MHT CET · Maths · Three Dimensional Geometry

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

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

Verified Solution

Answer & Solution

Correct Answer

(D) \(x-y-2 z+3=0\)

Step-by-step Solution

Detailed explanation

Since the plane bisects seg \(A B\), the plane meets the line \(A B\) at the mid point i.e. \(\left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right) \equiv(2,3,1)\)
Now line \(A B\) is \(\perp\) er to the plane
Direction ratios of plane are \(1-3,4-2,3+1\) i.e. \(-2,2,4\) i.e. \(-1,1,2\)
Equation of plane passing through \((2,3,1)\) and having d.r.s. \((-1,1,2)\) are
\(-(x-2)+(y-3)+2(z-1)=0 \Rightarrow-x+2\) \(+~y-3+2 z-2=0\)
\(\therefore x-y-2 z+3=0\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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

Step-by-step Solution

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