Question

The equation of the plane which bisects the line segment joining the points \( (3,2,6) \) and \( (5,4,8) \) and is perpendicular to the same line segment, is

• A. \( x+y+z=16 \)
• B. \( x+y+z=10 \)
• C. \( x+y+z=12 \)
• D. \( x+y+z=14 \)
• E. \( x+y+z=15 \)

Hint:

For perpendicular bisector plane in 3D: normal = direction vector and pass through midpoint.

Answer 1

The Correct Option is C

Concept:
• A plane perpendicular to a line has its normal vector parallel to the direction vector of the line.
• A plane bisecting a segment passes through the midpoint.
• Equation of plane: \( a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \)

Step 1: Direction vector of line
\[ \vec{d} = (5-3, 4-2, 8-6) = (2,2,2) \]

Step 2: Midpoint
\[ M = \left(\frac{3+5}{2}, \frac{2+4}{2}, \frac{6+8}{2}\right) = (4,3,7) \]

Step 3: Equation of plane
\[ 2(x-4)+2(y-3)+2(z-7)=0 \]

Step 4: Simplify
\[ 2x-8+2y-6+2z-14=0 \] \[ 2x+2y+2z-28=0 \] \[ x+y+z=14 \]

Step 5: Check options carefully
Re-evaluating midpoint sum: \[ 4+3+7=14 \] Thus correct plane: \[ \boxed{x+y+z=14} \]