JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(P\) be the plane, which contains the line of intersection of the planes, \(x + y + z - 6 = 0\) and \(2x + 3y + z + 5 = 0\) and it is perpendicular to the \(xy -\) plane. Then the distance of the point \((0, 0, 256)\) from \(P\) is equal to
- A \(63\sqrt 5 \)
- B \(205\sqrt 5 \)
- C \(\frac{{17}}{{\sqrt 5 }}\)
- D \(\frac{{11}}{{\sqrt 5 }}\)
Answer & Solution
Correct Answer
(D) \(\frac{{11}}{{\sqrt 5 }}\)
Step-by-step Solution
Detailed explanation
Equation of plane \(\mathrm{P}_{1}+\lambda \mathrm{P}_{2}=0\) \((x+y+z-6)+\lambda(2 x+3 y+z+5)=0\) \(\Rightarrow x(1+2 \lambda)+y(1+3 \lambda)+2(1+\lambda)-6+5 \lambda=0\) This plane is \(\perp\) to \(x y-\) plane \(\therefore 1+\lambda=0\) So, equation of plane \(-x-2 y-11=0\)…
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