Question

A triangular facet in a cad model has vertices: p1 (0,0,0); p2(1,1,0) and p3(1,1,1). The area of the facet is

Answer 1

Therefore the area of a triangle=2 square units

Step-by-step explanation:

Given ,a triangular  facet in a cab model has vertices $P_1 (0,0,0)$,  $P_2 (1,1,0)$ and $P_3 (1,1,1)$.

$\vec{P_1P_2}$= (1-0,1-0,0-0)=(1,1,0)     | $\vec{P_1P_2}$|=$\sqrt{1^2+1^2+0}$=$\sqrt{2}$

$\vec {P_2P_3}$=(1-1,1-1,1-0)=(0,0,1)         |$\vec {P_2P_3}$|=$\sqrt{0+0+1^2}$ = 1

$\vec{P_1P_2} \times\vec {P_2P_3} =\left|\begin{array}{ccc} \hat{i} &\hat{j}&\hat{k}\\1&1&0\\0&0&1\end{array}\right| = \hat{i}+\hat {j}$  

$|\vec{P_1P_2} \times \vec {P_2P_3}|=\sqrt{1^2 +1^2}$ =$\sqrt{2}$

Therefore the area of a triangle

    | $\vec{P_1P_2}$|.|$\vec {P_2P_3}$|.$|\vec{P_1P_2} \times \vec {P_2P_3}|$

=$\sqrt{2}.1.\sqrt{2}$

= 2 square units