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 -2n 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 ..."