Question

In Problems 49–52, find the area of the parallelogram with vertices$P_1,P_2,P_3,P_4$

role="math" localid="1646761740205" $$\begin{gathered} P_1=\left(1,2,-1\right) \\ {P}_2=\left(4,2,-3\right) \\ {P}_3=\left(6,-5,2\right) \\ {P}_4=\left(9,-5,0\right) \end{gathered}$$

Short answer

The area of the parallelogram is $\sqrt{998}$square units.

Step-by-step solution

Step 1. Given data

Vertices of the parallelogram are

$$\begin{gathered} P_1=\left(1,2,-1\right) \\ {P}_2=\left(4,2,-3\right) \\ {P}_3=\left(6,-5,2\right) \\ {P}_4=\left(9,-5,0\right) \end{gathered}$$

The area of the Parallelogram is the magnitude of the cross product of the adjacent sides.

$A=\left|\left|u\times v\right|\right|$

Step 2. Finding adjacent sides

Let's say the adjacent sides are

$$\begin{gathered} u=\stackrel{\rightharpoonup }{P_1{P}_2} \\ \Rightarrow u=3i+0j-2k \\ v=\stackrel{\rightharpoonup }{P_1{P}_3} \\ \Rightarrow v=5i-7j+3k \end{gathered}$$

Step 3. Finding the cross product

The cross product:

$$\begin{gathered} u\times v=\left|\begin{array}{ccc}i& j& k\\ 3& 0& -2\\ 5& -7& 3\end{array}\right| \\ u\times v=\left|\begin{array}{cc}0& -2\\ -7& 3\end{array}\right|i-\left|\begin{array}{cc}3& -2\\ 5& 3\end{array}\right|j+\left|\begin{array}{cc}3& 0\\ 5& -7\end{array}\right|k \\ u\times v=-14i-19j-21k \end{gathered}$$

Step 4. Finding the area

The area is:

$$\begin{gathered} A=\left|\left|u\times v\right|\right| \\ A=\left|\left|-14i-19-21k\right|\right| \\ A=\left|\sqrt{{(-14)}^2+{(-19)}^2+{(-21)}^2}\right| \\ A=\sqrt{998} \end{gathered}$$

Therefore the area is$\sqrt{998}$ square units.