What is the closest point to a plane?

Question

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. :(

Answer 1

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 $(1)$.

Answer 2

1. 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.
2. Compute their cross product to obtain a normal vector $v_3$ for the plane. $$v_3 = v_1\times v_2=\left| \begin{matrix} {\bf i} & {\bf j} & {\bf k} \\ 2&0&1\\-2&2&1 \end{matrix} \right| =-2{\bf i}-4{\bf j}+4{\bf k} $$
3. 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 $t$.
4. 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.
5. 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}