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Let's say that we have a plane in 3 dimensions and that we know 2 vectors that belong in this plane. I want to find a unit vector of this plane. My teacher proceeds to taking the cross product of these 2 vectors and then normalising the resulting vector by dividing it with it's norm.

But..Isn't the cross product of the two vectors, a vector perpendicular to the plane that our 2 initial vectors belong? If a vector is perpendicular to a plane, doesn't that mean that the vector is not in the plane?

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    $\begingroup$ What does "unit vector of a plane" mean? Is it not the unit normal projecting perpendicularly outward from the plane, since it uniquely identifies the plane? Once you clarify what this phrase means, then we can debate as to whether the given procedure is correct or not. EDIT : Looking at the answer below, I am inclined to say this approach is correct. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jun 6, 2018 at 14:09
  • $\begingroup$ What I mean is vector of length 1 that is in my plane. But yeah the comment below is correct, I just don't know what my teacher wanted to do. $\endgroup$
    – Thomas
    Jun 6, 2018 at 14:27
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    $\begingroup$ Hi Thomas, I think your question was very nicely written, and you have shown interest throughout. For this reason, +1. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jun 7, 2018 at 2:53
  • $\begingroup$ Thank you ! Have a nice day $\endgroup$
    – Thomas
    Jun 7, 2018 at 9:07
  • $\begingroup$ Same to you as well, friend. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jun 7, 2018 at 11:02

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Yes you are right, the unit vector obtained by the cross product is orthogonal to the plane but maybe that was exactly what your professor was looking for (otherwise we could simply normalize one of the two given vectors).

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  • $\begingroup$ Yes I realized that as well. I'm not sure what he is even looking for, he didn't explain it properly. It's an engineering subject. But yeah if he wanted just a vector of the plane he could just normalize one of the existing ones. $\endgroup$
    – Thomas
    Jun 6, 2018 at 14:26
  • $\begingroup$ @Thomas Yes, I suppose it is a matter of definition for the normal vector to the plane. $\endgroup$
    – user
    Jun 6, 2018 at 14:28
  • $\begingroup$ Can I ask you something else. The unit vector of a plane is always normal to the plane? $\endgroup$
    – Thomas
    Jun 6, 2018 at 14:57
  • $\begingroup$ @Thomas I don't use that definition but I suppose that with the term "unit vector of a plane" your professor means one of the 2 normal unitary vectors to the plane. That is $n=\pm(a,b,c)$ with $\sqrt{a^2+b^2+c^2}=1$ for the plane $ax+by+cz+d=0$. $\endgroup$
    – user
    Jun 6, 2018 at 15:02
  • $\begingroup$ Great. Also, is $n$ always perpendicular to the plane? I am from Greece so I don't know exactly how to call it in English. Can you tell me if there is a proper definition for it? $\endgroup$
    – Thomas
    Jun 6, 2018 at 15:51

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