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I have tried searching for a similar question but couldn't find any which helps me with my problem. I've got a point P on (1,2,3), A plane with a normal towards point (1,1,0) and a point Q on the plane which goes through (1,0,3). I have to calculate the distance between the point and the plane, and even though I even tried drawing in a 3d renderer, I can't seem to solve it.

Could anyone explain how this is done?

Thanks in advance

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  • $\begingroup$ ".. a normal towards the point $(1,1,0)$ .." ?? how many planes do you have ?? what is $Q$ ?? $\endgroup$
    – G Cab
    Jul 25, 2017 at 13:01

2 Answers 2

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Suppose that the normal vector to the plane is $\color {green}{(1,1,0)} $.

the cartesian equation of the plane which contains the point $Q (1,0,3) $ is

$$\color {green}{1 }.(x-1)+\color {green}{1} .(y-0)+\color {green}{0} .(z-3)=0$$ $$=x+y\color{red}{-1}$$

the distance from the point $P (1,2,3) $ to the plane is $$d=\frac {|1+2\color {red}{-1}|}{\sqrt {\color {green}{1^2+1^2+0^2}}}=\sqrt {2} $$

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This question was taken right from my exam. However, I just solved the answer myself. For people also looking for this problem:

The idea here is to project the line PQ onto the normal N. The formula for projection is: " Projection of v onto W = |v dotproduct w| / || v ||. So in this particular question; Projecting the line PQ onto the normal which is given by;

| 1 * ( Px - Qx) + 1 * (Py - Qy ) + 0 (Pz - Qz) | / ( sqrt(1^2 + 1^2 + 0^0)) =

| 1 * 0 + 1 * 2 + 0 | / sqrt 2 = sqrt 2

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