A sphere of radius 9 has its center at (3,-4,5). A ray emanates from the center & goes through the pt (4,-6,7)

Question

A sphere of radius 9 has its center at (3,-4,5). A ray emanates from the center & goes through the pt (4,-6,7). Find the exact pt coordinates where the ray punctures the sphere and rectangular equation (with integral coefficients) of the plane tangent to the sphere at that point.
so is this right...
sqrt (4-3)^2+(-6+4)^2+(7-5)^2=3 so x=3+(1/3)d y=-4-(2/3)d z=5+(2/3)d
so radius is 9 meaning d=9, then sub d into the 3 parametric equation above to get (6,-10,11)
next x-2y+2z=d so 4-(2*-6)+(2*7)=d thus d=30
concluding that x-2y+2z=30 is the rectangular equation of the plane tangent to the sphere

Answer 1

The equation of the sphere is:

(x - 3)² + (y + 4)² + (z - 5)² = 81

The vector from the center of the sphere to the point (4,-6,7) is:

v = <4-3,-6+4,7-5> = <1,-2,2>

v has magnitude √(1² + (-2)² + 2²) = √(1 + 4 + 4) = √9 = 3

We want a vector in this direction with magnitude 9 so we take.

3v = 3<1,-2,2> = <3,-6,6>

So the point on the sphere where the ray pierces the surface of the sphere is:

(3,-4,5) + <3,-6,6> = (6,-10,11)
_____________________

The vector v is the normal vector to the plane tangent to the sphere at that point. So the equation of the tangent plane is:

(x - 6) - 2(y + 10) + 2(z - 11) = 0
x - 6 - 2y - 20 + 2z - 22 = 0
x - 2y + 2z - 48 = 0