# Finding points of rectangular box in R3

• Sep 6th 2010, 04:00 AM
SyNtHeSiS
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?
• Sep 6th 2010, 04:25 AM
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?
• Sep 6th 2010, 05:13 AM
SyNtHeSiS
Quote:

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?
• Sep 6th 2010, 05:26 AM
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.
• Sep 6th 2010, 07:21 AM
SyNtHeSiS
Quote:

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)?)
• Sep 6th 2010, 07:28 AM
undefined
Quote:

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.
• Sep 6th 2010, 07:44 AM
SyNtHeSiS
Quote:

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.
• Sep 6th 2010, 07:48 AM
undefined
Quote:

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?
• Sep 6th 2010, 09:45 AM
Soroban
Hello, SyNtHeSiS!

Quote:

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

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:
. . $(1,2,3),\;(7,2,3),\;(1,5,3),\;(7,5,3)$

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

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

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 $(1,2,3)$

• Sep 6th 2010, 10:54 AM
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:confused:
• Sep 6th 2010, 11:37 AM
undefined
Quote:

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:confused:

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.