`p (2,3,-4) ,vecb =2 hati - hatj + 2 hatk`
Cartesian equation of a plane passing through the point with position vector b and perpendicular to the vector `vec(OPQ)` being the origin is :

Options

A. `2x -y + 2z + 7=0`
B. `2x -y + 2z -7 =0`
C. `2x + 3y-4 z + 7=0`
D. `2x + 3y - 4z -7=0`

Answer

To find the Cartesian equation of the plane passing through the point with position vector \( \vec{b} = 2\hat{i} - \hat{j} + 2\hat{k} \) and perpendicular to the vector \( \vec{OPQ} \) (where \( O \) is the origin), we can follow these steps:

### Step 1: Identify the Point and Normal Vector
The point \( P \) has coordinates (2, 3, -4), and its position vector is given as:
\[ \vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k} \]
The normal vector \( \vec{n} \) of the plane is given as:
\[ \vec{n} = 2\hat{i} - \hat{j} + 2\hat{k} \]

### Step 2: Write the Equation of the Plane
The general equation of a plane in vector form is given by:
\[ (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \]
where \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) is the position vector of any point on the plane.

### Step 3: Substitute the Values
Substituting \( \vec{r} \) and \( \vec{a} \) into the equation:
\[ (x\hat{i} + y\hat{j} + z\hat{k} - (2\hat{i} + 3\hat{j} - 4\hat{k})) \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 0 \]
This simplifies to:
\[ ((x - 2)\hat{i} + (y - 3)\hat{j} + (z + 4)\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 0 \]

### Step 4: Calculate the Dot Product
Now, we calculate the dot product:
\[ (x - 2) \cdot 2 + (y - 3) \cdot (-1) + (z + 4) \cdot 2 = 0 \]
This expands to:
\[ 2(x - 2) - (y - 3) + 2(z + 4) = 0 \]
Simplifying this gives:
\[ 2x - 4 - y + 3 + 2z + 8 = 0 \]
Combining like terms results in:
\[ 2x - y + 2z + 7 = 0 \]

### Step 5: Rearranging the Equation
Rearranging the equation, we can write it as:
\[ 2x - y + 2z + 7 = 0 \]

### Final Answer
Thus, the Cartesian equation of the plane is:
\[ 2x - y + 2z + 7 = 0 \]

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