Question

If \(A(3,2,-1)\) and \(B(1,4,3)\), then the equation of the plane which bisects the segment \(AB\) perpendicularly is

• A. \(x + y + 2z + 3 = 0\)
• B. \(x - y + 2z - 3 = 0\)
• C. \(x + y - 2z - 3 = 0\)
• D. \(x - y - 2z + 3 = 0\)

Hint:

For a perpendicular bisector plane, always use the midpoint of the segment and take the direction ratios of the segment as the normal vector of the plane.

Answer 1

The Correct Option is D

Step 1: Find the midpoint of \(AB\).
\[ M = \left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right) = (2,3,1) \] Step 2: Find the direction ratios of \(AB\).
\[ \vec{AB} = (1-3,\;4-2,\;3+1) = (-2,2,4) \] Step 3: Use perpendicular bisector condition.
The required plane is perpendicular to \(AB\), hence its normal vector is \((-2,2,4)\).
Equation of plane: \[ -2(x-2) + 2(y-3) + 4(z-1) = 0 \] Step 4: Simplify the equation.
\[ -2x + 4 + 2y - 6 + 4z - 4 = 0 \] \[ -2x + 2y + 4z - 6 = 0 \] Dividing by \(-2\): \[ x - y - 2z + 3 = 0 \]