Question

Problem Set 1, suplemental 1C-7
Has anybody think about a proof to the given hint. The volume of the tetrahedral is one sixth of the volume of the parallelepiped? I seams related to the preceding 1D-7 problem. (ps. last, is there a way to search previous questions in this forum using the provided tools?) Thanks, Nicolas

Answer 1

I assume you mean 1D-7 (not 1C-7 which does not reference tetrahedrons)

Are you asking for a proof of
Vol of tetrahedron = 1/6 Vol of parallelpiped
?

if we accept the formula
\[ V= \frac{1}{3} \ Area\ of\ base \cdot h \]
(I think we need to use integration to prove this)

and also, the volume of a parallelepiped defined by the 3 vectors A,B and C is
\[ V = ( A \times B) \cdot C \]



the tetrahedron formed by A, B, and C will have volume we can find using:
Area of base = \( | A \times B|/2\) (area of triangular base)
height= \( u \cdot C\)
where u is the unit vector perpendicular to the base formed by A and B
u is formed using
\[ u= \frac{ A \times B}{|A \times B|} \]
putting the pieces together we get
\[ V = \frac{1}{3} \cdot area \cdot h \\ = \frac{1}{3} \cdot \frac{|A \times B|}{2} \frac{ A \times B}{|A \times B|}\cdot C\\
= \frac{1}{6} (A \times B) \cdot C \\ = \frac{1}{6} \text{ vol of parallelepiped}
\]