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Question: Answered & Verified by Expert
Let P1:x+y+2z=3 and P2:x-2y+z=4 be two planes. Let A2,4,5 and B4,3,8 be two points in space. The equation of plane P3 through the line of intersection of P1 and P2 such that the length of the projection upon it of the line segment AB is the least, is
MathematicsThree Dimensional GeometryJEE Main
Options:
  • A 2x-y+3z=7
  • B 3y+z+1=0
  • C x+3y+z+2=0
  • D 3x-3y+4z-11=0
Solution:
2485 Upvotes Verified Answer
The correct answer is: 2x-y+3z=7
Equation of plane P3 is P1+λP2=0
x1+λ+y1-2λ+z2+λ=3+4λ
AB=<2,-1,3>
Length of projection of AB on plane is least
AB must be perpendicular to the plane
1+λ2=1-2λ-1=2+λ3λ=1
So, the equation of the plane is 2x-y+3z=7

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