# Finding Vertices of a Parallelepiped

• Jan 23rd 2009, 01:31 PM
veronicak5678
Finding Vertices of a Parallelepiped
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.
• Jan 23rd 2009, 03:00 PM
Mush
Quote:

Originally Posted by veronicak5678
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.

View my diagram and note that, by pythagorus:

\$\displaystyle |BH|^2 + |AH|^2 = |AB|^2 \$ (which you can see from the blue line!)

You know A and B, so you can most certainly find H from this, and then continue in that fashion to uncover the rest of the coordinates.
• Jan 23rd 2009, 03:31 PM
veronicak5678
Thanks for answering and for providing a diagram.
I understand what you mean, but don't know how to solve for |BH| and |AH| using the points I have.
• Jan 23rd 2009, 03:47 PM
Mush
Quote:

Originally Posted by veronicak5678
Thanks for answering and for providing a diagram.
I understand what you mean, but don't know how to solve for |BH| and |AH| using the points I have.

Show me your working for |AB|.
• Jan 23rd 2009, 03:56 PM
veronicak5678
|AB| = root ( (-1-3)^2 + (6-3)^2 + (7-4)^2 ) = root 34
• Jan 23rd 2009, 04:19 PM
Mush
Delete.