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 May 19th, 2017, 01:30 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 230 Thanks: 2 Planes The locus of a point $P$ which is at the same distance from two planes $x+y+z=1$ , $z=0$ is a) an unbounded set. b) a sphere. c) a pair of parallel planes. d) a pair of intersecting planes. Im guessing this could be intersecting of planes! Someone correct me If I'm wrong Thanks from agentredlum
 May 20th, 2017, 02:18 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 I think you are correct Thanks from Country Boy
 May 20th, 2017, 04:05 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,109 Thanks: 855 The two given planes intersect along a straight line. Take any point along that straight line and construct the plane containing that point perpendicular to the two given planes. The two given planes intersect in this third plane in two straight lines intersecting at that point. That makes four angles, two pair of "vertical angles". Bisecting each pair gives the two planes that are "at the same distance from the two planes". Thanks from Lalitha183
 May 20th, 2017, 05:27 AM #4 Member   Joined: Jan 2016 From: Athens, OH Posts: 87 Thanks: 45 Analytically, this is easy. Let P=(x,y,z) be equidistant from the 2 planes. Then $${|x+y+z-1|\over \sqrt3}=|z|$$ So $$(x+y+z-1)^2=3z^2$$ Thus $$(x+y+z-1-\sqrt3z)(x+y+z-1+\sqrt3z)=0$$ Hence $$P\text{ is on } x+y+(1-\sqrt3)z-1=0\text{ or P is on } x+y+(1+\sqrt3)z-1=0$$ These two planes are not parallel since they have non-parallel normals; so they intersect. Conversely, if P is on either plane, the above steps are reversible and so P is equidistant from the original two planes. Thanks from Lalitha183

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