If I want to find the distance from a point to a plane.(adsbygoogle = window.adsbygoogle || []).push({});

E.g. (2,1,-1) to the plane x+y+z=1

I know that distance from one point to another is given by:

[itex]\sqrt{(x-2)^{2}+(y-1)^{2}+(z+1)^{2}}[/itex]

And in this case the solution is to substitute in z=x+y-1 which fits nicely giving us:

distance=[itex]\sqrt{(x-2)^{2}+(y-1)^{2}+(x+y)^{2}}[/itex]

What I'm trying to work out in my own mind is why substituting that in will suddenly give us the distance from the plane to the point.

I think that x,y are still arbitrary points in 3D space but when we substitute in the stuff for Z (nice/formal way of saying this anyone?) now x,y are allowed to be arbitrary points in 3D space since if Z is forced to be on the plane then x,y should fall on the plane.

However, if you have a tilted plane I can visualize a case where z is on the level of the plane but x,y are running off somewhere else...

and btw for those of you guys answering questions here. I can't tell you how much more enjoyable this forum has made maths for me (I was starting to dislike it a lot). I'm able to delve into all the little details I wasn't before. Thank you!

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# Finding the distance from a point to a plane using substitution

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