`p (2,3,-4) ,vecb =2 hati - hatj + 2 hatk`
Vector equation of a plane passing through the point P perpendicular to the vector `vecb`

Options

A. `vecr, (2 hati - hatj + 2 hatk) =-7`
B. `vecr (2 hati - hatj+ 2 hatk)=7`
C. `vecr (2 hati + 3 hatj + 4 hatk)=-7`
D. `vecr ( 2hati + 3 hatj + 4 hatk)=7`

Answer

To find the vector equation of a plane passing through the point \( P(2, 3, -4) \) and perpendicular to the vector \( \vec{b} = 2\hat{i} - \hat{j} + 2\hat{k} \), we can follow these steps:

### Step 1: Identify the point and normal vector
The point \( P \) through which the plane passes is given as:
\[ P(2, 3, -4) \]
The normal vector \( \vec{n} \) to the plane is given as:
\[ \vec{b} = 2\hat{i} - \hat{j} + 2\hat{k} \]

### Step 2: Write the general equation of the plane
The vector equation of a plane can be expressed as:
\[ \vec{r} - \vec{a} \cdot \vec{n} = 0 \]
where \( \vec{r} \) is the position vector of any point on the plane, \( \vec{a} \) is the position vector of the given point \( P \), and \( \vec{n} \) is the normal vector.

### Step 3: Write the position vector of point \( P \)
The position vector \( \vec{a} \) corresponding to point \( P(2, 3, -4) \) is:
\[ \vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k} \]

### Step 4: Substitute into the plane equation
Now, substituting \( \vec{a} \) and \( \vec{n} \) into the plane equation:
\[ \vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = (2\hat{i} + 3\hat{j} - 4\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k}) \]

### Step 5: Calculate the right-hand side
Now, calculate the dot product on the right-hand side:
\[ (2\hat{i} + 3\hat{j} - 4\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 2 \cdot 2 + 3 \cdot (-1) + (-4) \cdot 2 \]
Calculating this gives:
\[ = 4 - 3 - 8 = -7 \]

### Step 6: Write the final equation of the plane
Thus, the equation of the plane becomes:
\[ \vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = -7 \]

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