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Find the volume of the parallelepiped with vertices (0,0,0), (3,0,0), (0,5,1), (3,5,1), (2,0,5), (5,0,5), (2,5,6), and (5,5,6).

 Jul 30, 2019
 #1
avatar+23891 
+2

Find the volume of the parallelepiped with vertices (0,0,0), (3,0,0), (0,5,1), (3,5,1), (2,0,5), (5,0,5), (2,5,6), and (5,5,6).

 

 

 

\(\text{Let $\vec{a}=\begin{pmatrix}0\\5\\1 \end{pmatrix}$ } \quad \text{Let $\vec{b}=\begin{pmatrix}3\\0\\0 \end{pmatrix}$ } \quad \text{Let $\vec{c}=\begin{pmatrix}2\\0\\5 \end{pmatrix}$ } \)

 

\(\begin{array}{|rcll|} \hline \mathbf{V} &=& \left| \begin{vmatrix}0&3&2\\5&0&0\\1&0&5 \end{vmatrix}\right| \\ &=& \left| 0 - 3\cdot \begin{vmatrix}5&0\\1&5 \end{vmatrix} +0 \right| \\ &=& \left| -3\cdot 25 \right| \\ &=& \left| -75 \right| \\ &=& \mathbf{75} \\ \hline \end{array}\)

 

laugh

 Jul 30, 2019
edited by heureka  Jul 30, 2019
 #2
avatar+106536 
+2

See the following image:

 

 

Let vectors u, v and w  be three edges of the paralleliped  which meet at  (0,0, 0)

 

So.....these vectors  can be :

 

  u  = (3, 0 , 0)     v  =  (2,0,5)     w  = (0,5, 1)

 

The volume of the  parallelpiped   =   the absolute value  of   the  determinant  of the following matrix

 

u1   u2   u3                3  0   0                  [ 0 5 ]              [ 2  5 ]           [ 2   0 ]

v1   v2    v3   =          2  0   5      =   3  *  [ 5  1 ]   +  0 * [ 0  1 ]   +  0 *[ 0   5 ]   =

w1  w2   w3               0   5  1

 

 

3 [ 0*1  - 5 *5 ] + 0 + 0   =   3 [ -25 ]   =   - 75   =  l -75 l   =  75  units^3

 

 

cool cool cool

 Jul 30, 2019
edited by CPhill  Jul 30, 2019

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