Given a unit cube in the first octant of a three dimensional coordinate system w

Question

Given a unit cube in the first octant of a three dimensional coordinate system with an axis of rotation passing through the points (1,0,0) and (0,1,1), find the equations of the TWO planes that contain the bases of the cones. Put your equations in the form

ax + by + cz = d, with a > 0. Hint: You can use the three-point method or use the fact that the axis of rotation is orthogonal to the planes.

I have already parameterized the axis of rotation.

r(t) = <1,0,0> + t<-1,1,1>

x = 1 - t

y = t

z = t

Explanation / Answer

The cube has vertices: (0,0,0), (0,0,1),(0,1,0),(0,1,1),(1,0,0), (1,0,1),(1,1,0),(1,1,1)

When we rotate this cube along (1,0,0) and (0,1,1), we have a shape like this: http://i.stack.imgur.com/jQN3Y.gif

We see the bases of the two cones passing through 3 vertices each on the cone, which are equidistant from the vertex of the cones, i.e. points of rotation.

Therefore, cone 1 has base passing through 3 vertices which are at an edge distance from (1,0,0) and cone 2 has base passing through 3 vertices which are at an edge distance from (0,1,1).

Cone1's base passes through: (1,0,1),(1,1,0),(0,0,0)

equation of cone 1's base: -x+y+z=0

Cone1's base passes through: (0,1,0),(0,0,1),(1,1,1)

equation of cone 1's base: -x+y+z-1=0

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