Perpendicular distance from origin to the triangle

Question

Can someone explain the following to me:

The vertices $(1, 0, 0)$, $(0, 1, 0)$ and $(0, 0, 1)$ form a triangle.

Find the perpendicular distance from the origin $(0, 0, 0)$ to this triangle using right triangles.

I'm having a hard time visualizing it.

Answer 1

Appealing to symmetry, the line through the origin $O$ and perpendicular to the triangle will meet it at the point $P=(\frac13,\frac13,\frac13)$. Drop a line from $P$ to the $xy$-plane to hit the point $Q=(\frac13,\frac13,0)$. Now use Pythagoras's Theorem twice: once to find the length $OQ$, and then again on the triangle $OPQ$ to find $OP$.

Answer 2

The distance of a point $(x',y',z')$ from the plane $Ax+By+Cz=D$ is given by $$\left| \frac{Ax'+By'+Cz'-D}{\sqrt{A^2+B^2+C^2}} \right|$$

The equation of a plane with $x$-, $y$- and $z$-intercepts $a$, $b$ and $c$ respectively is $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$

Now the $x$, $y$ and $z$ intercepts are $1$. Can you proceed?