Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

3 spheres are on $z=0$ plane and touch each other as shown in the picture. Coordinates of their centers are $O_1=(0,0,5),O_2=(0,y_2,3),O_3=(x_3,y_3,2)$. What is the tangent plane equation on 3 spheres? ($ax+by+cz=d$)

Thanks for answers.

share|improve this question
    
You're going to need to give the radii too... –  Peter Taylor Apr 9 '12 at 20:34
    
The radii are already given by the coordiantes of the centres. Interesting question.... One thing to note is that $(a,b,c)$ is the normal vector to the plane $ax+by+cz=d$, so the problem might be simplified by trying to find the normal vector first. –  you Apr 9 '12 at 20:35
1  
Reminds me of this proof without words on MathOverflow. Which implies that if you extend a line passing through the centers of any two of the spheres, the point where it meets the $z=0$ plane must also lie on the desired plane. Not sure if that leads anywhere, but it's interesting. –  Rahul Apr 9 '12 at 20:46
    
@you, I missed the "touch each other". Thanks. –  Peter Taylor Apr 10 '12 at 10:22
add comment

3 Answers

up vote 4 down vote accepted

First solve some quadratic equations to find $y_2$, $x_3$, and $y_3$.

Then, since each of the spheres is tangent to $z=0$ as well as to the mystery plane, its center must lie on the plane that bisects the angle made between these two planes. We can compute an equation for bisecting plane, because that is defined by the centers.

Now reflect the plane $z=0$ about the bisecting plane.

share|improve this answer
add comment

Isn't $z=0$ a tangent plane, by construction?

share|improve this answer
1  
I'm going to go out on a limb and guess that they want the other one, like the diagram suggests. –  Rahul Apr 9 '12 at 20:27
    
I need to find the other one on them as shown picture –  Mathlover Apr 9 '12 at 20:27
    
Oh, I see. I'm being dim today. –  Johann Hibschman Apr 9 '12 at 20:33
add comment

I get: -150.235 x-89.3299 y-128.83 z+1729.87=0 for the tangent plane equation. (Close approximation.)

Tangent points:

Sphere #1: {3.4594896600074643,2.05701926455194,7.966581035101458`}

Sphere #2: {2.0756918905582737,8.980177236680767,4.779953859745328`}

Sphere #3: {5.375454589335715,5.728584817849962,3.1866359029829208`}

Sphere center points:

Sphere #1 center point: {0, 0, 5}

Sphere #2 center point: {0., 7.74597, 3.}

Sphere #3 center point: {3.99166, 4.90578, 2.}

I calculated the tangent plane equation by using a plane thru 3 points. The first 2 points are apexes of 2 cones that envelope sphere #1 and sphere #2, then sphere #1 and sphere #3. Think of the line that connects the apexes as a hinge line that lies in the tangent plane. Select a third point whose x and y coordinates will be close to the tangent point on sphere #1. Reduce or add to the z coordinate as required to lower or raise the tangent plane. Plot each step as a visual aid. Using Mathematica, I used NMinimize to tell me how close the tangent plane was to sphere #1. When I was satisfied with the results, I took the coordinates as the tangent point. Repeat this with the other 2 sphere equations and the tangent plane equation to obtain the other 2 tangent points.

Regards,

Bill W.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.