation of the plane
The vector equation of a plane can be expressed as:
\[ \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} \]
where \( \mathbf{r} \) is the position vector of any point on the plane, \( \mathbf{n} \) is the normal vector, and \( \mathbf{a} \) is the position vector of the given point.

### Step 3: Calculate \( \mathbf{a} \cdot \mathbf{n} \)
First, we need to calculate the dot product \( \mathbf{a} \cdot \mathbf{n} \):
\[ \mathbf{a} \cdot \mathbf{n} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k}) \]
Calculating this dot product:
\[ = 2 \cdot 2 + 3 \cdot (-1) + (-4) \cdot 2 \]
\[ = 4 - 3 - 8 \]
\[ = 4 - 11 = -7 \]

### Step 4: Substitute into the plane equation
Now, substituting back into the equation of the plane:
\[ \mathbf{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = -7 \]

### Step 5: Write the final vector equation of the plane
The final vector equation of the plane can be expressed as:
\[ \mathbf{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = -7 \]

### Summary
The vector equation of the plane is:
\[ \mathbf{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = -7 \]