Chapter 11.3, Problem 58E

To determine

To calculate: The volume of parallelepiped.

  

Answer to Problem 58E

The volume of parallelepiped is9.

Explanation of Solution

Given information:

The parallelepiped is provided below,

  

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 parallelepiped which is provided below,

  

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, $u=\langle 1,1,3\rangle ,v=\langle 0,3,3\rangle ,w=\langle 3,0,3\rangle$

  $\begin{array}{c}u\cdot \left(v\times w\right)=\begin{array}{ccc}1& 1& 3\\ 0& 3& 3\\ 3& 0& 3\end{array}\\ =1\begin{array}{cc}3& 3\\ 0& 3\end{array}-1\begin{array}{cc}0& 3\\ 3& 3\end{array}+3\begin{array}{cc}0& 3\\ 3& 0\end{array}\\ =1\left(9\right)-1\left(-9\right)+3\left(-9\right)\\ =-9\end{array}$

Now, volume of parallelepiped is,

  $\begin{array}{c}V=\left|u\cdot \left(v\times w\right)\right|\\ =\left|-9\right|\\ =9\end{array}$

Thus, the volume of parallelepiped is9.