Question

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

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

$P1=\left(1,2,0\right),P2=\left(-2,3,4\right),P3=\left(0,-2,3\right)$

Short answer

The area of the parallelogram is :$\sqrt{555}$

Step-by-step solution

Step1. Giving information 

three points$P1\left(1,2,0\right),P2\left(-2,3,4\right),P3\left(0,-2,3\right)$

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

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

Step2. Calculation of cross product 

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

$$\begin{gathered} u\times v=\left|\begin{array}{ccc}i& j& k\\ -3& 1& 4\\ -1& -4& 3\end{array}\right| \\ u\times v=\left[i\left|\begin{array}{cc}1& 4\\ -4& 3\end{array}\right|+j\left|\begin{array}{cc}-3& 4\\ -1& 3\end{array}\right|+k\left|\begin{array}{cc}-3& 1\\ -1& -4\end{array}\right|\right] \\ u\times v=19i+5j+13k \end{gathered}$$

Step3. Area of parallelogram

$$\begin{gathered} \left|\left|u\times v\right|\right|=\left|19i+5j+13k\right| \\ \left|\left|u\times v\right|\right|=\sqrt{361+25+169} \\ \left|\left|u\times v\right|\right|=\sqrt{555} \end{gathered}$$