Consider a prism with a base given by $A(7,10,0) , B(8,18,0) , C(14,12,0) $ and a height of $10$.Determine the true size of the section made through the prism by a plane given with its frontal and horizontal traces.The frontal and horizontal traces intersect the ground line at $\alpha(0,11,0)$ where the angle between the ground line and the frontal trace ( horizontal trace ) is $30$ degrees ($-60$ degrees),respectively.
Could anyone help me with this?
Thanks in advance!
HINT.
The equation of the plane is: $x+\sqrt3 y-3z=11\sqrt3$. Intersecting that with the lateral edges of the prism gives the vertices $G$, $H$, $I$ of the section. Its area can then be computed as $$ {1\over2}\big|(G-H)\times(I-H)\big| $$ and turns out to be $9\sqrt{13}$.
EDIT.
There is a much simpler solution: once you have found a normal vector $\vec n=(-1,-\sqrt3,3)$ to the intersecting plane, then you also know the angle $\theta$ formed by $\vec n$ with the normal $\vec z=(0,0,1)$ to the base of the prism: $$ \cos\theta={\vec n\cdot\vec z\over|\vec n|}={3\over\sqrt13}. $$ As the base area is $27$, we get the area $A$ of the intersection: $$ A={27\over\cos\theta}=9\sqrt13. $$