Vectors: check if coordinates are in the same plane

[terhje]

Hello guys,

How can i check if coordinantes A,B,C and D are in the same plane? 3D space(x,y,z)

Can i take the cross product: AB x AC and check if its perpendicular to for example DC x DB. and then
check if the crossproducts are parallell? but i guess this can give me two parallell vectors in two different planes.

sorry to bother, but I am kinda lost.
Terhje

 

[member 587159]

Hello guys,

How can i check if coordinantes A,B,C and D are in the same plane? 3D space(x,y,z)

Can i take the cross product: AB x AC and check if its perpendicular to for example DC x DB. and then
check if the crossproducts are parallell? but i guess this can give me two parallell vectors in two different planes.

sorry to bother, but I am kinda lost.
Terhje

Well, the easiest approach is to consider the plane ABC (there is only one plane that goes through these three points) and then verify whether the point D satisfies the equation of the plane ABC.

 

[terhje]

Thanks for the help, Math_QED,

I think I am at a solution. I found the equation of the ABC plane by taking the cross product of AB and AC vectors. Then taking the dot product of D and the plane equation. and checking if i get the same number as with the other points in the ABC plane.

 

[SammyS]

Thanks for the help, Math_QED,

I think I am at a solution. I found the equation of the ABC plane by taking the cross product of AB and AC vectors. Then taking the dot product of D and the plane equation. and checking if i get the same number as with the other points in the ABC plane.

How can you take the dot product of D with any vector?

Is D a Vector? No. It is a point.

 

[terhje]

Sorry, math is not my strong suit, but I am trying to learn.
I saw them doing it here: https://www.math.washington.edu/~king/coursedir/m445w04/notes/vector/normals-planes.html#cross
under: Normal Vectors and Cross Product

Did i miss something?

 

[SammyS]

Sorry, math is not my strong suit, but I am trying to learn.
I saw them doing it here: https://www.math.washington.edu/~king/coursedir/m445w04/notes/vector/normals-planes.html#cross
under: Normal Vectors and Cross Product

Did i miss something?

After taking the cross product ##\ \vec{AB}\times\vec{AC} \,,\ ## what can you then do with ##\ \vec{AD} \, ?\ ##

 

[terhje]

I can check if AD is perpendicular to ABxAC, by checking if the dot product goes to zero. I was thinking it would leave room for D beeing in a parallel plane, but AD can't be in a parallell plane cause A already is in the ABC plane. Meaning AB and AD can't both go to zero if D was in a parallel plane. Correct?

Appreciate the help.
Terje

 

[SammyS]

I can check if AD is perpendicular to ABxAC, by checking if the dot product goes to zero. I was thinking it would leave room for D being in a parallel plane, but AD can't be in a parallel plane cause A already is in the ABC plane. Meaning AB and AD can't both go to zero if D was in a parallel plane. Correct?

Appreciate the help.
Terje

You have some run-on thoughts/questions. Let's separate them somewhat.

I can check if AD is perpendicular to ABxAC, by checking if the dot product goes to zero.

That works provided that the cross product is not zero. ( * More on this at the end of this post.)

I was thinking it would leave room for D being in a parallel plane, but AD can't be in a parallel plane cause A already is in the ABC plane.

I'm not sure what you mean by this. Problem states that all four points lie in the same plane.

Meaning AB and AD can't both go to zero if D was in a parallel plane. Correct?

It's fair to assume that A, B, C, and D are all distinct points.

* Back to the first item above:
What is implied if ##\ \vec{AB}\times\vec{AC} \ ## is zero ?

 

[terhje]

What is implied if →AB×→AC \ \vec{AB}\times\vec{AC} \ is zero ?

That AB or AC or both are zero.

Ill just show you what i got so far, which i think is the correct solution.
Question:
Check if the following points, A(-1,3,4), B(0,5,7), R(0,3,6) and S(1,5,9) are in the same plane.

My solution:
making vectors, AB, AC and AD
AB= i+2j+3k
AC=i+2k
AD=2i+2j+5k
defining the ABC plane normal:
ABxAC = matrix(2,3;0,2)i - matrix(1,3;1,2)j + matrix(1,2;1,0)k
= 4i+j-2k = E

Checking if E*AD= 0.
4*2+1*2-2*5 = 8+2-10 = 0.

A,B,C and D are all in the same planeWhat i ment was that if you have a new vector EF, and EF*E = 0, then EF is perpendicular to E, but it can can still be in a parallel plane.
but since the plane is defined by ABC and AD*E=0, then it must be in the same plane.

Thank you for the help.
Terhje

 

[SammyS]

That AB or AC or both are zero.

No !
If A, B, and C are distinct points, then neither AB or AC will be zero.

What can be concluded if the cross product of two non-zero vectors is zero ?

So, I ask again, what is implied if AB × AC = 0 ?

(I'll address the rest of your post when I get more time. There are inconsistencies and errors in it.)

 

[terhje]

lol. AB x AC = 0 implies that the vectors are parallel! my norwegian mathbook said nothing about it. Had to look in a calculus book i ordered from the states that i haven't had time to go through yet.
fixed some typos.

 

[SammyS]

lol. AB x AC = 0 implies that the vectors are parallel! my norwegian mathbook said nothing about it. Had to look in a calculus book i ordered from the states that i haven't had time to go through yet.
fixed some typos.

Yes. Specifically, the vectors point in the same direction or in opposite directions.

Since these are vectors construed with a point in common, the three points, A, B, and C must be co-linear.

In this case, what can you say about the four points?

 

[SammyS]

Ill just show you what i got so far, which i think is the correct solution.
Question:
Check if the following points, A(-1,3,4), B(0,5,7), R(0,3,6) and S(1,5,9) are in the same plane.

In the following, you use the coordinates of R and S as coordinates of points C and D respectively.

My solution:
making vectors, AB, AC and AD
AB= i+2j+3k
AC=i+2k
AD=2i+2j+5k
defining the ABC plane normal:
ABxAC = matrix(2,3;0,2)i - matrix(1,3;1,2)j + matrix(1,2;1,0)k
= 4i+j-2k = E

Those should actually determinants of those matrices:

AB×AC = det(matrix(2,3;0,2))i - det(matrix(1,3;1,2))j + det(matrix(1,2;1,0))k​

Then you use E as a vector. That's perfectly acceptable as is what follows, up to the following.

What i meant was that if you have a new vector EF, and EF*E = 0, then EF is perpendicular to E, but it can can still be in a parallel plane.
but since the plane is defined by ABC and AD*E=0, then it must be in the same plane.

You have a vector called EF, which presumably is constricted from two new points, E and F. However, you are previously using E as the normal vector. Don't now also use E to refer to a point.

Even if you fix this double use of E, it's not clear what you mean.