Find Equations for 2 spheres that are tangent to the planes

[Easy_as_Pi]

Homework Statement

Find the equations for two spheres that are tangent to the plane x+y-z=3 and x+y+z=9 and the line x=t , y=2t, z=3t passes through its center.

Preface: This problem was on a test I took yesterday. My professor handed it back today. The relevant equations and work are what he put on my test while grading it. So, I'm more trying to see why the method he chose to use is correct. This problem really threw me for a loop.

Homework Equations

He has the distance from a point to a plane equation: d= |Ax1+By1+Cz1-D|/sqrt(A^2+B^2+C^2). I'm not sure why this equation was selected- I think it has to do with the fact that we know the center of both spheres, so it would be our point, and the distance from it to the plane would be the radius for each sphere?

The Attempt at a Solution

His work is as follows:
x=t, y=2t, z=3t
(t,2t,3t) = center

t+2t-3t-3/sqrt3 = 3/sqrt3...not sure where the 3/sqrt3 on the right side came from. I know he just plugged in the t's to x+y-z=3, and then used the distance equation.

next I have: |6t-9|=3, solving for t we get t=2,1

plugging those into our center equation, we get (1,2,3) and (2,4,6)

Then, he just used the equation of a sphere, but had it equal to 3 for some reason.
(x-1)^2+(y-2)^2+(z-3)^2=3
(x-2)^2+(y-4)^2+(z-6)^2=3

Basically, if someone could guide me through this work a little bit, I would greatly appreciate it. The gears are turning, and the problem is starting to click, but I need a little push in the right direction.

Thanks!

 

[lanedance]

first geometrically, a plane tangent to a sphere, will have a normal pointing towards (or away from) the centre of the sphere, and the centre will be a distance r away.

now you know the centre of the sphere is some where on the line given by the parameter t
(t,2,3t)

for any given point on the line what is the minimum distance form that point to arbitrary plane?
ax+by+cz=d

 

[lanedance]

basically i think what your professor is doing is writing the distance from each plane for an arbitrary point on the line.

When they are equal, it is the centre point and you can solve for both the centrepoint and the radius

 

[lanedance]

so one way to find the minimum distance is to start at (t,2t,3t) and project a line in the direction of the plane which will be along the normal to the plane

the normal of the first plane is (normalising to length 1)
$$\frac{1}{\sqrt{3}}(1,1,-1)$$

the equation of the line that will intersect the plane at the minimum point is
$$(t,2t,3t)+ \frac{r}{\sqrt{3}}(1,1,-1)$$

the intersection point must lie on the plane
$$(t+2t-3t)+ \frac{r}{\sqrt{3}}(1+1-1)= 3$$
$$0 + \frac{r}{\sqrt{3}}= 3$$
$$r= 3\sqrt{3}$$

note the t's cancel so in fact the line is parallel to the first plane and you can solve directly for the radius, and use that with the other plane. we could have checked this directly with a dot product to start

 

[vela]

His work is as follows:
x=t, y=2t, z=3t
(t,2t,3t) = center

t+2t-3t-3/sqrt3 = 3/sqrt3...not sure where the 3/sqrt3 on the right side came from. I know he just plugged in the t's to x+y-z=3, and then used the distance equation.

Your professor did the same thing for the other plane to get the value on the RHS. The t's conveniently cancel out, leaving just the constant.