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 \(d=\)
- A ab
- B \(-2 \mathrm{ab}\)
- C \(4 a b\)
- D \(-3 \mathrm{ab}\)
Answer & Solution
Correct Answer
(D) \(-3 \mathrm{ab}\)
Step-by-step Solution
Detailed explanation
\(2 a-3 b+11=0\) \(3 a+b-11=0 \implies b=11-3 a\) \(2 a-3(11-3 a)+11=0 \implies 2 a-33+9 a+11=0 \implies 11 a-22=0 \implies a=2\) \(b=11-3(2)=5\) \(\frac{|d|}{\sqrt{a^2+b^2+11^2}}=\sqrt{6} \implies \frac{|d|}{\sqrt{2^2+5^2+11^2}}=\sqrt{6}\)…
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