Finding the 3rd axis in 3d space

Question

I want to draw 3 perpendicular axis in 3d space, for example like this

I figured out this much:

• start by drawing the first axis (x). Any line will do

• draw the second axis (y). Any line that intersects the x axis should do (because in 3d space we can find an infinite ammount of lines perpendicular to x, and it is impossible to know if a line is perpendicular or not without the 3rd axis)

• make a ritual sacrifice to the gods of math

• ???

The first two axis can be basically any lines that intersect, as there is an infinite ammount of 2d planes in 3d space. But the third axis has to be perpendicular to the first two, and there is exactly ONE such line in 3d space. I just don't know how to find it.

Answer 1

You can use the cross product.

Given two vectors

$$v_1=(x_1,x_2,x_3)\;,\;\;\;v_2=(y_1,y_2,y_3)$$

the perpendicular vector is

$$v_3=v_1\times v_2:=\begin{vmatrix}e_1&e_2&e_3\\x_1&x_2&x_3\\y_1&y_2&y_3\end{vmatrix}=(x_2y_3-x_3y_2\,,\,x_3y_1-x_1y_3\,,\,x_1y_2-x_2y_1)$$

the direction of $v_3$ is given by the right hand rule.

Answer 2

First of all it is not true that ANY line intersecting the $x$ axis is perpendicular to it: you must choose a plane $\alpha$ containing the $x$ axis and in that plane construct, as usual, a line perpendicular to the $x$ axis at some point $O$, which will be the the $y$ axis.

You can then choose a point $P$ outside $\alpha$. In the plane containing point $P$ and the $x$ axis, construct a line $a$ perpendicular to the $x$ axis at $A$. In the plane containing point $P$ and the $y$ axis, construct a line $b$ perpendicular to the $y$ axis at $B$. Now construct, in plane $\alpha$, the line perpendicular to the $x$ axis at $A$ and the line perpendicular to the $y$ axis at $B$: those two lines will meet at a point $H$, which is the perpendicular projection of $P$ onto plane $\alpha$.

Finally, in the plane containing points $P$, $H$ and $O$, construct the midpoint $M$ of $OP$ and produce line $HM$ to a point $H'$ such that $MH'=MH$: line $OH'$ is perpendicular to plane $\alpha$ and consequently to both $x$ and $y$ axes.

Answer 3

Done a lot of research and came up with this:

1. Draw any line. This will be our x axis.

2. Draw any line intersecting with x. This will be our y axis.

This is where it gets interesting. I thought that two lines fully define a plane. No, when projecting from 3d to 2d space they actualy don't. This is because the plane can also be tilted. If you only projected the axis onto the picture plane, then there can be an infinite ammount of planes with different degrees of tilt that offer the same projection, for example

and (purple is an example rectangular shape on the infinite planes)

okay so here's how we go about it: 3. Draw a point on each axis and connect them. The distance from the center of the coordinate system and a given point will represent one unit of length on that axis IT IS IMPORTANT to make sure that the angle shown on the image is at least 90*. If it isn't you have to put your points elsewhere.

4. Draw a line going through the center of the coordinate system and perpendicular to the line you've just drawn

And there you have it! That's your z axis.

Now, you may ask "but what about the z axis metric? We have a metric for x and y, we could use a metric for z". Here's how:

BONUS STEP: Draw a line going through one of the points you've marked earlier and perpendicular to either x or y. Wherever that line intersects z is your metric. You can double check by drawing the other line and making sure that it intersects in the same place.