What is the closest point to a plane? I can't solve this question: Let $\mathcal{P}$ be the plane containing the points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$.
Find the point in this plane that is closest to $(0,3,-1)$.
I don't know how to do this question. Any hints/solutions? If it is vectors, then I am pretty sure I don't know how to do it. :(
 A: It is the orthogonal projection $H$  of the $M(0,3,-1)$ onto the plane defined by the points $A(-3,4,-2)$, $B(1,4,0)$ and $C(3,2,-1)$.  


*

*If the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are orthogonal, it is given by the formula
$$\overrightarrow{AH}=\frac{\overrightarrow{AM}\cdot\overrightarrow{AB}}{\overrightarrow{AB}\cdot\overrightarrow{AB}}\,\overrightarrow{AB}+\frac{\overrightarrow{AM}\cdot\overrightarrow{AC}}{\overrightarrow{AC}\cdot\overrightarrow{AC}}\,\overrightarrow{AC}.\tag{1}$$
To end determining the point $H$, just calculate 
$$\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{AH}.$$

*If the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are not orthogonal, you first have to deduce an orthogonal basis of the plane $\bigl\langle\overrightarrow{AB}, \overrightarrow{AC}\bigr\rangle$ by Gram-Schmidt process: if $$\overrightarrow{AC'}=\overrightarrow{AC}-\frac{\overrightarrow{AM}\cdot\overrightarrow{AB}}{\overrightarrow{AB}\cdot\overrightarrow{AB}}\,\overrightarrow{AB},$$
the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC'}$ are orthogonal, and you can apply  formula  $(1)$.
A: *

*Find two vectors lying on the plane: $v_1=(-3,4,-2)-(1,4,0)=(-4,0,-2)$ and $v_2=(1,4,0)-(3,2,-1)=(-2,2,1)$.  Rescale $v_1$ to $(2,0,1)$ for simplicity.

*Compute their cross product to obtain a normal vector $v_3$ for the plane.
$$v_3 = v_1\times v_2=\left|
\begin{matrix}
{\bf i} & {\bf j} & {\bf k} \\
2&0&1\\-2&2&1
\end{matrix}
\right|
=-2{\bf i}-4{\bf j}+4{\bf k}
$$

*Parametrize the straight line $L$ passing through $(0,3,-1)$ with the direction $v_3$ (i.e. normal to the plane) with $(0,3,-1)+tv_3 \,\forall t\in\Bbb R$.
$$(0,3,-1)+tv_3=(-2t,3-4t,-1+4t) \tag{$L$}\label{L}$$
Each point on $L$ corresponds to a value of $t$. 

*Find the intersection of $L$ with the plane.  Let $x\in L \cap{\cal P}$.  Since $x \in L$, express $x$ in terms of $t$: $x=(-2t,3-4t,-1+4t)$.  It's given that $(-3,4,-2)$ lies on the plane, so
\begin{align}
v=x-(-3,4,-2)&=(-2t,3-4t,-1+4t)-(-3,4,-2)\\
&=(3-2t,-1-4t,1+4t)
\end{align}
is parallel to the plane.

*Compute the dot product $v\cdot v_3 = 0$ in terms of $t$.  ($v$ is parallel to the plane, while $v_3$ is normal to the plane.)
\begin{align}
v\cdot v_3&=(3-2t,-1-4t,1+4t)\cdot(-2,-4,4)\\
&=-6+4t+4+16t+4+16t\\
&=36t+2 = 0 \\
\therefore t &=-\frac{1}{18} \\
x&=\left(-2\left(-\frac{1}{18}\right),3-4\left(-\frac{1}{18}\right),-1+4\left(-\frac{1}{18}\right)\right) \\
&= \left(\frac{1}{9},\frac{29}{9},-\frac{11}{9}\right)
\end{align}
So the required distance is
\begin{align}
{\rm dist}((0,3,-1),x) &= \sqrt{\left(\frac19\right)^2+\left(\frac{29}{9}-3\right)^2+\left(-\frac{11}{9}+1\right)^2} \\
&= \frac{\sqrt{1^2+2^2+(-2)^2}}{9} = \frac13
\end{align}

