Question

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

$P1=\left(1,1,2\right),P2=\left(1,2,3\right),P3=\left(-2,3,0\right),P4=\left(-2,4,1\right)$

Short answer

The area of the parallelogram is :$\sqrt{34}$

Step-by-step solution

Step1. Given information 

four points$P1=\left(1,1,2\right),P2=\left(1,2,3\right),P3=\left(-2,3,0\right),P4=\left(-2,4,1\right)$

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

$$\begin{gathered} U=\overrightarrow{P1P2} \\ V=\overrightarrow{P1P3} \\ U=\left(1-1\right)i+\left(2-1\right)j+\left(3-2\right)k \\ U=j+k \\ likethatV=-3i+2j-2k \end{gathered}$$

Step2. Calculating the orthogonal vector of u and v 

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

$$\begin{gathered} U=j+k \\ V=-3i+2j-2k \\ u\times v=\left|\begin{array}{ccc}i& j& k\\ 0& 1& 1\\ -3& 2& -2\end{array}\right| \\ u\times v=\left[i\left|\begin{array}{cc}1& 1\\ 2& -2\end{array}\right|-j\left|\begin{array}{cc}0& 1\\ -3& -2\end{array}\right|+k\left|\begin{array}{cc}0& 1\\ -3& 2\end{array}\right|\right] \\ u\times v=-4i-3j+3k \end{gathered}$$

Step3. Calculating the resultant of orthogonal vector 

$$\begin{gathered} \left|\left|u\times v\right|\right|=\left|-4i-3j+3k\right| \\ \left|\left|u\times v\right|\right|=\sqrt{16+9+9} \\ \left|\left|u\times v\right|\right|=\sqrt{34} \end{gathered}$$

resultant of orthogonal vector = area of given parallelogram