Question

In Problem, find the area of the parallelogram with one corner at P1 and adjacent sides

$\overrightarrow{P1P2}and\overrightarrow{P1P3}$

$P1=\left(-2,0,2\right),P2=\left(2,1,-1\right),P3=\left(2,-1,2\right)$

Short answer

The area of the parallelogram is :$\sqrt{217}$

Step-by-step solution

Step1.Giving information 

three points$P1\left(-2,0,2\right),P2\left(2,1,-1\right),P3\left(2,-1,2\right)$

We know that the vector is formed by two points like

$A=\left(x_2-x_1\right)i+\left(y_2-y_1\right)j+\left(z_2-z_1\right)k$

To given question

$$\begin{gathered} U=\overrightarrow{P1P2} \\ U=(2-(-2\left)\right)i+\left(1-0\right)j+\left(-1-2\right)k \\ U=4i+j-3k \\ likethatv=4i-j \end{gathered}$$

Step2. calculation cross product of vectors 

We know that to find the area of a parallelogram by two vectors. it's necessary to find an orthogonal vector, for that

$$\begin{gathered} u\times v=\left|\begin{array}{ccc}i& j& k\\ 4& 1& -3\\ 4& -1& 0\end{array}\right| \\ u\times v=\left[i\left|\begin{array}{cc}1& -3\\ -1& 0\end{array}\right|-j\left|\begin{array}{cc}4& -3\\ 4& 0\end{array}\right|+k\left|\begin{array}{cc}4& 1\\ 4& -1\end{array}\right|\right] \\ u\times v=-3i-12j-8k \end{gathered}$$

Step3. calculations of area 

area of the parallelogram is :$\left|\left|u\times v\right|\right|$

$$\begin{gathered} \left|\left|u\times v\right|\right|=\left|-3i-12j-8k\right| \\ \left|\left|u\times v\right|\right|=\sqrt{9+144=64} \\ \left|\left|u\times v\right|\right|=\sqrt{217} \end{gathered}$$