JEE Mains · Maths · STD 12 - 11. three dimension geometry
If \({L_1}\) is the line of interseciton of the planes \(2x - 2y + 3z - 2 = 0,\) \(x - y + z + 1 = 0\) and \({L_2}\) is the line of intersection of the planes \(x + 2y - z - 3 = 0,\) \(3x - y + 2z - 1 = 0\) , then the distance of the origin from the plane , containing the line \({L_1}\) and \({L_2}\) is:
- A \(\frac{1}{{3\sqrt 2 }}\)
- B \(\frac{1}{{2\sqrt 2 }}\;\;\;\;\)
- C \(\frac{1}{{\sqrt 2 }}\)
- D \(\frac{1}{{4\sqrt 2 }}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{{3\sqrt 2 }}\)
Step-by-step Solution
Detailed explanation
Equation of plane passing through the line of intersection of first two planes is: \((2 x-2 y+3 z-2)+\lambda(x-y+z+1)=0\) or \(x(\lambda+2)-y(2+\lambda)+z(\lambda+3)+(\lambda-2)=0 \ldots(1)\) is having infinite number of solution with \(x+2 y-z-3=0\) and \(3 x-y+2 z-1=0,\) then…
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