Minimizing distance from point to intersection of two planes

  • Thread starter icesalmon
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  • #1
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Homework Statement


identify the point on the line of intersection of the two planes that is nearest to the point (2,1,1) not on this line
p1: x + 2y - z - 1 = 0
p2: x + y + z - 3 = 0


Homework Equations




The Attempt at a Solution


I think I can find the line of intersection by crossing the two normal vectors from p1 and p2
so N3 = <1,2,-1> x <1,1,1> and then trying to minimize the distance from the point (2,1,1) to N3 by differentiating the distance formula using the x,y,z parameters.
 

Answers and Replies

  • #2
HallsofIvy
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I would NOT find the line of intersection that way. Instead, solve the two equations simultaneously: adding the two equations gives 2x+ 3y- 4= 0 so y= -(2/3)x- 4/3. Then we can write the second equation as x+ y+ z- 3= x- (2/3)x- 4/3+ z- 3= 0 so z= -(1/3)x+ 13/3. We could take "x" as the parameter or, to reduce the number of fractions, x= 3t, y= -2t- 4/3, z= -t+ 13/3. Now, you could minimize the distance by differentiating the distance formula but I think it would be simpler to use the geometric fact that the line giving the shortest distance to a line must be perpendicular to that line. And that means that the line giving the shortest distance form (2, 1, 1) to the line must line in the plane perpendicular to the line containing the point (2, 1, 1). A vector in the direction of this line is <3, -2, -1> and is perpendicular to that plane so that plane is 3(x- 2)- 2(y- 1)- (z- 1)= 0. Determine the point where the line x= 3t, y= -2t- 4/3, z= -t+ 13/3 crosses that plane and find the distance from that point to (2, 1, 1).
 
  • #3
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how did you get a negative 4/3 on the RHS after 2x + 3 y - 4 = 0 shouldn't it be y = 4/3 -(2x/3)
 
  • #4
HallsofIvy
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Yes, I wrote it wrong.
 

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