Chapter 9, Problem 44RE

To determine

To find : The volume of parallelepiped.

Explanation of Solution

Given: The points are $P\left(0,2,2\right)$ , $Q\left(3,1,-1\right)$ and $R\left(1,4,1\right)$ .

The vectors of the parallelepiped are,

$\begin{array}{l}u=2j+2k\\ v=3i+j-k\\ w=i+4j+k\end{array}$

The volume of parallelepiped is given by,

$V=\left|u.\left(v\times w\right)\right|$

The cross product is calculated as,

$\begin{array}{c}v\times w=\left|\begin{array}{ccc}i& j& k\\ 3& 1& -1\\ 1& 4& -1\end{array}\right|\\ =\left(1+4\right)i-\left(3+1\right)j+\left(2-1\right)k\\ =5i-4j+11k\end{array}$

The volume of the parallelepiped is calculated as,

$\begin{array}{c}V=\left(2j+2k\right)\cdot \left(5i-4j+11k\right)\\ =-8+22\\ =16\text{cubic}\text{unit}\end{array}$

Therefore, the volume of the parallelepiped is $16\text{cubic}\text{unit}$ .