Distance from a point to a plane (in the parametric form not general form)

Question

I have a point $Q = [7 2 1]$ and I need to find the distance to the plane $P = [1 2 0] + s[1 -1 1] + t[1 0 3]$. I'm not sure how to approach this question.I know to do this question if the equation of the plane was given in the form $Ax + By + Cz = D$ but I'm not sure how to do it in this form I have thought of two methods but I'm not sure which is correct. This is the first methodand this is the second method

Could someone please advise me if either method is correct and if not the correct method. Thanks

Answer 1

A point in the plane is $(1,2,0)$ and a vector normal to the plane is

\begin{align*}(1,-1,1) \times (1,0,3) & = (-3,-2,1). \end{align*}

So the equation of the plane is $$-3(x-1) -2(y-2) +z = 0$$ which simplifies to $$3x +2y - z = 7.$$ You can take it from here. Edit: both of your methods are equivalent, but you made a mistake in the second attempt.

Answer 2

The $Ax+By+Cz+D$ approach is correct, here's another one:
Let $Q=(7,2,0)^T$, $P=a+sv_1+tv_2$ where $a=(1,2,0)^T$, $v_1=(1,−1,1)^T$, $v_2=(1,0,3)^T$ then we find minimal distance $|PQ|=\sqrt{(Q-P)^2}$ over all possible values of $s,\,t$ as the distance.
Let's rearrange things a bit first, find $v_2'=v_2+uv_1$ such that $v_2'.v_1=0$: $(v_2+uv_1).v_1=0\Leftrightarrow$ $v_2.v_1+uv_1.v_1=0\Rightarrow$ $u=-\frac{v_2.v_1}{v_1.v_1}\Leftrightarrow$ $v_2'=v_2-\frac{v_2.v_1}{v_1.v_1}v_1$.
Let's say now $sv_1+tv_2=sv_1+t(v_2'-uv_1)=(s-tu)v_1+tv_2'=s'v_1+tv_2'$ then $(P-Q)^2=$ $(a-Q+s'v_1+tv_2')^2=$ $(a-Q)^2+s'\cdot(2(a-Q).v_1)+t(2(a-Q).v_2')+{s'}^2v_1^2+t^2{v_2'}^2$ it's obvious that $(P-Q)^2$ have the minimal value when $s'=-\frac{(a-Q).v_1}{v_1^2}$, $t=-\frac{(a-Q).v_2'}{{v_2'}^2}$, hence the desired result.

And another possible approach is finding such $s,\,t$ that $(P-Q).v_i=0$ for $i=1,2$: $$\begin{cases} (a-Q+sv_1+tv_2)v_1=0\\ (a-Q+sv_1+tv_2)v_2=0 \end{cases}$$ $$\begin{cases} (a-Q).v_1+sv_1^2+tv_2.v_1=0\\ (a-Q).v_2+sv_1.v_2+tv_2^2=0 \end{cases}$$ is linear w.r.t. $s,\,t$ so hence finding $s,\,t$ we find $|PQ|$ as the desired distance.

However, these two approaches are useful when one can't take $v_1\times v_2$ cross product, e.g. working in $4$ or more dimentions.