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Let's say that three consecutive edges of a parallelepiped be a , b , c . Then how to show that volume is = [a b c] Or = a.(b × c) ?

Also how to prove this this too?

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    see also here https://en.wikipedia.org/wiki/Parallelepiped2017-02-27
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    Depending on how rigorous you want the proof to be, you need to say what you mean by volume first.2017-02-27

2 Answers 2

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Let $\vec a$ and $\vec b$ form the base. Then the area of the base is,

$$|\vec a \times \vec b|$$

The height of the parallelogram is orthogonal to the base, so it is the component of $\vec c$ onto $\vec a \times \vec b$ which is perpendicular to the base,

$$\text{comp}_{\vec a \times \vec b}\vec c=\frac{|c. (\vec a \times \vec b)|}{|\vec a \times \vec b|}$$

Multiplying the two together gives the desired result.

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area of base of parallelepiped (parallelogram) = $\mathbf b \times \mathbf c$

the vector $\mathbf b \times \mathbf c$ will be perpendicular to base

therefore: $$ \begin{align} \text{volume of parallelopiped} &= \text{area of base} \times \text{height}\\ &= (\mathbf b \times \mathbf c) \times A \cos \theta\\ &= \mathbf a\cdot(\mathbf b \times \mathbf c) \end{align} $$

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    I don't think it's correct _"the vector **b** x **c** will be perpendicular to vector **a**"_2017-02-27
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    oh sorry i meant base2017-02-27