Question

Find the area of the parallelogram with vertices$P_1=\left(1,1,1\right),P_2=\left(2,3,4\right),P_3=\left(6,5,2\right),and{P}_4=\left(7,7,5\right).$

Short answer

The area of the parallelogram is$18.2$square units.

Step-by-step solution

Step 1. Given Information

The given vertices are$P_1=\left(1,1,1\right),P_2=\left(2,3,4\right),P_3=\left(6,5,2\right),and{P}_4=\left(7,7,5\right).$

Step 2. Finding the area of the parallelogram

To find the area of the parallelogram we will use the formula $\mathrm{Area}=\left|\left|\mathrm{u}\times \mathrm{v}\right|\right|.$

Let's find the two adjacent sides of a parallelogram

$$\begin{gathered} \mathrm{u}=\overrightarrow{{\mathrm{P}}_1{\mathrm{P}}_2} \\ \mathrm{u}=\left(2-1\right)\mathrm{i}+\left(3-1\right)\mathrm{j}+\left(4-1\right)\mathrm{k} \\ \mathbf{u}\mathbf{=}\mathbf{i}\mathbf{+}\mathbf{2}\mathbf{j}\mathbf{+}\mathbf{3}\mathbf{k} \\ \mathrm{v}=\overrightarrow{{\mathrm{P}}_1{\mathrm{P}}_3} \\ \mathrm{v}=\left(6-1\right)\mathrm{i}+\left(5-1\right)\mathrm{j}+\left(2-1\right)\mathrm{k} \\ \mathbf{v}\mathbf{=}\mathbf{5}\mathbf{i}\mathbf{+}\mathbf{4}\mathbf{j}\mathbf{+}\mathbf{k} \end{gathered}$$

Step 3. Solve 

Let's find the cross product of the adjacent sides, the formula of the cross product is$v\times w=\left(b_1{c}_2-b_2{c}_1\right)\mathbf{i}\mathbf{}-\left({\mathrm{a}}_1{\mathrm{c}}_2-{\mathrm{a}}_2{\mathrm{c}}_1\right)\mathbf{j}\mathbf{}+\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)\mathbf{k}\mathbf{.}$

So,

$$\begin{gathered} \mathrm{u}\times \mathrm{v}=\left(\mathrm{i}+2\mathrm{j}+3\mathrm{k}\right)\times \left(5\mathrm{i}+4\mathrm{j}+\mathrm{k}\right) \\ \mathrm{u}\times \mathrm{v}=\left(2·1-4·3\right)\mathbf{i}\mathbf{}-\left(1·1-5·3\right)\mathbf{j}\mathbf{}+\left(1·4-5·2\right)\mathbf{k} \\ \mathrm{u}\times \mathrm{v}=-10\mathbf{i}+14\mathbf{j}-6\mathbf{k} \end{gathered}$$

Now, use the formula of area of a parallelogram

$$\begin{gathered} \mathrm{Area}=\left|\left|\mathrm{u}\times \mathrm{v}\right|\right| \\ \mathrm{Area}=\left|\left|-10\mathrm{i}+14\mathrm{j}-6\mathrm{k}\right|\right| \\ \mathrm{Area}=\sqrt{{\left(-10\right)}^2+{\left(14\right)}^2+{\left(-6\right)}^2} \\ \mathrm{Area}=\sqrt{332} \\ \mathrm{Area}\approx 18.2\mathrm{square}\mathrm{units} \end{gathered}$$