TS EAMCET · Maths · Three Dimensional Geometry
The equation of the plane passing through the line of intersection of planes \(\pi_1=2 x+6 y+4 z-7=0\), \(\pi_2=x-y-2 z-2=03\) and perpendicular to the plane \(x+y+2 z-5=0\) is
- A \(3 x+y-2 z=0\)
- B \(6 x+2 y-4 z+55=0\)
- C \(6 x+2 y-4 z-15=0\)
- D \(3 x+y-2 z-15=0\)
Answer & Solution
Correct Answer
(C) \(6 x+2 y-4 z-15=0\)
Step-by-step Solution
Detailed explanation
Equation of plane passes through the point of intersection of planes \(\pi_1=2 x+6 y+4 z-7=0\) and \(\pi_2=x-y-2 z-2=0\) is \((2 x+6 y+4 z-7)+\lambda(x-y-2 z-2)=0\) \(\Rightarrow(2+\lambda) x+(6-\lambda) y+(4-2 \lambda) z-(7+2 \lambda)=0 \ldots(\mathrm{i})\) \(\because\) Plane…
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