# Extremely hard vectors/geometry problem: Find the centre of a sphere

1. Dec 6, 2011

### Nikitin

1. The problem statement, all variables and given/known data
Plane α: x-2y+z=2. Plane β: 2x+y+z=5

Point P= (0,1,4) is on the line of intersection, with directional vector L, between the two planes.

A sphere is touching plane alpha at point A and plane beta at point B. Show that PB is perpendicular to L and the normal vector of plane Beta, and that PA is perpendicular to L and the normal vector of plane Alpha. Find the centre of the sphere, point S.

point A=(11,9,9) point B= (4,-14,-1).

2. Relevant equations
I found the line of intersection, which contains Point P. I also confirmed that vectors PB & PA are both perpendicular to the line of intersection, and PB is perpendicular to the normal vector of plane alpha while PA is perpendicular to the normal vector of plane beta. This I was ordered to do by the assignment.

3. The attempt at a solution

I set up a parametric function. The centre coordinates are: x,y,z. Using some geometry I set up a parametric function where x= 7t y=21t+1 z=4.

I then figured that the radius of the circle equals= (11-x)^2 + (9-y)^2 + (9-z)^2 = |SA| = |SB| = (-4-x)^2 + (14-y)^2 + (-1-<)^2

Well, I tried to do the algebra, substituting x with 7t, y with 21t+1 and z with 4 but I just ended up with 0t=0.

Can you guys help me?

2. Dec 6, 2011

### Nikitin

uhh, sorry a few typos:

1) radius of the circle = (11-x)^2 + (9-y)^2 + (9-z)^2 = |SA| = |SB| = (-4-x)^2 + (14-y)^2 + (-1-z*)^2

* replaced < with z.

2) I put the coordinates of point S, the centre of the sphere, as X, Y, Z.

3. Dec 6, 2011

### Nikitin

bump? anyone knows? and why was the post editing feature removed?