Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian



Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.

IIT JEE SOLVED PAPERS ON CO-ORDINATE GEOMETRY(3D)

Here is some solved problems from IIT-JEE Previous Year Papers.


Q. [IIT-JEE 2006] A plane passes through (1,-2,1) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4, then the distance of the plane from the point (1,2,2) is
(a) 0     (b) 1     (c) √2      (d) 2√2
Solution :(d) Let equation of the plane through (1,-2,1) be            a(x-1) + b(y+2) + c(z-1)=0 ----------------(1)
Since plane (1) is perpendicular to 2x – 2y + z =0 & x – y +2z =4.
2a – 2b + c =0&a – b+2c=0
Solving we have a=k, b=k, c=0.
Substituting in (1), we have x + y + 1=0-----------(2)
Let p be perpendicular distance of the plane (2) from the point (1,2,2). Then   p = (1+2+3)/√(12+12)=2√2.
Q. [IIT-JEE 2003] The value of k (x-4)/1 = (y-2)/1 = (z-k)/2 lies in the plane 2x - 4y + z = 7, is
(a) 7      (b) -7      (c) no real value      (d) 4
Solution : (a) The given line passes through the point P(4,2,k) & lies in plane  2x - 4y + z = 7 . So, P must lie on the plane.
  8 – 8 + k = 7     k=7.