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Is it necessary that in 3 dimensions, two vectors will always be coplanar ? My teacher told the class that in 3 dimensions two vectors are always coplanar. But what if we consider two non parallel and non intersecting vectors ? like the ones parallel to 2 skew lines how can these be co planar ?

Another example is consider a cube. Now take one vector that is along the face diagonal of the upper face of the cube and another one that is along the face diagonal of the bottom face (the diagonal that is not in the same direction as the other one). These 2 are not coplanar as well.

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  • $\begingroup$ What are intersecting vectors? $\endgroup$ Aug 4, 2021 at 10:17
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    $\begingroup$ Vectors are not the same thing as segments. $\endgroup$ Aug 4, 2021 at 10:30
  • $\begingroup$ @JoséCarlosSantos Intersecting vectors from what I have been told by my high school teacher are vectors that you know intersect $\endgroup$ Aug 4, 2021 at 10:36
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    $\begingroup$ This is analogous to two points being collinear. Include the origin and now the two vectors from the origin are coplanar. $\endgroup$
    – Somos
    Aug 4, 2021 at 12:42

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You're confusing vectors with line segments and linear spaces with affine spaces.

When you take the example of the cube, you describe line segments. Those are indeed not coplanar. But those are also not vectors.

If you want to visualize (not a good mathematical word by the way...) coplanar vectors, you have to position the origin of those vectors at a common point. If the other vertices of those vectors and the common points are coplanar, then you can say that those vectors are coplanar.

When you apply this logic, you'll get the conclusion that any two vectors are always coplanar. Which by the way is not only true in dimension $3$ but for any dimension $n \ge 2$.

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  • $\begingroup$ since I am in Highschool, I don't know what affine spaces or linear spaces are. $\endgroup$ Aug 4, 2021 at 10:37
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    $\begingroup$ Then forget this part of the sentence and concentrate on vectors and line segments. $\endgroup$ Aug 4, 2021 at 10:38
  • $\begingroup$ Also why is it necessary for the vectors to have the same point as the end of their tail $\endgroup$ Aug 4, 2021 at 10:40
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    $\begingroup$ @KoustubhJain This is by definition of a vector. I imagine that your teacher provided you with a definition of a vector. Or at least I hope so. It would be interesting that you update your question with this definition. I also imagine that you were provided with a definition for two vectors to be coplanar. $\endgroup$ Aug 4, 2021 at 10:42
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    $\begingroup$ I don't know what a localized vector is... What's the definition of that? $\endgroup$ Aug 4, 2021 at 10:57
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I think you, your professor and the other answer mixed different definition of vectors used in physics and in mathematics - which is understandable since this forum is about mathematics.

In physics is often useful to use free vectors (or Euclidian vectors) which have a direction and magnitude, and localized vectors (or bounded vectors or affine vectors) which also have a point of application. This way, localized vectors are defined by two points while free vectors are defined by just one point. In fact, free vectors are like localized vectors with a common start. A summary of why physicists often use localized vectors may be found on https://physics.stackexchange.com/questions/139824/why-is-force-a-localized-vector-and-not-a-free-vector.

On the other hand, in mathematics, when talking about vectors and vector spaces, we nearly always mean free vectors.

Then, it's true that two free vectors are always coplanar (as the other answer says), but two localized vectors may not be coplanar (as the question says).

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  • $\begingroup$ This is why I pushed the OP in the comments to write down carefully the definitions. Especially when a term, i.e. vector is used improperly to cover different notions. Beeches and oaks are trees. But don't use the word tree if you want to deal specifically with an oak!!! $\endgroup$ Aug 4, 2021 at 19:35
  • $\begingroup$ In my experience, physics teachers (and even more in high school) aren't as rigorous about definitions as Linear Algebra 101 teachers would be. In fact, I don't remember being told a definition of vector in the physics context, neither in high school nor at Engineering school. I think I never needed such a definition, but in maths it's just the opposite. $\endgroup$
    – Pere
    Aug 4, 2021 at 20:14
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I had the exact same doubt but the following way of visualisation clears it up!

Take two pens with their lengths representing the magnitude and the tips representing the direction of the respective free vectors.

Then Orient them any way you think they should be non coplanar and since they are supposed to be free have them share a common tail.

In the end just imagine a plane (remember there can be 2 such planes but choose the one which is somewhat along the length of the vectors) which passes through the head of the vectors(which you can think about as points) and since we always have a plane passing through 2 points in a three dimensional plane you will see the vectors sharing a common plane...

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