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How this shape called and why this is not a pyramid? Here link to Desmos3D. I wanted to find a volume of shape on 3D-vector $ v=(v_1, v_2, v_3) $ and thought it's a pyramid with volume $ \frac{1}{3} |v_1 v_2 v_3| $ but actually volume is equal to $ \frac{1}{4} |v_1 v_2 v_3| $ (made additional geometric constructions and subtracted volumes).

Edit:

How to find volume of this shape built on some vector $ \vec{v} = (v_1, v_2, v_3) $?

I firstly got $ \frac{1}{3} |v_1 v_2 v_3| $ via formula for pyramid's volume. But I can make additional constructions: parallelepiped with sides $ (v_1, v_2, v_3) $ and volume $|v_1 v_2 v_3|$, same pyramid but with different orientation (above vector $\vec{v}$) and volume $V$ and two truncated prisms with rigth triangles as bases and their volumes equal $\frac{1}{2} \cdot \frac{1}{2}|v_2 v_3| \cdot |v_1| = \frac{|v_1 v_2 v_3|}{4}$ for both. For truncated prisms volume can be found as $A_{base} \cdot h_{mean}$ where $h_{mean} = \frac{v_i + 0}{2}$ since one of the heights is n-th component of vector $\vec{v}$, which is $v_i$, and other height is $0$ because prism truncated from top to bottom (from one base to another).

So now I can solve equation for $V$:

$|v_1 v_2 v_3| = 2V + 2 \cdot \frac{|v_1 v_2 v_3|}{4} $

$\therefore V=\frac{|v_1 v_2 v_3|}{4} $

Picture of shape on one 3-dimensional vector

Picture of same shape but above this vector

First truncated prism

Second truncated prism

This 4 bodies form parallelepiped with sides equal to components of vector $\vec{v}$ and with this vector as a diagonal.

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$ Commented Oct 15 at 6:22
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    $\begingroup$ Since you showed none of your work, we can't tell what you did right or wrong. The four solids you constructed do not all have the same volume, so if that's how you concluded $\frac{1}{4} |v_1 v_2 v_3|$, that's your mistake. Why do you think that's the volume of anything in your figure? $\endgroup$ Commented Oct 15 at 6:53
  • $\begingroup$ About naming : you can say it's a square base (regular) pyramid : see here where it is said that not all people agree on some definitions. $\endgroup$ Commented Oct 15 at 7:35
  • $\begingroup$ I want to find volume of this shape built on vector. If I use formula for volume of pyramid I get $ V = \frac{1}{3} |v_1 v_2 v_3| $. But if I make additional constructions I can say that volume of bigger parallelepiped built on some vector is equal to two volumes of this pyramids (first base on top parallel to Oxy and second on bottom which is on Oxy) added to volumes of pair of truncated prisms with right triangles as their bases: $ v_1 v_2 v_3 = 2V + \frac{1}{4} v_1 v_2 v_3 + \frac{1}{4} v_1 v_2 v_3 $; $ v_1 v_2 v_3 = 2V + \frac{1}{2} v_1 v_2 v_3 $; $ V = \frac{1}{4}v_1 v_2 v_3 $ $\endgroup$ Commented Oct 15 at 8:36
  • $\begingroup$ I've edited question. Is this now understandable? $\endgroup$ Commented Oct 15 at 9:52

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Your "truncated prisms" are actually tetrahedra, also known as pyramids with triangular bases. Count the vertices: there are only four, and every vertex is connected to every other vertex by an edge of the figure. That's a tetrahedron.

The volume of a tetrahedron is one-third of the product of the area of a base and the length of an altitude on that base. Looking at your first "truncated prism" we can take the triangular base to be in the $y,z$ plane. The area of the base is $\frac12|v_2v_3|$ where $v_2$ and $v_3$ are the $y$ and $z$ coordinates of your vector. The altitude is $|v_1|$, the $x$ coordinate of your vector. Hence the volume of the tetrahedron is

$$ \frac13 \cdot \left(\frac12 |v_2v_3|\right) \cdot |v_1| = \frac16 |v_1v_2v_3|.$$

The "shape on one three-dimensional vector" and the "same shape but above this vector" are simply pyramids on rectangular bases as you initially thought, and the volume of each of these pyramids is $\frac{1}{3} |v_1 v_2 v_3|.$

The sum of volumes of your four solid figures is therefore

$$ \frac13 |v_1 v_2 v_3| + \frac13 |v_1 v_2 v_3| + \frac16 |v_1v_2v_3| + \frac16 |v_1v_2v_3| = |v_1v_2v_3|, $$

that is, the volumes of the four shapes add up to exactly the volume of the rectangular box that they compose, so there is no discrepancy in the volume measurements at all.

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  • $\begingroup$ Thanks. I noticed that I also used formula of volume for truncated prism incorrectly (should've been added one more zero and divide by 3 because heights are $(v_i, 0, 0)$ and then it will be exactly one-third of base area times height how you written). If you're interested then I have another question about volume of different body: link. And since I firstly used correct volume for pyramid with right coefficient of $\frac{1}{3}$ then I have mistake in it... $\endgroup$ Commented Oct 17 at 12:36

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