Decide if two points are located on the same side of the plane

[Randyyy]

Decide if the two points, (1,2-1) and (4,1,2) are located on the same side of the plane [MATH]\pi : x+2y+z=3[/MATH].

If I plug in I get that (1,2-1) gives 4 and (4,1,2) gives 8. and since 4 > 3 and 8 > 3 they are both on the positive side of the plane.
Is this reasoning correct?

Thanks in advance!

 

[pka]

Decide if the two points, (1,2-1) and (4,1,2) are located on the same side of the plane [MATH]\pi : x+2y+z=3[/MATH].

Is there a point of the plane between \((1,2,-1)~\&~(4,1,2)~?\)

 

[Dr.Peterson]

Decide if the two points, (1,2-1) and (4,1,2) are located on the same side of the plane [MATH]\pi : x+2y+z=3[/MATH].

If I plug in I get that (1,2-1) gives 4 and (4,1,2) gives 8. and since 4 > 3 and 8 > 3 they are both on the positive side of the plane.
Is this reasoning correct?

Thanks in advance!

Yes, I think the reasoning and the answer are right. pka is suggesting another approach, for which there are several ways you might carry it out. My first thought had been using vector projections, until I saw what you were doing and realized it was a waste of effort!

 

[Randyyy]

Ah, I see, maybe then the alternate method could be from both your inputs:

(0,0,0) is a point on the plane between both of them. Maybe I should first realize that the normal is [MATH]\vec{N}=(1,2,1)[/MATH] and then if I suppose P: (1,2-1) and Q: (4,1,2), and O: (0,0,0) I can then check whether or not [MATH]\vec{PO}.\vec{N}>0[/MATH] or [MATH]<0[/MATH].

[MATH](P-O).\vec{N}=4[/MATH], [MATH](Q-O).\vec{N}=8[/MATH] and since they are both >0 they are on the positive side of the plane.

 

[Dr.Peterson]

Ah, I see, maybe then the alternate method could be from both your inputs:

(0,0,0) is a point on the plane between both of them. Maybe I should first realize that the normal is [MATH]\vec{N}=(1,2,1)[/MATH] and then if I suppose P: (1,2-1) and Q: (4,1,2), and O: (0,0,0) I can then check whether or not [MATH]\vec{PO}.\vec{N}>0[/MATH] or [MATH]<0[/MATH].

[MATH](P-O).\vec{N}=4[/MATH], [MATH](Q-O).\vec{N}=8[/MATH] and since they are both >0 they are on the positive side of the plane.

Huh? (0,0,0) doesn't satisfy x+2y+z=3, so it isn't on the plane. And it isn't on the line PQ, either.

So it's not that your dot products 4 and 8 are both positive. It's that both are greater than the dot product for any point on the plane, which is 3. And that is essentially the same thing you did before. Again, this has nothing to do with a point on the plane that lies between the given points.

If I wanted to find that point, I would write the parametric equation of the line through P and Q such that for t=0 we get P and for t=1 we get Q. Find the intersection of this line with the plane x+2y+z=3, and see whether 0<t<1. If it is, then it is a point on the plane between P and Q, and they must be on opposite sides. (Then you can ponder whether this method has any connection to your two methods.)

 

[Randyyy]

Huh? (0,0,0) doesn't satisfy x+2y+z=3, so it isn't on the plane. And it isn't on the line PQ, either.

So it's not that your dot products 4 and 8 are both positive. It's that both are greater than the dot product for any point on the plane, which is 3. And that is essentially the same thing you did before. Again, this has nothing to do with a point on the plane that lies between the given points.

If I wanted to find that point, I would write the parametric equation of the line through P and Q such that for t=0 we get P and for t=1 we get Q. Find the intersection of this line with the plane x+2y+z=3, and see whether 0<t<1. If it is, then it is a point on the plane between P and Q, and they must be on opposite sides. (Then you can ponder whether this method has any connection to your two methods.)

I just tried with 2 points on the plane, (1,0,2) and (0,0,3) and it is just like you said that their dot product is equal to 3, this I did not know!
So just to be clear, they are both on the same side of the plane because their dot product is larger than that of the plane for any given point on the plane? So had for example the dot product of [MATH] \vec{N}. \vec{P} < 3 [/MATH] and [MATH] \vec{N}. \vec{Q}>3 [/MATH] They would have been on the opposite sides of the plane.

I think I would need to do a little bit more research on the topic to grasp your idea fully, I´m still new to the ideas of linear algebra and so far I feel like a lot of things are magical because the concepts are so different from the math I have done prior. I get what you are saying, just not exaclty how to achieve it so I´ll do a bit of research and see if I can come back with a solution that follows you lead (just for practice sakes).

 

[Dr.Peterson]

I just tried with 2 points on the plane, (1,0,2) and (0,0,3) and it is just like you said that their dot product is equal to 3, this I did not know!
So just to be clear, they are both on the same side of the plane because their dot product is larger than that of the plane for any given point on the plane? So had for example the dot product of [MATH] \vec{N}. \vec{P} < 3 [/MATH] and [MATH] \vec{N}. \vec{Q}>3 [/MATH] They would have been on the opposite sides of the plane.

If you are learning about both vectors and equations, then I would think they would have introduced the equation of a plane by pointing out that it is really a dot product, [MATH]\vec{N}\cdot\vec{P} = k[/MATH]. So what you discovered (which is correct) is not just magic!

See here: https://tutorial.math.lamar.edu/classes/calciii/eqnsofplanes.aspx

 

[Randyyy]

If you are learning about both vectors and equations, then I would think they would have introduced the equation of a plane by pointing out that it is really a dot product, [MATH]\vec{N}\cdot\vec{P} = k[/MATH]. So what you discovered (which is correct) is not just magic!

See here: https://tutorial.math.lamar.edu/classes/calciii/eqnsofplanes.aspx

Very useful link, that actually clears up so much of my confusion. Thank you very much Dr.Peterson and thanks to you too pka for the help! I appreciate it.