Eqn Of Lines: x=2s y=s z=2-2s for -1 ≤ s ≤ 0 defines one edge of cube drawn in space

[baseballblondie03]

Eqn Of Lines: x=2s y=s z=2-2s for -1 ≤ s ≤ 0 defines one edge of cube drawn in space

So here's the problem...

The equation L_AB: x=2s y=s z=2-2s for -1 ≤ s ≤ 0 defines one edge of a cube drawn in space. If A and B are the endpoints of this edge, then X(-3,3,5) is a third, non-adjacent vertex of the cube. Give the equations of the three edges are adjacent to X. Give one equation in vector form, one in parametric form and one in symmetric form.

 

[Deleted member 4993]

So here's the problem...

The equation L_AB: x=2s y=s z=2-2s for -1 ≤ s ≤ 0 defines one edge of a cube drawn in space. If A and B are the endpoints of this edge, then X(-3,3,5) is a third, non-adjacent vertex of the cube. Give the equations of the three edges are adjacent to X. Give one equation in vector form, one in parametric form and one in symmetric form.

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[baseballblondie03]

I have started to graph the points that I know.... When s= 0, Point A has the coordinates of (0,0,2), and when s=-1, Point B has the coordinates of (-2,-1,4). I also know the direction vector of AB, which is (2,1,-2), and thus the magnitude is 3. With the direction vector, I (think) I found a coordinate of (-1,4,3) because I know the edges of cubes must be all be the same