3D Geometry 3 Question 41

41. The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and xy+z=3 and at a distance 2/3 from the point (3,1,1) is (2012)

(a) 5x11y+z=17

(c) x+y+z=3

(b) 2x+y=321

(d) x2y=12

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Answer:

Correct Answer: 41. (c)

Solution:

  1. Key Idea

(i) Equation of plane through intersection of two planes,

 i.e. (a1x+b1y+c1z+d1)+λ(a2x+b2y+c2z+d2)=0

(ii) Distance of a point (x1,y1,z1) from

ax+by+cz+d=0=|ax1+by1+cz1+d|a2+b2+c2

Equation of plane passing through intersection of two planes x+2y+3z=2 and xy+z=3 is

(x+2y+3z2)+λ(xy+z3)=0

(1+λ)x+(2λ)y+(3+λ)z(2+3λ)=0 whose distance from (3,1,1) is 23.

|3(1+λ)+1(2λ)1(3+λ)(2+3λ)|(1+λ)2+(2λ)2+(3+λ)2=23

|2λ|3λ2+4λ+14=23

3λ2=3λ2+4λ+14

λ=72

172x+2+72y+372z2212=0

5x2+112y12z+172=0

or5x11y+z17=0



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