Consider a cube whose faces are given by
$x+y+z=3\sqrt{3}$,
$x+y+z=2\sqrt{3}$,
$4x-5y+z=\sqrt{42}$,
$4x-5y+z=2\sqrt{42}$,
$2x+y-3z=\sqrt{14}$,
$2x+y-3z=2\sqrt{14}$,
and a triangle whose vertices are $(2, 1, 3)$, $(1, 1, 1)$, $(3, 1, 0)$,
then find the number of point of intersection of cube and triangle.
My work
I checked position of each point w.r.t 6 planes, then found that first point is outside the cube, second point is on one of the face and third point is inside the cube. Thus there should be 2 intersection, Is there some other way of solving ?
$\endgroup$ 2 Reset to default
