If O,P are the points (0,0,0), (2,3,-1) respectively. Then what is the equation to the plane through P at right angles to OP?

Options

A. `2x+3y+z=16`
B. `2x+3y-z=14`
C. `2x+3y+z=14`
D. `2x+3y-z=0`

Answer

To find the equation of the plane through point P (2, 3, -1) that is perpendicular to the line segment OP (where O is the origin (0, 0, 0)), we can follow these steps:

### Step 1: Identify the points O and P
- O = (0, 0, 0)
- P = (2, 3, -1)

### Step 2: Determine the direction vector OP
The direction vector OP can be calculated by subtracting the coordinates of O from P:
\[ \text{OP} = P - O = (2 - 0, 3 - 0, -1 - 0) = (2, 3, -1) \]

### Step 3: Write the normal vector of the plane
Since the plane is perpendicular to the vector OP, the normal vector \( \mathbf{n} \) of the plane is the same as the vector OP:
\[ \mathbf{n} = (2, 3, -1) \]

### Step 4: Use the point-normal form of the plane equation
The equation of a plane in point-normal form is given by:
\[ \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} \]
Where:
- \( \mathbf{r} = (x, y, z) \)
- \( \mathbf{a} = (2, 3, -1) \) (the point P)
- \( \mathbf{n} = (2, 3, -1) \)

### Step 5: Substitute into the equation
Substituting the values into the equation:
\[ (x, y, z) \cdot (2, 3, -1) = (2, 3, -1) \cdot (2, 3, -1) \]
Calculating the right side:
\[ (2, 3, -1) \cdot (2, 3, -1) = 2 \cdot 2 + 3 \cdot 3 + (-1) \cdot (-1) = 4 + 9 + 1 = 14 \]
So, the equation becomes:
\[ 2x + 3y - z = 14 \]

### Final Equation
The equation of the plane through point P at right angles to OP is:
\[ 2x + 3y - z = 14 \]