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Let's say we have a xyz system in 3D and there lies a random plane with its normal vector. We can find the unit vector in the direction of the normal vector by dividing with its magnitude. Could we express this unit vector in i (unit vector on x axis), j(unit vector on y axis) and k(unit vector on z axis) components? If yes, what will the coefficients be in each axis? If the unit vector in the direction of the normal vector has magnitude 1, then the coefficients must be less than 1 (pythagorean theorem), right? I'm talking about vector projections on the x, y and z axis of a unit vector that it is in the direction of the normal vector of a surface. Can we do that, can we split a unit vector even further and find its coefficients on the i, j and k main unit vectors? Why am I thinking about it? Let's say we have a tetrahedron with its 3 sides on the xy, yz, xz plane (they all meet at (0,0,0)). The 3 sides are the projection of the fourth one which has a normal vector n and makes an angle θ with the bottom plane. My teacher said that : dot product of n and k unit vectors equals (cosθ)=(the projection of n on the z axis). And then I wondered how can we measure the projection of the unit vector on the z axis. But, I guess that we are talking about the NORMALIZED vector, so the k coefficient is indeed the projection on the z axis.

Sok
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    Does this answer your question? Uniform distribution on the surface of unit sphere If not., look further for uniform distribution on sphere. – Ethan Bolker Feb 12 '23 at 14:35
  • And what exactly should I try to look for? – Sok Feb 12 '23 at 14:55
  • If the plane is given as ax+by+cz+d=0 it is simple to find two linearly independent vectors that lie in the plane. Then by taking their cross product you can find the components of a normal vector to the plane, hence obviously also a unit vector. – Amit Feb 12 '23 at 14:56
  • Oh yes, I know that. But my question was how we can express the unit vector with the main unit vectors of the coordinates system. – Sok Feb 12 '23 at 14:58
  • I am assuming here x,y,z are the axes of the coordinate system. Hence all calculations I mentioned will naturally be expressed by their unit vectors. – Amit Feb 12 '23 at 14:58
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    So if for example you get (1,2,3) as a result those are the $\hat{i}, \hat{j} , \hat{k}$ components in that order... – Amit Feb 12 '23 at 15:00
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    I think there's a basic misunderstanding here. Can you explicitly write out the numbers and ordered triples of a particular case? I think the things you're asking for are already there, you just need them pointed out to you. – JonathanZ Feb 12 '23 at 15:01
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    Please [edit] the question to clarify what you want - dDon't reply in the comments..See @JonathanZsupportsMonicaC 's comment. – Ethan Bolker Feb 12 '23 at 15:03
  • (@Amit, they seem to want a normalized normal vector. Maybe normalize your example? Also, what an unfortunate fact that "normalized" and "normal" have totally different meanings but sound so similar!) – JonathanZ Feb 12 '23 at 15:04
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    @JonathanZsupportsMonicaC I know, it was just an example to clarify that the vectors are already wrt unit vectors of the axes he wants. Anyway I can't edit, the ship has sailed :) – Amit Feb 12 '23 at 15:06
  • @Amit, Yes, I see, but an already confused person can get hung up on the most trivial side issues. Anyways, I think we're going to need more/better responses from the OP. I'm going to try to restate what you said, and then I'm off. – JonathanZ Feb 12 '23 at 15:08
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    Sok - @amit's point is that for any vector, $(a,b,c)$ is just a shorthand (or you could say "equivalent to") $a\hat i + b\hat j +c \hat k$. The "coefficients in each axis" are exactly what make up the ordered triple. – JonathanZ Feb 12 '23 at 15:10
  • @JonathanZsupportsMonicaC Roger, thanks. – Amit Feb 12 '23 at 15:14
  • @Sok Please be explicit: from exactly what representation of the vector do you want to convert to the xyz representation? It's not clear without the concrete details – Amit Feb 12 '23 at 15:24

1 Answers1

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The plane $2x+3y+6z =24$ has normal vector $(2,3,6)$. Normalizing it to a unit vector yields

$$\hat n =\frac {(2,3,6)}{\|(2,3,6)\|} =(2/7,3/7,6/7)$$

$2/7$ is the coefficient of $\hat n$ along the $\hat i$ main unit vector.

JonathanZ
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  • I posted the whole case, but I think I can see it clearly now. I think I did not understand the difference between normalized vector and normal vector. Thank you all! – Sok Feb 12 '23 at 17:54
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    @Sok - Note that in the above answer, both the components of the normal vector and the normalized vector are with respect to the $\hat{i} , \hat{j} , \hat{k}$ basis. Just to be sure you got that at no part of the calculation did he use another basis. – Amit Feb 12 '23 at 17:56