Finding Vertices of a Parallelepiped

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monomoco

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Nov 3, 2008
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I will try to describe the picture:

The figure is a rectangular box, with the near lower left point A labeled (3,3,4) and the far top right point B labeled (-1,6,7).

I am to find the coordinates of the remaining 6 vertices. There are no examples in my book, and I'm not sure how to solve. I have found |A|= root 34, |B| = root 86, and |AB| = root 34.
 


I'm not sure what you're doing with the symbols A and B, since they are both nothing more than the names of the two given points.

For example, you wrote |AB|.

Are you trying to multiply two points together? I don't understand.

What I did was start by taking a bird's-eye view looking down on the xy-plane from above.

(I'm totally ignoring the z-dimension, for now.)

From the given information, two opposite corners of the bottom of the box are located at the following.

(3, 3, z)

(-1, 6, z)

Think of a light shining down on the box from directly above. It casts a rectangular-shaped shadow onto the xy-plane. This rectangular-shaped shadow has the same measurements as the bottom of the box.

If we were to plot these two opposite corners of the rectangle on the xy-plane at (3, 3, 0) and (-1, 6, 0), then it would be clear that the remaining two corners are located at the following.

(3, -1, 0)

(3, 6, 0)

This tells us that the box's bottom has corners at the following.

(3, 3, z)

(3, -1, z)

(-1, 6, z)

(3, 6, z)

Now we consider the levels of the box's top and bottom above the xy-plane. We know that the bottom of the box is 4 units above the xy-plane. We know that the top of the box is 7 units above the xy-plane.

Replace the z-coordinates in the four sets of ordered triples above with both 4 and 7, and you'll have the coordinates of all eight vertices.

MY EDITS: Corrected misstatements about box having square-shaped bottom
 
monomoco said:
With |AB|, I was trying to denote the distance between the two vectors.
Click to expand...


What vectors?

(The term "vector" does not appear in your original post.)

If you would like to do something with vectors, that's fine. But you may not name vectors using symbols A and B because those symbols are already in use as names of two given points.

Pick other names for your vectors, and explain what you're tying to do.

 
Sorry for the confusion. I was trying to use vectors to find the distances between the known points.
 
monomoco said:
… I [am] trying to use vectors to find the [distance] between the known points.
Click to expand...


Okay. I got the impression that you were trying to find the coordinates of the remaining six vertices, instead. I think I'm starting to get it, but it seems like you already found the distance between the two known points.

First, some comments about vector notation.

You posted |A| = sqrt(34).

So, I think that you found the magnitude of the position vector from the origin to point A, and that's what you were trying to say by typing |A|. Is that right?

If we want to use the names of points to denote vectors here, then we need to state what we're doing so that everybody understands the notation.

(Or, we could use LaTex coding to render standard vector symbolism.)

I'll use bold-face type to denote vectors. The name of the origin is point O.

So, the typed notation would be:

|OA| = sqrt(34)

|OB| = sqrt(86)

The known points are A and B, so there is only one distance to find.

|AB| is the magnitude of the vector from point A to point B.

You posted this magnitude as sqrt(34). You did not show your work, but sqrt(34) is correct.

Is this all that you wanted to know? You're trying to ask for confirmation that the distance between the two known points is sqrt(34)?



 
Yes, I meant the distance from the origin. I'm sorry about this. I have never seen vectors before and am (obviously) confused.
What I wanted to know is how to find the coordinates of the other vertices. I was thinking I could use the Pythagorean Th. to make triangles out of the vectors.
 
Hello, monomoco!

I don't suppose you made a sketch . . .

Given: a rectangular box, near lower-left point A(3,3,4), and the far top-right point B(-1,6,7).
Find the coordinates of the remaining 6 vertices.
Click to expand...

I must assume that the faces are parallel to the coordinate planes.

Code: 
                      (-1,6,7)
          * - - - - - - - * B
         /               /|
        /               / |
       /               /  |
    C * - - - - - - - *   |
      |               |   |
      |               |   *
      |               |  /
      |               | /
      |               |/
    A * - - - - - - - * D
  (3,3,4)

\(\displaystyle \text{From the }x\text{-coordinates, the box is: }|\text{-}1-3| = 4\text{ units deep (front to back).}\)

\(\displaystyle \text{From the }y\text{-coordinates, the box is: }|6-3| = 3\text{ units wide (left to right).}\)

\(\displaystyle \text{From the }z\text{-coordinates, the box is: }|7-4| = 3\text{ units high (top to bottom).}\)


\(\displaystyle \text{Vertex }C\text{ is directly above }A\text{, with }z\text{-coordinate }7.\)
. . \(\displaystyle \text{Hence: }\:C(3,3,7)\)

\(\displaystyle \text{Vertex }D\text{ is directly to the right of }A\text{, with }y\text{-coordinate }6.\)
. . \(\displaystyle \text{Hence: }\:D(3,6,4)\)


Get the idea?
You should be able to find the other vertices now . . .

 
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