Question

find the area of the parallelogram with vertices P1 , P2 , P3 , and P4 .

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

Short answer

$Area=5squnits$

Step-by-step solution

Step 1. Area of parallelogram

We know that area of parallelogram is given as :

$\overrightarrow{u}\times \overrightarrow{v}$

where ;

u and v being the vector of adjacent sides of parallelogram

Step 2. Solving for vectors u→ & v→

$$\begin{gathered} \overrightarrow{u}=P_1{P}_2 \\ \overrightarrow{u}=(2,3,1)-(2,1,1) \\ \overrightarrow{u}=(0,2,0) \\ \overrightarrow{v}=\overrightarrow{P_1{P}_4} \\ \overrightarrow{v}=(-2,4,1)-(2,1,1) \\ \overrightarrow{v}=(-4,3,0) \end{gathered}$$

We can write these vectors as:

$$\begin{gathered} \overrightarrow{u}=0\hat{i}+2\hat{j}+0\hat{k} \\ \overrightarrow{v}=-4\hat{i}+3\hat{j}+0\hat{k} \end{gathered}$$

Step 3. Area is equals to

$$\begin{gathered} Area=|\overrightarrow{u}\times \overrightarrow{v}| \\ \overrightarrow{u}\times \overrightarrow{v}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& 0\\ -4& 3& 0\end{array}\right| \\ \overrightarrow{u}\times \overrightarrow{v}=\hat{i}\left(0\right)-\hat{j}\left(0\right)+\hat{k}\left(8\right) \\ \overrightarrow{u}\times \overrightarrow{v}=8\hat{k} \\ \Rightarrow Area=|\overrightarrow{u}\times \overrightarrow{v}|=8squnits \end{gathered}$$