How do I find the perpendicular distance of a point from a plane?

Question

Find the perpendicular distance of the point $(3,1,-2)$ from the plane $2x+y-2z = 8$. (The answer is $1$). I keep getting $3$ :( .. Thanks in advance.

Answer 1

$$d=\frac{|2*3+1*1+(-2)*(-2)-8|}{\sqrt{2^2+1^2+(-2)^2}}$$ $$d=\frac{|3|}{\sqrt{9}}=\frac{3}{3}=1$$

Answer 2

You may want to note that the normal vector $\langle 2,1,-2\rangle$ has length 3 here.

Answer 3

The perpendicular distance is $$\frac{2(3)+1-2(-2)-8}{\sqrt{2^2+1^2+(-2)^2}}=1$$
In general, the perpendicular distance $D$ from a point $P(x_0,y_0,z_0)$ to a plane $ax+by+cx=d$ is derived like so.

Define $\vec{u}$ to be the vector that connects any arbitary point on the plane $(x,y,z)$ to $P$. It is not hard to visualise that perpendicular distance is the magnitude of the projection of $\vec{u}$ onto $\vec{n}$, the normal vector to the plane. Hence \begin{align} D &=\|\operatorname{proj}_{\vec{n}}\vec{u}\|\\ &=\frac{\vec{u} \cdot \vec{n}}{\|\vec{n}\|}\\ &=\frac{a(x_0-x)+b(y_0-y)+c(z_0-z)}{\sqrt{a^2+b^2+c^2}}\\ &=\frac{ax_0+by_0+cz_0-d}{\sqrt{a^2+b^2+c^2}}\\ \end{align}

Answer 4

Let $\vec{z}=\vec{b}+\lambda\vec{a}$ be the projection of $\vec{b}$ on plane $P:=\left\{ \vec{x}\in\mathbb{R}^{3}\mid\left(\vec{a},\vec{x}\right)=c\right\} $.

Then $\left(\vec{a},\vec{b}+\lambda\vec{a}\right)=c$ leads to: $$d\left(\vec{b},P\right)=\left\Vert \lambda\vec{a}\right\Vert =\frac{\left|c-\left(\vec{a},\vec{b}\right)\right|}{\left\Vert \vec{a}\right\Vert }$$

In your case you can take $\vec{a}=\left(2,1,-2\right)$, $\vec{b}=\left(3,1,-2\right)$ and $c=8$.