Plane through three points and distance

Equation of plane

Two vectors in the plane:

$$\mathbf{AB}=(-1,2,0),\qquad \mathbf{AC}=(-1,0,3)$$

Normal:

$$\mathbf n=\mathbf{AB}\times\mathbf{AC}= \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-1&2&0\\-1&0&3\end{vmatrix} =(6-0)\mathbf i-(-3-0)\mathbf j+(0+2)\mathbf k=(6,\ 3,\ 2)$$

Plane: $6x+3y+2z=d$. Using A: $d=6$.

$$6x+3y+2z=6\quad\Longleftrightarrow\quad \frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1$$

Check B: $6(0)+3(2)+0=6$. Check C: $0+0+2(3)=6$.

Unit normal

$|\mathbf n|=\sqrt{36+9+4}=7$, so $\hat{\mathbf n}=(0.857,\ 0.429,\ 0.286)$.

Parametric form

$$\mathbf P(u,w)=\mathbf A+u\,\mathbf{AB}+w\,\mathbf{AC}=(1-u-w,\ 2u,\ 3w)$$

where $u,w$ in $[0,1]$ give the part of the plane bounded by the parallelogram on $\mathbf{AB}$ and $\mathbf{AC}$.

Distance of Q(2, 2, 2)

$$D=\frac{|6(2)+3(2)+2(2)-6|}{7}=\frac{16}{7}=2.286$$

Answer: plane 6x + 3y + 2z = 6; unit normal (0.857, 0.429, 0.286); distance of Q = 2.286 units.