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anonymous
 one year ago
Problem Set 1, suplemental 1C7
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 1D7 problem. (ps. last, is there a way to search previous questions in this forum using the provided tools?) Thanks, Nicolas
anonymous
 one year ago
Problem Set 1, suplemental 1C7 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 1D7 problem. (ps. last, is there a way to search previous questions in this forum using the provided tools?) Thanks, Nicolas

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phi
 one year ago
Best ResponseYou've already chosen the best response.2I assume you mean 1D7 (not 1C7 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you buddy, we will come back to this when we study integration then.
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