# Vectors: check if coordinates are in the same plane

1. Sep 7, 2016

### terhje

• Member warned that the homework template is required
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 im kinda lost.
Terhje

2. Sep 7, 2016

### Math_QED

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.

3. Sep 7, 2016

### terhje

Thanks for the help, Math_QED,

I think im 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.

4. Sep 7, 2016

### SammyS

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

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

5. Sep 7, 2016

6. Sep 7, 2016

### SammyS

Staff Emeritus
Last edited: Sep 7, 2016
7. Sep 8, 2016

### 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 cant be in a parallell plane cause A already is in the ABC plane. Meaning AB and AD cant both go to zero if D was in a parallel plane. Correct?

Appreciate the help.
Terje

Last edited: Sep 8, 2016
8. Sep 8, 2016

### SammyS

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

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

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

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 ?

9. Sep 8, 2016

### terhje

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
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

4*2+1*2-2*5 = 8+2-10 = 0.

A,B,C and D are all in the same plane

What 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

Last edited: Sep 8, 2016
10. Sep 8, 2016

### SammyS

Staff Emeritus
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.)

Last edited: Sep 8, 2016
11. Sep 8, 2016

### 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 havent had time to go through yet.
fixed some typos.

12. Sep 8, 2016

### SammyS

Staff Emeritus
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?

13. Sep 8, 2016

### SammyS

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

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.