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NDA 2017 Maths · Mathematics · Three Dimensional Geometry · medium

The plane through (1,-1,2) perpendicular to the vector 3hat i-2hat k has the equation

Single correct option

The plane through (1,−1,2)(1,-1,2) perpendicular to the vector 3i^−2k^3\hat i-2\hat k has the equation
  1. A. 3x−2z−1=03x-2z-1=0
  2. B. 3x+2z=73x+2z=7
  3. C. 3x−2y−2z=−13x-2y-2z=-1
  4. D. 3x−2z+1=03x-2z+1=0
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Answer: (D) 3x−2z+1=03x-2z+1=0

The normal is (3,0,−2)(3,0,-2), so the plane is 3x−2z=d3x-2z=d. Substituting (1,−1,2)(1,-1,2) gives d=3−4=−1d=3-4=-1, i.e. 3x−2z+1=03x-2z+1=0. The wrong options show a sign slip in dd, a sign change in the normal, and an extra yy-term that the normal does not allow.
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Original question written in the NDA 2017 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake