3D vectors: lines and planes

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SHORTEST DISTANCE (POINT TO PLANE) 
\(=\left| \overrightarrow{AP}\cdot \mathbf{\hat{n}} \right|\) 
Hence, Shortest Distance to Origin \({\left(\overrightarrow{OP} = \mathbf{O}\right) \Rightarrow \left| \mathbf{a} \cdot \hat{\mathbf{d}} \right|}\)
FINDING THE COORDINATES OF FOOT OF PERPENDICULAR, \(\overrightarrow{ON}\)
1.	Find line equation \({PN:\ \overrightarrow{PN} = \mathbf{p} + \lambda \mathbf{n}}\) for some \({\lambda \in \mathbb{R}}\). 
2.	Solve \(\lambda \) by finding the intersection point of line \(PN\) and the plane. 
3.	Substitute \({\lambda}\) into \({\overrightarrow{PN} = \mathbf{p} + \lambda n\mathbf{P}}\) to find \({\overrightarrow{ON}}\). 
\(A\) is any point on the plane, it can be obtained by substituting values which satisfy the equation of plane. 
For example, \(\prod :\mathbf{r}\cdot \left( \begin{matrix}
 2 \\
 -1 \\
 3 \\
\end{matrix} \right)=4\Rightarrow \)\(A\) can be \(\left( \begin{matrix}
 2 \\
 0 \\
 0 \\
\end{matrix} \right)\), \(\left( \begin{matrix}
 0 \\
 -4 \\
 0 \\
\end{matrix} \right)\) or \(\left( \begin{matrix}
 1 \\
 -2 \\
 0 \\
\end{matrix} \right)\).

For \(\pi:(\mathbf r-\mathbf a)\cdot\mathbf n=0\), substitute \(\mathbf r=\mathbf p\) to test whether \(P\) lies on the plane.

\[\mathbf n_P=\mathbf p-\lambda\mathbf n,\qquad (\mathbf n_P-\mathbf a)\cdot\mathbf n=0,\qquad \operatorname{dist}(P,\pi)=\frac{|(\mathbf p-\mathbf a)\cdot\mathbf n|}{|\mathbf n|}.\]

Foot and distance

For \(ax+by+cz=d\), this becomes \(\displaystyle\frac{|ap_1+bp_2+cp_3-d|}{\sqrt{a^2+b^2+c^2}}\).

Point-Plane Relationships

Point-Plane Relationships

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