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I have a reference plane formed by $3$ points in $\mathbb{R}^3$$A, B$ and $C$. I have a $4$th point, $D$. I would like to project the vector $\vec{BD}$ onto the reference plane as well as project vector $\vec{BD}$ onto the plane orthogonal to the reference plane at vector $\vec{AB}$. Ultimately, I need the angle between $\vec{AB}$ and $\vec{BD}$ both when the vectors are projected on to the reference plane as well as the orthogonal plane. I have completed tutorials on projecting a vector onto a line in $\mathbb{R}^2$ but haven't figured out how to translate that to $\mathbb{R}^3$...

Please note the diagram only shows the reference plane as parallel to the $xy$ plane for the sake of convenience. In my examples, the reference plane could be at any orientation. I am using $3$D coordinate data from an electromagnetic motion tracker and the reference plane will be constantly moving. I understand the cross product of the two vectors $\vec{AB} \times \vec{BC}$ results in the normal vector to their plane. I have $2$ different methods to calculate that but am a little lost once I get to this point. I have seen both unit vector notation and column vector notation but am confused by moving between the different styles. It would be most helpful if you could tell me the formal name of the notation/equations you use. I know the scalar equation of a plane through point $(a,b,c)$ with normal $\hat{n} = [n_1, n_2, n_3]$ is:
$$ n_1(x-a) + n_2(y-b) +n_3(z-c) = 0 $$ and the standard linear equation definition is: $$ Ax + By + Cz = D $$ but I could use some tips on when the equation is $=D$ and when it is $=0$ as well as any additional equations for a plane and in which circumstances the different forms are appropriate. I hope I've made sense here. Thanks for any help you can provide. enter image description here

Some Guy
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MsHF
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3 Answers3

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If $A$,$B$,$C$ are not on the same line, then $\vec{AB}\times \vec{BC}$ will give you the direction of your reference plane normal vector $\hat{n}_1$. I think you should know how to do the normalization so that $|\hat{n}_1|=1$

Then the projection of $\vec{BD}$ on the reference plane is $\vec{BD}-(\vec{BD}\cdot \hat{n}_1)\hat{n}_1$

Another plane $\hat{n}_2$ orthogonal to the reference plane ABC can be found as $\vec{AB}\times\hat{n}_1$ (again you need to normalize it). Then the projection of $\vec{BD}$ on that plane can alsow be found in a similar way as shown for the first plane.

MoonKnight
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    Thank you @MoonKnight - I had the inlaws staying for a few days so I'm just getting back to this. I'm a little fuzzy on normalizing - are you referring to the unit vector with length =1? I'm not sure how my results are supposed to look for that. I believe I'm supposed to multiple the inverse of the length of the vector by the vector itself. That would give me another vector with 3 components, correct? Again, notation is tripping me up. Can you confirm that your formula reads vector BD minus [the dot product of (BD⋅n) multiplied by n]? Thank you. – MsHF Jan 13 '14 at 20:24
  • Another thing I'm hoping to clarify: Does your method first find the normal vector to the vectors formed by my sensor coordinates THEN find the actual reference plane created by the sensors? I'm confused by your saying in the 2nd paragraph "the projection of BD on the reference plane" when it seems you just had me find n1 as the normal vector, NOT the reference. To be clear, I am referring to the reference plane as the plane formed by points ABC and the plane orthogonal to that as the normal vector. But how do you get from a vector to a plane? Is it really just the same coefficients – MsHF Jan 13 '14 at 20:47
  • @MsHF 1) Yes the normalization means to make your vector has a length of $1$. For example $\hat{n}_A=\vec{A}/|\vec{A}|$ is the normalized vector in the direction of vector $A$. – MoonKnight Jan 13 '14 at 21:27
  • your understanding about $\left( \vec{BD}\cdot\hat{n}_1 \right)\hat{n}_1$ is correct.
  • – MoonKnight Jan 13 '14 at 21:28
  • No, the normal vector $\hat{n}1$ is the vector normal to the reference plane $ABC$. Basically $\left(\vec{BD}\cdot\hat{n}_1\right)\hat{n}_1$ is the $\vec{BD}\parallel$ mentioned in @Berci's post. And the projection on plane $ABC$ is the $\vec{BD}_\perp$ he mentioned.
  • – MoonKnight Jan 13 '14 at 21:33
  • I have calculated my cross product and found the normalized unit vector. Do I need to normalize the vectors created by my coordinates as well or just the normal vectors I find using the cross product? Also, I'm having trouble writing out the calculations for →BD–(→BD⋅n1)n1. Do I want to use column vector notation and subtract across the x, y and z values, respectively? As an aside, what app or program do forum members use to enter arithmetic notation? It would certainly make my questions clearer. thanks again. – MsHF Jan 14 '14 at 00:33
  • @MsHF 1) just the normal vector need to be normalized. 2) Yes, subtract across the x,y and z values respectively. 3) it is latex code we use to write arithmetic notation, see http://math.stackexchange.com/editing-help#latex – MoonKnight Jan 14 '14 at 00:36