Formula to project a vector onto a plane

Solution 1:

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.

Solution 2:

1. Obtain the equation of the reference plane by ${\bf n}:=\vec{AB}\times\vec{AC}$, the left hand side of equation will be the scalar product ${\bf n}\cdot{\bf v}$ where $\bf v$ is the (vector from origin to the) variable point of the equation, and the right hand side is a constant, such that e.g. ${\bf v}=A$ has to satisfy it, that is, the equation will be

${\bf n}\cdot{\bf v}={\bf n}\cdot A\,$. 2. Calculate orthogonal and parallel components of $\vec{BD}$ relative to $\bf n$: we look for vectors $\vec{BD}_\perp$ and $\vec{BD}_\| $ such that $\vec{BD}=\vec{BD}_\perp+\vec{BD}_\| $, $\ \vec{BD}_\perp\ \perp\ {\bf n}$ and $\ \vec{BD}_\| \,\parallel\,{\bf n}$.
That means, $\vec{BD}_\|=\lambda{\bf n}$ for some $\lambda\in\Bbb R$ and $\vec{BD}_\perp\cdot{\bf n}=0$. Using this and scalar multiplying by $\bf n$: $$\vec{BD}\cdot{\bf n}=0+\lambda\cdot{\bf n}^2$$ this enables you to compute $\lambda$, hence $\vec{BD}_\|$ then $\vec{BD}_\perp:=\vec{BD}-\vec{BD}_\|$. 3. Do the same for the other reference plane, (one of) its normal vector will be ${\bf n}\times\vec{AB}$. 4. The angle $\varphi$ of vectors ${\bf u}$ and ${\bf v}$ can be calculated by the following formula: $$\cos\varphi=\frac{{\bf u}\cdot{\bf v}}{\Vert{\bf u}\Vert\,\Vert{\bf v}\Vert} \,.$$

Solution 3:

Let the displacement of $\mathbf{\vec{{v}_{0}{v}_{1}}}$ be a directed vector from $\mathbf{\vec{v}_{0}}$ to $\mathbf{\vec{v}_{1}}$ defined as $\left(\mathbf{\vec{v}_{1}}-\mathbf{\vec{v}_{0}}\right)$.

$$\mathtt{\operatorname{Let }}\mathbf{V=\left\{\vec{a},\vec{b}\right\}};$$

"Let $\operatorname{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)$ be the orthogonalization of $\mathbf{\vec{a}}$ compared to $\mathbf{\vec{b}}$ about $\mathbf{\vec{0}}$."

$ \mathbf{\forall\vec{a}\forall\vec{b}}\left(\text{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)=\mathbf{\vec{a}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\implies\mathbf{\forall\vec{v}}\left(\pm\left(\mathbf{\vec{a}}\times\mathbf{\vec{b}}\right)\cdot\mathbf{\vec{v}}=\mathrm{0}\land\mp\left(\mathbf{\vec{b}}\times\mathbf{\vec{a}}\right)\cdot\mathbf{\vec{v}}=\mathrm{0}\land\mathbf{\vec{v}}\in\mathbf{V}\implies\text{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\cdot\mathbf{\vec{v}}=0\land\left(\mathbf{\vec{v}}\in\mathbb{R}^3\implies\text{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\in\mathbb{R}^3\right)\right)\right) $

"Let $\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)$ be the (linear) projection of $\mathbf{\vec{a}}$ onto $\mathbf{\vec{b}}$ about $\mathbf{\vec{0}}$."

$ \mathbf{\forall\vec{a}\forall\vec{b}}\left(\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)=\frac{\mathbf{\vec{a}}\cdot\mathbf{\vec{b}}}{\mathbf{\vec{b}}\cdot\mathbf{\vec{b}}}\mathbf{\vec{b}}\land\mathbf{\vec{b}}\not=\mathbf{\vec{0}}\land\frac{\pi}{2}= \arccos⁡\left(\frac{\left(\mathbf{\vec{a}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\right)\cdot\left(\mathbf{\vec{b}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\right)}{\Vert\mathbf{\vec{a}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\Vert \Vert\mathbf{\vec{b}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\Vert}\right)\land\left(\mathbf{\vec{b}}\in\mathbb{R}^n\land\mathbf{\vec{a}}\in\mathbb{R}^n\land\mathbf{\vec{0}}\in\mathbb{R}^n\implies\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\in\mathbb{R}^n\right)\right) $

The formula to project a vector onto a plane.

"Let $\vec{F}\left(\mathbf{\vec{v}},i,\mathbf{\vec{X}}_{i,j}\right)$ be the orthogonalization of $\mathbf{\vec{v}}$ compared to $\mathbf{\vec{X}}_{i,1}$ about $\mathbf{\vec{X}}_{i,2}$."

$ \therefore\forall \mathbf{\vec{v}}\exists\mathbf{\vec{X}}\left(\mathbf{\vec{X}}\in\left\{\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)},\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}\right\}\implies\vec{F}\left(\mathbf{\vec{v}},i,\mathbf{\vec{X}}_{i,j}\right)=\mathbf{\vec{v}}-\text{Proj}_{\left(\mathbf{\vec{X}}_{i,1}-\mathbf{\vec{X}}_{i,2}\right)}\left(\mathbf{\vec{v}}-\mathbf{\vec{X}}_{i,2}\right)\right) $

the traced codomain of $\vec{F}$ of a traced random $\mathbf{\vec{v}}$ about one instance of $\mathbf{\vec{X}}$.

"A reference plane formed by 3 points in $\mathbb{R}^3$."

Let plane $\mathbf{P}$ be a binary tuple of a normal constructor $\vec{\text{C}}$ using planer point(s) $\mathbf{\vec{p}}\in\mathbf{V}$:

$\mathbf{P}= \left(\vec{\text{C}}\left(\mathbf{\vec{p}}\right),\mathbf{\vec{p}}\right)\iff\left(\mathbf{\vec{n}}=\vec{\text{C}}\left(\mathbf{\vec{p}}\right)\implies \mathbf{P}\overset{\text{def}}{=}\forall\mathbf{\vec{v}}\left(\left(\mathbf{\vec{n}}-\mathbf{\vec{0}}\right)\cdot\left(\mathbf{\vec{v}}-\mathbf{\vec{p}}\right)=0\right)\right);$

$\mathtt{\text{Let }}\mathbf{V}=\left\{\vec{A},\vec{B},\vec{C}\right\};$

Let $\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}$ is an array data structure that contains all possible implementations of planes using all vectors in $\mathbf{V}$ given $\mathbf{P}$. $$ \mathtt{\text{Let }}\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}= \begin{bmatrix} \left(\frac{(\vec{A}-\vec{C}) \times (\vec{B}-\vec{C})}{\Vert(\vec{A}-\vec{C}) \times (\vec{B}-\vec{C})\Vert}+\vec{C}, \vec{C}\right) \\ \left(\frac{(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})}{\Vert(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\ \left(\frac{(\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})}{\Vert(\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})\Vert}+\vec{A}, \vec{A}\right) \\ \left(\frac{(\vec{B}-\vec{C}) \times (\vec{A}-\vec{C})}{\Vert(\vec{B}-\vec{C}) \times (\vec{A}-\vec{C})\Vert}+\vec{C}, \vec{C}\right) \\ \left(\frac{(\vec{C}-\vec{B}) \times (\vec{A}-\vec{B})}{\Vert(\vec{C}-\vec{B}) \times (\vec{A}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\ \left(\frac{(\vec{C}-\vec{A}) \times (\vec{B}-\vec{A})}{\Vert(\vec{C}-\vec{A}) \times (\vec{B}-\vec{A})\Vert}+\vec{A}, \vec{A}\right) \end{bmatrix}; $$

Example: $$\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2}=\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,j}=\left(\frac{(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})}{\Vert(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})\Vert}+\vec{B}, \vec{B}\right);$$ $$\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,1}=\frac{(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})}{\Vert(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})\Vert}+\vec{B};$$

$$\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,2}=\vec{B};$$

"The plane orthogonal to the reference plane at $\vec{AB}$."

Let $\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}$ inherit the definition of $\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}$, but redefines the normal constructor $\vec{\text{C}}\left(\mathbf{\vec{p}}\right)$ implemented in $\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}$. $$ \mathtt{\text{Let }}\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)} = \begin{bmatrix} \left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,1}-\vec{B}\right)\times(\vec{A}-\vec{B})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,1}-\vec{B}\right)\times(\vec{A}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\ \left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{3,1}-\vec{A}\right)\times(\vec{B}-\vec{A})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{3,1}-\vec{A}\right)\times(\vec{B}-\vec{A})\Vert}+\vec{A}, \vec{A}\right) \\ \left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{5,1}-\vec{B}\right)\times(\vec{C}-\vec{B})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{5,1}-\vec{B}\right)\times(\vec{C}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\ \left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{6,1}-\vec{A}\right)\times(\vec{C}-\vec{A})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{6,1}-\vec{A}\right)\times(\vec{C}-\vec{A})\Vert}+\vec{A}, \vec{A}\right) \end{bmatrix}; $$

Project $\vec{BD}$ onto the reference plane for any $i \in [1,6]$ at $R\left(i\right)$:

$$ \mathtt{\text{Let }}R\left(i\right)=\begin{Bmatrix} \vec{D}_r=\vec{F}\left(\vec{D},i,\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\ \vec{B}_r=\vec{F}\left(\vec{B},i,\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\ \vec{A}_r=\vec{F}\left(\vec{A},i,\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\ \vec{BD}_r=\vec{D}_r-\vec{B}_r, \\ \vec{AB}_r=\vec{B}_r-\vec{A}_r \end{Bmatrix}; $$

Project $\vec{BD}$ onto the orthogonal plane for any $i \in [1,4]$ at $O\left(i\right)$:

$$ \mathtt{\text{Let }}O\left(i\right)=\begin{Bmatrix} \vec{D}_{o}=\vec{F}\left(\vec{D},i,\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\ \vec{B}_{o}=\vec{F}\left(\vec{B},i,\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\ \vec{A}_{o}=\vec{F}\left(\vec{A},i,\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\ \vec{BD}_{o}=\vec{D}_{o}-\vec{B}_{o}, \\ \vec{AB}_{o}=\vec{B}_{o}-\vec{A}_{o} \end{Bmatrix}; $$

The angle between the projection of $\vec{AB}$ and $\vec{BD}$.

while projected onto the orthogonal plane: $\theta_{A_{o},D_{o}}=\arccos⁡\left(\frac{\vec{AB}_{o}\cdot\vec{BD}_{o}}{\Vert\vec{AB}_{o}\Vert\Vert\vec{BD}_{o}\Vert}\right)\implies\theta_{a_{o},d_{o}}= \arccos⁡\left(\frac{\left(\vec{B}_{o}−\vec{A}_{o}\right)\cdot\left(\vec{D}_{o}−\vec{B}_{o}\right)}{\Vert\vec{B}_{o}-\vec{A}_{o}\Vert\Vert\vec{D}_{o}−\vec{B}_{o}\Vert}\right);$

while projected onto the reference plane: $ \theta_{A_{r},D_{r}}=\arccos⁡\left(\frac{\vec{AB}_{r}\cdot\vec{BD}_{r}}{\Vert\vec{AB}_{r}\Vert\Vert\vec{BD}_{r}\Vert}\right)\implies\theta_{a_{r},d_{r}}= \arccos⁡\left(\frac{\left(\vec{B}_{r}−\vec{A}_{r}\right)\cdot\left(\vec{D}_{r}−\vec{B}_{r}\right)}{\Vert\vec{B}_{r}-\vec{A}_{r}\Vert\Vert\vec{D}_{r}−\vec{B}_{r}\Vert}\right); $