# Check if a point is inside a rectangular shaped area (3D)?

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.

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?

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.

• 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
– user297514
Commented Sep 8, 2018 at 10:10
• 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) Commented Dec 14, 2019 at 19:11
• @Empy2 do I need to normalize the direction vectors u,v,w ?
– Kong
Commented Aug 9, 2021 at 14:04
• @Kong No, $ku.x$ is between $ku.P_1$ and $ku.P_2$ whenever $u.x$ is between $u.P_1$ and $u.P_2$ Commented Aug 9, 2021 at 14:26
• $P1=(x_1,y_1,z_1), P2=(x_2,y_2,z_2)$ and so on. Commented Jul 25, 2022 at 9:24

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.

• I like this approach. Simple and clever.
– Omid
Commented Jul 14, 2022 at 15:25
• 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 ? Commented Jul 25, 2022 at 0:43
• 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 Commented Jul 25, 2022 at 4:36
• @TreyReynolds thanks alot ! This helped :D Commented Jul 25, 2022 at 11:38
• Can you please give some sources or explain, what is the reason behind these equations working the way they are working ? @TreyReynolds Commented Aug 5, 2022 at 8:05

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.