Chapter 9, Problem 56RE

To determine

To find: The area of the parallelogram.

Answer to Problem 56RE

The area of the parallelogram is $18.2\text{square}\text{units}$ .

Explanation of Solution

Given information:

The vertices of the parallelogram are

  $P_1=\left(1,1,1\right)$

  $P_2=\left(2,3,4\right)$

  $P_3=\left(6,5,2\right)$

  $P_4=\left(7,7,5\right)$

Calculation:

Calculate vector $u$ .

  $\begin{array}{c}u=\overrightarrow{P_1{P}_2}\\ =\left(2-1\right)i+\left(3-1\right)j+\left(4-1\right)k\\ =i+2j+3k\end{array}$

Calculate vector $v$ .

  $\begin{array}{c}v=\overrightarrow{P_1{P}_3}\\ =\left(6-1\right)i+\left(5-1\right)j+\left(2-1\right)k\\ =5i+4j+k\end{array}$

Calculate the cross product of vector $u$ and $v$ .

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 3\\ 5& 4& 1\end{array}\right|\\ =\left|\begin{array}{cc}2& 3\\ 4& 1\end{array}\right|i-\left|\begin{array}{cc}1& 3\\ 5& 1\end{array}\right|j+\left|\begin{array}{cc}1& 2\\ 5& 4\end{array}\right|k\\ =-10i+14j-6k\end{array}$

Calculate the area of the parallelogram.

  $\begin{array}{c}A=\Vert u\times v\Vert \\ =\Vert -10i+14j-6k\Vert \\ =\sqrt{{\left(-10\right)}^2+{\left(14\right)}^2+{\left(-6\right)}^2}\\ =18.2\text{square}\text{units}\end{array}$

Therefore, the area of the parallelogram is $18.2\text{square}\text{units}$ .