Question:
The equation of the plane that perpendicularly bisects the line segment joining the points \(A(2,-1,3)\) and \(B(3,1,0)\) is:
The equation of the plane that perpendicularly bisects the line segment joining the points \(A(2,-1,3)\) and \(B(3,1,0)\) is:
Show Hint
The plane passes through the midpoint of AB and has AB as its normal vector.
Updated On: Nov 12, 2026
- \(x+2y-3z+2=0\)
- \(x+2y-3z-2=0\)
- \(x-3z+2=0\)
- \(x+2y-3z=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The plane that perpendicularly bisects AB passes through the midpoint of AB and has \(\overrightarrow{AB}\) as its normal.
Step 2: Find the normal and midpoint
\(\overrightarrow{AB}=(1,2,-3)\).
Midpoint \(M=\left(\frac{5}{2},0,\frac{3}{2}\right)\).
Step 3: Write the plane
\(1\left(x-\frac{5}{2}\right)+2(y-0)-3\left(z-\frac{3}{2}\right)=0\)
Step 4: Simplify
\(x+2y-3z-\frac{5}{2}+\frac{9}{2}=0\Rightarrow x+2y-3z+2=0\).
Final Answer:
The plane is x + 2y - 3z + 2 = 0. \[ \boxed{\text{(A) }x+2y-3z+2=0} \]
The plane that perpendicularly bisects AB passes through the midpoint of AB and has \(\overrightarrow{AB}\) as its normal.
Step 2: Find the normal and midpoint
\(\overrightarrow{AB}=(1,2,-3)\).
Midpoint \(M=\left(\frac{5}{2},0,\frac{3}{2}\right)\).
Step 3: Write the plane
\(1\left(x-\frac{5}{2}\right)+2(y-0)-3\left(z-\frac{3}{2}\right)=0\)
Step 4: Simplify
\(x+2y-3z-\frac{5}{2}+\frac{9}{2}=0\Rightarrow x+2y-3z+2=0\).
Final Answer:
The plane is x + 2y - 3z + 2 = 0. \[ \boxed{\text{(A) }x+2y-3z+2=0} \]
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