# Hard geometry question need help (vectors planes spheres)

In a space with an orthonormal coordinate system consider

the points

P(0, - 1, 1) and Q(3, 0, - 3) .

the line

d:

x = 2t

y = t , t included in R

z = 2 + 2t

the sphere

S: x ^ 2 + y ^ 2 + z ^ 2 - 2x - 2y + 2z - 6 = 0

and the plane

sigma : 2 * x - y + 4 = 0

a) The plane pi contains the point P and the line d Show that x + 2y - 2z + 4 + 0 is an equation of pi

b) Find the coordinates of the centre C and the radius R of the sphere S.

c) Find an equation for each of the two spheres with radius r = 3 which are tangential to at the point P. Verify that one of these spheres is S.

## 1 Answer

a) It's easy to check that the coordinates of P and also (x,y,z) of the line d, for any t, satisfy the equation (with =0 instead of +0)

b) it's easy to find the center and radius of S by completing the squares: (x-1)² + (y-1)² + (z+1)² = 6+1+1+1 = 9 = 3²

c) Completing with "... to pi at ..." we must add a multiple of a normal vector of pi, e.g., **n** = (1 ,2, -2), to the coordinates of P to find the line on which are the centers of the two spheres; and the length of the vector added must be 3 so **n** with length sqrt(1+4+4) = sqrt(9) = 3 is OK and gives one solution and the opposite, -**n**, gives the other solution. Indeed, we get the center C = (1,1,-1) found in (b) and the other center is (-1,-3,3) (obtained by adding **-n** to P or **-**2**n** to C), the equation of the 2nd sphere is (x+1)²+(x+3)²+(x-3)² = 3².

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This is a time consuming problem. I suggest you to offer a bounty.

The equation of the plane is ... = 0, not ... + 0.

and I guess in (c) it should read "...to *pi* at ..."