Chapter 11.3, Problem 62E

To determine

To calculate: The volume of parallelepiped with vertices, $A\left(0,0,0\right),B\left(1,1,0\right),C\left(1,0,2\right),D\left(0,1,1\right),E\left(2,1,2\right),F\left(1,1,3\right),G\left(1,2,1\right),H\left(2,2,3\right)$

Answer to Problem 62E

The volume of parallelepiped is3.

Explanation of Solution

Given information:

The vertices of parallelepiped are,

  $A\left(0,0,0\right),B\left(1,1,0\right),C\left(1,0,2\right),D\left(0,1,1\right),E\left(2,1,2\right),F\left(1,1,3\right),G\left(1,2,1\right),H\left(2,2,3\right)$ .

Formula used:

For the set of vectors, $u=u_{1}i+u_{2}j+u_{3}k$ , $v=v_{1}i+v_{2}j+v_{3}k$ and $w=w_{1}i+w_{2}j+w_{3}k$ , which represent the adjacent edges of a parallelepiped, then it’s volume is denoted by, $V=\left|u\cdot \left(v\times w\right)\right|$ , where, $u\cdot \left(v\times w\right)$ denotes the scalar triple product.

  $u\cdot \left(v\times w\right)=\begin{array}{ccc}{u}_1& u_2& u_3\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}$

Calculation:

Consider the vertices of parallelepiped,

  $A\left(0,0,0\right),B\left(1,1,0\right),C\left(1,0,2\right),D\left(0,1,1\right),E\left(2,1,2\right),F\left(1,1,3\right),G\left(1,2,1\right),H\left(2,2,3\right)$ .

Adjacent sides of the parallelepiped are $\stackrel{\rightharpoonup }{AB}$ , $\stackrel{\rightharpoonup }{AD}$ and $\stackrel{\rightharpoonup }{AH}$

Now, the value of $\stackrel{\rightharpoonup }{AB}$ is evaluated as, subtract the coordinate of B from A and write it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}=\langle 1-0,1-0,0-0\rangle \\ =\langle 1,1,0\rangle \end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{AD}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AD}=\langle 0-0,1-0,1-0\rangle \\ =\langle 0,1,1\rangle \end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{AH}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AH}=\langle 2-0,2-0,3-0\rangle \\ =\langle 2,2,3\rangle \end{array}$

Recall that for the set of vectors, $u=u_{1}i+u_{2}j+u_{3}k$ , $v=v_{1}i+v_{2}j+v_{3}k$ and $w=w_{1}i+w_{2}j+w_{3}k$ , which represent the adjacent edges of a parallelepiped, then it’s volume is denoted by, $V=\left|u\cdot \left(v\times w\right)\right|$ , where, $u\cdot \left(v\times w\right)$ denotes the scalar triple product.

  $u\cdot \left(v\times w\right)=\begin{array}{ccc}{u}_1& u_2& u_3\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}$

Apply it for provided set of vectors, $u=\langle 1,1,0\rangle ,v=\langle 0,1,1\rangle ,w=\langle 2,2,3\rangle$

  $\begin{array}{c}u\cdot \left(v\times w\right)=\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 2& 2& 3\end{array}\\ =1\begin{array}{cc}1& 1\\ 2& 3\end{array}-1\begin{array}{cc}0& 1\\ 2& 3\end{array}+0\begin{array}{cc}0& 1\\ 2& 2\end{array}\\ =\left(3-2\right)-1\left(-2\right)\\ =3\end{array}$

Now, volume of parallelepiped is,

  $\begin{array}{c}V=\left|u\cdot \left(v\times w\right)\right|\\ =\left|3\right|\\ =3\end{array}$

Thus, the volume of parallelepiped is3.