# Thread: Finding points of rectangular box in R3

1. ## Finding points of rectangular box in R3

The faces of a rectangular box are parallel to the coordinate planes. Two of the vertices are at the points (1,2,3) and (7,5,4).

*Draw the box and label the vertices

Since it said the faces of the box are parallel to the coordinate planes, I tried finding 2 more vertices parallel to the z-axis. I said the point (1,2,3) on the other side of the z-axis = (1,-2,3) (red) since all y points are negative there and used the same reasoning for the point (7,5,4) (blue). However if I try to join the points (1,2,3), (1, -2, 3), (7,5,4) and (7,-5,4) together, I dont get a rectangular face on the z-axis (yellow). The orange and green line shows how I drew the points (7,-5,4) and (1,-2,3) respectively.

What have I done wrong?

2. Co-ordinates of four points on the lower surface of the recanguler box are

(1,2,3), (1,5,3), (7,5,3) and (7,2,3)

Similarly can you find the co-ordinates of the four points on the upper surface of the rectangular box?

3. Originally Posted by sa-ri-ga-ma
Co-ordinates of four points on the lower surface of the recanguler box are

(1,2,3), (1,5,3), (7,5,3) and (7,2,3)

Similarly can you find the co-ordinates of the four points on the upper surface of the rectangular box?
I am not sure how to find the correct points.What method did you use to find those 4 points?

4. In lower surface of the rectangular box, z co-ordinate of all the four points must remain constant and that is equal to z co-ordinate of the given point. Similarly in the case of the upper surface.

5. Originally Posted by sa-ri-ga-ma
In lower surface of the rectangular box, z co-ordinate of all the four points must remain constant and that is equal to z co-ordinate of the given point. Similarly in the case of the upper surface.
Oh ok I understand that. I made a rough sketch of the rectangular box and got the 4 points for the one face. Like you said for the z-coordinates on the other side of the top surface = 4 (blue x), but how would you determine its x and y coordinate (also for the bottom surface with constant z = 3 (yellow x)?)

6. Originally Posted by SyNtHeSiS
The faces of a rectangular box are parallel to the coordinate planes. Two of the vertices are at the points (1,2,3) and (7,5,4).

*Draw the box and label the vertices
Might help to think this way: there are 8 vertices. For each point, the x-coordinate is either 1 or 7; the y coordinate is either 2 or 5; the z-coordinate is either 3 or 4. There are 8 possibilities (2^3). These are the 8 vertices we're after.

7. Originally Posted by undefined
Might help to think this way: there are 8 vertices. For each point, the x-coordinate is either 1 or 7; the y coordinate is either 2 or 5; the z-coordinate is either 3 or 4. There are 8 possibilities (2^3). These are the 8 vertices we're after.
I understand what you saying, but when I try this by drawing a diagram to find the other vertices, I dont know what to do . I could only find the one face (shaded in grey) in the attachment above.

8. Originally Posted by SyNtHeSiS
I understand what you saying, but when I try this by drawing a diagram to find the other vertices, I dont know what to do . I could only find the one face (shaded in grey) in the attachment above.
If you understand what I'm saying, then you know the coordinates of all 8 vertices. Are you saying you don't know how to plot them?

9. Hello, SyNtHeSiS!

The faces of a rectangular box are parallel to the coordinate planes.
Two of the vertices are at the points (1,2,3) and (7,5,4).

Draw the box and label the vertices.

Don't try to make a sketch now . . . wait until the end.

The two points are diametrically opposite vertices.

. . Since $\displaystyle \,x$ goes from 1 to 6, the length of the box is 5.
. . Since $\displaystyle \,y$ goes from 2 to 5, the width of the box is 3.
. . Since $\displaystyle \,z$ goes from 3 to 4, the height of the box is 1.

Consider the "bottom" of the box.
All four vertices of this rectangle have $\displaystyle \,z= 3.$

Plot the points on a two-dimensional graph.

Code:
      |
|(1,5,3)        6        (7,5,3)
|   * → → → → → → → → → → → *
|   ↑                       ↑
|   ↑                       ↑
|  3↑                       ↑3
|   ↑                       ↑
|   ↑                       ↑
|   * → → → → → → → → → → → *
|(1,2,3)        6        (7,2,3)
|
|
- - + - - - - - - - - - - - - - - - - - -
|

We start at (1,2,3).

We know the length is 6, so we move right 6 units . . . to (7,2,3).

We know the width is 3, so we move up 3 units . . . to (1,5,3).

And we do both: move right 6 units, up 3 units . . . to (7,5,3).

And we have the coordinates of the bottom of the box:
. . $\displaystyle (1,2,3),\;(7,2,3),\;(1,5,3),\;(7,5,3)$

Since the height of the box is 1, add 1 to the $\displaystyle \,z$-coordinates.

And we have the corordinates of the top of the box:
. . $\displaystyle (1,2,4),\;(7,2,4),\;(1,5,4),\;(7,5,4)$

Now I can draw the box . . . and then label the coordinates.

Code:
             (1,2,4)     (1,5,4)
* - - - - - *
/           /|
/           / |
/           /  *(1,5,3)
/           /  /
/           /  /
/           /  /
/           /  /
(7,2,4)* - - - - - *(7,5,4)
|           | /
|           |/
* - - - - - *
(7,2,3)     (7,5,3)

And the unseen vertex (in the back) is $\displaystyle (1,2,3)$

10. Thanks a lot, that post helped a lot. I tried to draw the box, but I am having a problem projecting the surfaces through the z-axis. Referring to the attachment I drew the bottom surface in R3 with the coordinates: (1,5,0), (7,5,0), (1,2,0) and (7,2,0), then I tried projecting these points up by 3 to get (1,5,3), (7,5,3), (1,2,3) and (7,2,3), but they all lie on the line z = 3 (in blue) and look like a line as opposed to a plane

11. Originally Posted by SyNtHeSiS
Thanks a lot, that post helped a lot. I tried to draw the box, but I am having a problem projecting the surfaces through the z-axis. Referring to the attachment I drew the bottom surface in R3 with the coordinates: (1,5,0), (7,5,0), (1,2,0) and (7,2,0), then I tried projecting these points up by 3 to get (1,5,3), (7,5,3), (1,2,3) and (7,2,3), but they all lie on the line z = 3 (in blue) and look like a line as opposed to a plane
z=3 is a plane, not a line.

To plot a point (a,b,c) in R3, start at the origin, go a units in the x-direction, b units in the y-direction, and c units in the z-direction.