Find the area of the triangle with the vertices $A = ( 0,0,0 ) , B = ( 3,1,2 ) , \text { and }$ $C = ( 1,1,2 ) .$ Round your answer to two decimal places.
A) $20.00$
B) $8.00$
C) $6.00$
D) $2.24$
E) $4.47$

Question

Quizplus · Accepted Answer

The area of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ can be found using the formula: $\frac{1}{2} \| \vec{AB} \times \vec{AC} \|$. Here, $\vec{AB} = (3, 1, 2)$ and $\vec{AC} = (1, 1, 2)$. The cross product $\vec{AB} \times \vec{AC} = (0, -4, 2)$, and its magnitude is $\sqrt{0^2 + (-4)^2 + 2^2} = \sqrt{20}$. Thus, the area is $\frac{1}{2} \times \sqrt{20} = \frac{\sqrt{20}}{2} \approx 2.24$.

Find the Area of the Triangle with the Vertices A=(0,0,0),B=(3,1,2), and A = ( 0,0,0 ) , B = ( 3,1,2 ) , \text { and }A=(0,0,0),B=(3,1,2), and

Find the Area of the Triangle with the Vertices $A = ( 0,0,0 ) , B = ( 3,1,2 ) , \text { and }$