• anonymous
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
OCW Scholar - Multivariable Calculus
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • phi
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 \[ V = ( A \times B) \cdot C \] |dw:1441640875133:dw| 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 \[ V = \frac{1}{3} \cdot area \cdot h \\ = \frac{1}{3} \cdot \frac{|A \times B|}{2} \frac{ A \times B}{|A \times B|}\cdot C\\ = \frac{1}{6} (A \times B) \cdot C \\ = \frac{1}{6} \text{ vol of parallelepiped} \]
  • anonymous
Thank you buddy, we will come back to this when we study integration then.

Looking for something else?

Not the answer you are looking for? Search for more explanations.