15
$\begingroup$

I am having a hard time figuring out if a 3D point lies in a cuboid (like the one in the picture below). I found a lot of examples to check if a point lies inside a rectangle in a 2D space for example this on but none for 3D space.

enter image description here

I have a cuboid in 3D space. This cuboid can be of any size and can have any rotation. I can calculate the vertices $P_1$ to $P_8$ of the cuboid.

Can anyone point me in a direction on how to determine if a point lies inside the cuboid?

$\endgroup$

3 Answers 3

Reset to default
21
$\begingroup$

The three important directions are $u=P_1-P_2$, $v=P_1-P_4$ and $w=P_1-P_5$. They are three perpendicular edges of the rectangular box.

A point $x$ lies within the box when the three following constraints are respected:

  • The dot product $u.x$ is between $u.P_1$ and $u.P_2$
  • The dot product $v.x$ is between $v.P_1$ and $v.P_4$
  • The dot product $w.x$ is between $w.P_1$ and $w.P_5$

EDIT:
If the edges are not perpendicular, you need vectors that are perpendicular to the faces of the box. Using the cross-product, you can obtain them easily:
$$u=(P_1-P_4)\times(P_1-P_5)\\ v=(P_1-P_2)\times(P_1-P_5)\\ w=(P_1-P_2)\times(P_1-P_4)$$ then check the dot-products as before.

$\endgroup$
9
  • $\begingroup$ In the case of perpendicular edges the constraints are that all the three dot products must be valid right not just any one of them? Superb answer by the way $\endgroup$ Commented Sep 8, 2018 at 10:10
  • 3
    $\begingroup$ According to my testing in unity, I had to check if ux is between uP2 & uP1; if vx is between vP4 & vP1; and if wx is between wP5 & wP1 (the inverse order) $\endgroup$ Commented Dec 14, 2019 at 19:11
  • $\begingroup$ @Empy2 do I need to normalize the direction vectors u,v,w ? $\endgroup$ Commented Aug 9, 2021 at 14:04
  • $\begingroup$ @Kong No, $ku.x$ is between $ku.P_1$ and $ku.P_2$ whenever $u.x$ is between $u.P_1$ and $u.P_2$ $\endgroup$ Commented Aug 9, 2021 at 14:26
  • 1
    $\begingroup$ $P1=(x_1,y_1,z_1), P2=(x_2,y_2,z_2)$ and so on. $\endgroup$ Commented Jul 25, 2022 at 9:24
10
$\begingroup$

Given $p_1,p_2,p_4,p_5$ vertices of your cuboid, and $p_v$ the point to test for intersection with the cuboid, compute: $$\begin{matrix} i=p_2-p_1\\ j=p_4-p_1\\ k=p_5-p_1\\ v=p_v-p_1\\ \end{matrix}$$

then, if $$\begin{matrix} 0<v\cdot i<i\cdot i\\ 0<v\cdot j<j\cdot j\\ 0<v\cdot k<k\cdot k \end{matrix}$$ The point is within the cuboid.

$\endgroup$
5
  • $\begingroup$ I like this approach. Simple and clever. $\endgroup$ Commented Jul 14, 2022 at 15:25
  • $\begingroup$ for a non-math guy, are these p1 p2 etc the Position vectors of the co-ordinates ? How can i translate my x,y,z co-ordinate into these P1, P2 , etc ? $\endgroup$ Commented Jul 25, 2022 at 0:43
  • 1
    $\begingroup$ p1 is a vector which represents the position of the front bottom left point of the box. the front bottom left point of the box, p1, should have x y z coordinates. p1 = (x, y, z). You can perform b - a by doing (bx - ax, by - ay, bz - az). The dots represent dot product. a dot b = (ax, ay, az) dot (bx, by, bz) = axbx + ayby + az*bz $\endgroup$ Commented Jul 25, 2022 at 4:36
  • $\begingroup$ @TreyReynolds thanks alot ! This helped :D $\endgroup$ Commented Jul 25, 2022 at 11:38
  • $\begingroup$ Can you please give some sources or explain, what is the reason behind these equations working the way they are working ? @TreyReynolds $\endgroup$ Commented Aug 5, 2022 at 8:05
0
$\begingroup$

Since 2-D problem is known, divide the problem into 3 parts.

Disregard z-coordinate. If any point is within box bounds $ x_2-x_1,y_2-y_1, $ select it and all other such points. Similarly y-z and z-x boxes.

Next choose points that the are common to three selections.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.