Question

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

• A. \(x+y+z+15=0\)
• B. \(x+y+z-15=0\)
• C. \(x-y+z-15=0\)
• D. None of these

Hint:

Right bisector plane passes through midpoint and is perpendicular to joining line.

Answer 1

The Correct Option is B

Step 1: Midpoint of the given points: \[ \left(\frac{2+6}{2},\frac{3+7}{2},\frac{4+8}{2}\right)=(4,5,6) \]
Step 2: Direction vector of the segment is \((4,4,4)\), hence normal to plane.
Step 3: Equation: \[ 4(x-4)+4(y-5)+4(z-6)=0 \Rightarrow x+y+z-15=0 \]