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AP EAMCET · Maths · Three Dimensional Geometry

A plane \(\pi\) given by \(a x+b y+11 z+d=0\) is perpendicular to the planes \(2 x-3 y+z=4,3 x+y-z=5\) and the perpendicular distance from the origin to the plane \(\pi\) is \(\sqrt{6}\) units. If all the intercepts made by the plane \(\pi\) on the coordinate axes are positive, then \(d=\)

  1. A ab
  2. B \(-2 \mathrm{ab}\)
  3. C \(4 a b\)
  4. D \(-3 \mathrm{ab}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-3 \mathrm{ab}\)

Step-by-step Solution

Detailed explanation

\(2 a-3 b+11=0\) \(3 a+b-11=0 \implies b=11-3 a\) \(2 a-3(11-3 a)+11=0 \implies 2 a-33+9 a+11=0 \implies 11 a-22=0 \implies a=2\) \(b=11-3(2)=5\) \(\frac{|d|}{\sqrt{a^2+b^2+11^2}}=\sqrt{6} \implies \frac{|d|}{\sqrt{2^2+5^2+11^2}}=\sqrt{6}\)…