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).
+26367 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).
\(\text{Let $\vec{a}=\begin{pmatrix}0\\5\\1 \end{pmatrix}$ } \quad \text{Let $\vec{b}=\begin{pmatrix}3\\0\\0 \end{pmatrix}$ } \quad \text{Let $\vec{c}=\begin{pmatrix}2\\0\\5 \end{pmatrix}$ } \)
\(\begin{array}{|rcll|} \hline \mathbf{V} &=& \left| \begin{vmatrix}0&3&2\\5&0&0\\1&0&5 \end{vmatrix}\right| \\ &=& \left| 0 - 3\cdot \begin{vmatrix}5&0\\1&5 \end{vmatrix} +0 \right| \\ &=& \left| -3\cdot 25 \right| \\ &=& \left| -75 \right| \\ &=& \mathbf{75} \\ \hline \end{array}\)
+128631 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
