The equation of the right bisector plane of the segment joining (2, 3, 4) and (6, 7, 8) is

Options

A. x + y + z + 15 = 0
B. x + y + z – 15 = 0
C. x – y + z – 15 = 0
D. None of these

Answer

To find the equation of the right bisector plane of the segment joining the points \( A(2, 3, 4) \) and \( B(6, 7, 8) \), we can follow these steps:

### Step 1: Find the Midpoint of the Segment
The midpoint \( M \) of the segment joining points \( A \) and \( B \) can be calculated using the midpoint formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
Substituting the coordinates of points \( A \) and \( B \):
\[ M = \left( \frac{2 + 6}{2}, \frac{3 + 7}{2}, \frac{4 + 8}{2} \right) = \left( \frac{8}{2}, \frac{10}{2}, \frac{12}{2} \right) = (4, 5, 6) \]

### Step 2: Find the Direction Vector of the Segment
The direction vector \( \vec{AB} \) from point \( A \) to point \( B \) is given by:
\[ \vec{AB} = B - A = (6 - 2, 7 - 3, 8 - 4) = (4, 4, 4) \]

### Step 3: Find the Normal Vector to the Plane
The normal vector \( \vec{n} \) to the right bisector plane is the same as the direction vector \( \vec{AB} \):
\[ \vec{n} = (4, 4, 4) \]

### Step 4: Write the Equation of the Plane
The equation of a plane can be expressed in the form:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane (which we found to be \( M(4, 5, 6) \)) and \( (n_1, n_2, n_3) \) are the components of the normal vector \( \vec{n} \).

Substituting the values:
\[ 4(x - 4) + 4(y - 5) + 4(z - 6) = 0 \]

### Step 5: Simplify the Equation
Distributing the terms:
\[ 4x - 16 + 4y - 20 + 4z - 24 = 0 \]
Combining like terms:
\[ 4x + 4y + 4z - 60 = 0 \]
Dividing the entire equation by 4:
\[ x + y + z - 15 = 0 \]

### Final Answer
Thus, the equation of the right bisector plane of the segment joining the points \( (2, 3, 4) \) and \( (6, 7, 8) \) is:
\[ x + y + z = 15 \]