Question

In Problem, find the area of the parallelogram with one corner at P1 and adjacent sides$\overrightarrow{P1P2}and\overrightarrow{P1P3}$

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

Short answer

The area of the parallelogram is$\sqrt{166}$

Step-by-step solution

Step1. Given information 

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

We know that the vector is formed by two points like

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

To given question

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

step2. Calculations of the 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

$$\begin{gathered} u\times v=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 3\\ -2& 3& 0\end{array}\right| \\ u\times v=\left[i\left|\begin{array}{cc}2& 3\\ 3& 0\end{array}\right|-j\left|\begin{array}{cc}1& 3\\ -2& 0\end{array}\right|+k\left|\begin{array}{cc}1& 2\\ -2& 3\end{array}\right|\right] \\ u\times v=-9i-6j+7k \end{gathered}$$

Step3. Calculations of the area of a parallelogram 

The area of parallelogram = $\left|\left|u\times v\right|\right|$

$$\begin{gathered} \left|\left|u\times v\right|\right|=\left|-9i-6j+7k\right| \\ \left|\left|u\times v\right|\right|=\sqrt{81+36+49} \\ \left|\left|u\times v\right|\right|=\sqrt{166} \end{gathered}$$