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JEE Mains · Maths · STD 12 - 11. three dimension geometry
The equation of a plane through the line of intersection of the planes \(x + 2y = 3,y-2z + 1= 0\), and perpendicular to the first plane is
- A \(2x -y- 10z = 9\)
- B \(2x -y +7z = 11\)
- C \(2x -y + 10z = 11\)
- D \(2x -y- 9z = 10\)
Answer & Solution
Correct Answer
(C) \(2x -y + 10z = 11\)
Step-by-step Solution
Detailed explanation
Equation of a plane through the line of intersection of the planes \(x+2 y=3, y-2 z+1=0\) is \((x+2 y-3)+\lambda(y-2 z+1)=0\) \(\Rightarrow x+(2+\lambda) y-2 \lambda(z)-3+\lambda=0\) ....\((i)\) Now, plane \((i)\) is \(\perp\) to \(x+2 y=3\) \(\therefore \quad\) Their dot…
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