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JEE Mains · Maths · STD 12 - 11. three dimension geometry

Let the equation of plane passing through the line of intersection of the planes \(x+2 y+a z=2\) and \(x-y+z=3\) be \(5 x-11 y+b z=6 a-1\). For \(c \in Z\), if the distance of this plane from the point \((a,-c, c)\) is \(\frac{2}{\sqrt{a}}\), then \(\frac{a+b}{c}\) is equal to

  1. A \(-2\)
  2. B \(2\)
  3. C \(-4\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-4\)

Step-by-step Solution

Detailed explanation

\((x+2 y+a z-2)+\lambda(x-y+z-3)=0\) \(\frac{1+\lambda}{5}=\frac{2-\lambda}{-11}=\frac{ a +\lambda}{ b }=\frac{2+3 \lambda}{6 a -1}\) \(\lambda=-\frac{7}{2}, a =3, b =1\) \(\frac{2}{\sqrt{a}}=\left|\frac{5 a +11 c + bc -6 a +1}{\sqrt{25+121+1}}\right|\) \(c =-1\)…
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