0
$\begingroup$

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.

  • 1
    shouldn't the end of the sentence be "using right *angles*"? – 2017-01-19
  • 0
    If the problem is visualization, imagine a plastic equilateral triangle leant on a corner of your room. Well, do it only if your walls are perpendicular :) – 2017-01-19
  • 0
    @ajotatxe Yes that helped I guess, but what do we mean by perpendicular distance from the origin to the triangle? – 2017-01-19
  • 0
    @ajotatxe We can find the distance of each vertex, but what would be the distance between the origin and the triangle? – 2017-01-19
  • 0
    @RSerrao That's what the book says. – 2017-01-19
  • 0
    @dud3 The "perpendicular distance" is the shortest distance from the origin to the triangle. If $L$ is the shortest line segment joining that point to the triangle, then $L$ will be perpendicular to the surface of the triangle. – 2017-01-19
  • 0
    @dud3 If you had to measure the distance from your hand to the floor, for example, you would drop a line straight down, and it would be perpendicular to the floor. Your problem is the same, except that the "floor" is at a different angle. – 2017-01-19
  • 0
    @Théophile I see, so it's a segment that is perpendicular to the surface of the triangle from the origin of the coordinate system. Would it mean that it would land in the middle of the triangle, I'm still having a hard time visualizing it, sorry. ;( – 2017-01-19
  • 0
    @dud3 Yes, that's right. In this case, it's in the middle of the triangle, which is because of the symmetry of the problem. – 2017-01-19

2 Answers 2

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 $OP$.

  • 0
    I guess, visualizing a triangular based(having a triangular base) pyramid helps. I've ended up with: `sqrt(3) / 3` – 2017-01-19
  • 0
    @dud3 That's the right answer. As a further exercise, use the same technique to find the distance from $(0,0,0)$ to an arbitrary point $(a,b,c)$. – 2017-01-20
1

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?

  • 0
    yes I think I'm nearly getting it. A, B and C are vertices of the triangle, right? x`, y`, z` is the point from the origin? How did we derive to this formula? – 2017-01-19
  • 0
    Be careful, $Ax+By+Cz=D$ is just the general form of a plane. It ***doesn't mean*** that $A=(1,0,0)$ etc. In fact, $D=1$ and $A=\dfrac{1}{a}$, etc. In your case $(x',y',z')=(0,0,0)$ and $a=b=c=1$. – 2017-01-19