what 3d geometric shape has 3 faces and 2 sides?
The general method is perhaps best illustrated by working through an example. Find the shortest distance between the given lines, and the points of closest approach on each line. x y-3 z x-5 y-8 z-2 — = — = — = s and —– = —– = —— = t 1 1 -1 3 7 -1 The common perpendicular is obtained from the vector product | i j k | = i(6) -j(2) + k(4) | 1 1 -1 | | 3 7 -1 | So common perpendicular is the vector (3, -1, 2) which can be written as a unit vector in the form 1/(sqrt(14)[3, -1, 2] The vector connecting the given point on line (1) with the given point on line (2) is [(5-0), (8-3), (2-0)] = (5, 5, 2) The scalar product of this with the common perpendicular in unit vector form is 5 x 3 + 5 x (-1) + 2 x 2 14 ————————- = ——- = sqrt(14) sqrt(14) sqrt(14) So the shortest distance is sqrt(14). Now to find the points of closest approach, we have: line (1) is r = (0, 3, 0) + s(1, 1, -1) = [s, (3+s), -s] line (2) is r = (5, 8, 2) + t(3, 7, -1) = [(5+3t), (8+7t), (2
