Spheres/Will this work?/In Class #3

answered 2014-07-01 12:09:42 -0600

Justin
1009 ●2 ●20 ●38

updated 2014-07-01 16:53:11 -0600

I agree with Tiffany about everything except the radius and $\vec u$ and $\vec v$. Just for the sake of completion, I will go over all my steps.

The general equation for a parametrized circle is as follows:

$$ \vec p(t) = \vec c + r \cos(t) \vec u + r \sin(t) \vec v $$

The first piece of information that we must find is an equation describing the plane that the circle lies in. We do this by setting the equations of the spheres equal to each other. Since the radii are both equal to 1, we don't have to wrry about adding or subtracting constants.

$$ x^2 + y^2 + z^2 = (x-1)^2 + (y-1)^2 + (z-1)^2 $$ $$ x^2 + y^2 + z^2 = (x^2 - 2x + 1) + (y^2 - 2y + 1) + (z^2 - z + 1) $$

After rearranging the terms, we obtain:

$$ - 2x - 2y - 2z = -3 $$

This is the equation for our plane. This problem can now be thought of as the intersection of the unit sphere and this plane.

The next piece of information we need is the normal vector to the plane. This is easy!

$$ \vec n = \langle -2, -2, -2 \rangle $$

Now we can find $\vec u$ and $\vec v$. To find $\vec u$, I will choose a random vector that is perpendicular to $\vec n$ (i.e. one that lies on our plane).

$$ \vec u = \langle 0, 1, -1 \rangle $$

Note that $\vec n \cdot \vec u = 0$. We will now cross $\vec n$ with $\vec u$ to find a vector that is perpendicular to both of them.

$$ \vec v = \vec n \times \vec u = \langle 4, -2, -2 \rangle $$

Note that if you take the dot product between any of these two vectors, you will get $0$. Next, we must normalize these vectors (divide them by their magnitude).

$$ \vec u = \langle 0, 1, -1 \rangle / \sqrt{2} $$ $$ \vec v = \langle 4, -2, -2 \rangle / \sqrt{24} $$

Next we will find the center of this circle. The center of the circle must lie on our plane, so we will find when the line that has a direction vector as the normal of the plane intersects the plane. This point of intersection is our center. Note that in our line, $p_0 = \langle 0, 0, 0 \rangle$.

$$ \vec l(t) = \vec n t = \langle -2, -2, -2 \rangle t = \langle -2t, -2t, -2t \rangle $$

If we plug in the components of this vector into our plane, we will find the $t$ value of the intersection.

$$ -2 (-2t) - 2(-2t) - 2(-2t) = -3 $$ $$ 4t + 4t + 4t = -3 $$ $$ t = -\frac{1}{4} $$

We will now substitute in $-\frac{1}{4}$ for $t$ in our line.

$$\vec c = \vec l \left(-\frac{1}{4} \right) = \left\langle \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right\rangle $$

Now we have everything except the radius. I admit, I am completely drawing a blank on how to find the radius. I've thought about it for a while, but I just have no idea. Experimental evidence (Grapher on Mac OS X) has concluded that the radius is equal to $\frac{1}{2}$, but why? If someone could clarify, I would be so happy!

This is the circle with a radius of $0.5$:

As it stands, here is our equation:

$$ \vec p(t) = \left\langle \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right\rangle + r\cos(t)\frac{\langle 0, 1, -1 \rangle}{\sqrt{2}} + r\sin(t)\frac{\langle 4, -2, -2 \rangle}{\sqrt{24}} $$

So if anyone knows how to find the radius... please let me know!

Update

cbass, you are amazing on so many levels. I now see why the radius is $\frac{1}{2}$. This has been haunting me for hours haha! Thanks a ton!

answered 2014-07-01 14:54:50 -0600

cbass
96 ●4 ●5

I agree with your solution Justin, and your computer program got the same answer as my 3d drafting program, lol. To find it mathematically we can deduce that the plane is halfway between the two circles, and the distance between the centers of the circles is the magnitude of the vector pointing from the center of the first to the center of the second. let $\vec w$ be equal to this vector and we find $$\vec w= \langle 1,1,1\rangle$$ Now we can find the distance from the center of the unit circle to the plane by taking half the magnitude of $\vec w$$$\frac{1}{2}\|\vec w\|=\frac{1}{2}\sqrt{1^2+1^2+1^2}=\frac{1}{2}\sqrt{3}$$
We also know that the distance form the origin to anywhere on the unit circle is the radius 1. puting these values into the graph below we see

I didnt have a 3d program i could easly write over, so you have to ignore the 2 dimensional part, but it still allows us to see that to compute the radius we can use pythagorean theorem$$a^2+b^2=c^2\to a^2=c^2-b^2\to a=\sqrt{c^2-b^2}$$ Now substituting $a=r,b=\frac{\sqrt{3}}{2},c=1$ we get$$ r=\sqrt{1^2-\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{1-\frac{3}{4}}=\sqrt\frac{1}{4}=\frac{1}{2}$$ Plugging this into your equation above we get an answer of $$\vec p(t) = \left\langle \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right\rangle + \frac{1}{2}\cos(t)\frac{\langle 0, 1, -1 \rangle}{\sqrt{2}} + \frac{1}{2}\sin(t)\frac{\langle 4, -2, -2 \rangle}{\sqrt{24}}$$

answered 2014-07-01 10:22:44 -0600

Tiffany
656 ●3 ●18 ●34

updated 2014-07-01 12:17:14 -0600

I'm not sure if thats actually how its done or not, I haven't worked farther than setting them equal as you just did, however I came up with

-2x -2y -2z = -3

or 2x +2y+2z = 3

when you square out the right side of your equation, you should get +1 +1 +1 which would put the +3 on the right, and subtract it to the left should either give you -3 with all others being negative, or +3 if you move the xyz to the left.

I'm going to continue to work on this problem, and will update from here.

So far what i've found is that subtracting the two equations gives the plane of intersection. Subtracting the two equations stills gives the 2x+2y+2z=3 . So from notes in class on 6-30-14, he showed us how to describe the intersection between the unit sphere and a plane. First we need to find the center. To find the center you take a normal vector from the plane, which in this case is $\langle 2,2,2\rangle$ now we set $$\vec l(t) = \langle2,2,2\rangle t$$ multiplying out, we get $$\langle 2t,2t,2t\rangle$$

plugging this back into the equation of the plane and solving for t we get: $$ 2(2t)+2(2t)+2(2t)=3$$ $$= 4t+3t+3t=3$$ $$=12t=3$$ $$=t=\frac{3}{12}$$ $$=t=\frac 14$$

plugging this back into the equation of the plane, we can find the center of the circle of intersection.

$$ \vec c = \langle2(\frac14),2(\frac14),2(\frac14)\rangle $$ which simplifies to $$ \vec c = \langle \frac12, \frac12, \frac12\rangle$$

so now we have the center of the intersection. So from here we need to find the radius, since the radius of both spheres is 1, the radius of the circle that they form when intersecting will also be 1.

As far as $\vec u$ and $\vec v$. From what I understood from the in class problem yesterday in my group, they follow the vectors $\vec i, \vec j, \vec k$. So this would lead me to believe that the equation we are looking for is...

$$ \vec p (t) = \vec c + r\cos(t)\vec u + r\sin(t) \vec v $$ $$ \vec p (t) = \langle \frac12, \frac12, \frac12 \rangle + 1 \cos(t) \langle 1,0,0 \rangle +1\sin(t) \langle 0,1,0 \rangle$$ $$ \vec p (t) = \langle \frac12, \frac12, \frac12\rangle + \cos(t) \langle 1,0,0 \rangle + \sin(t) \langle 0,1,0 \rangle $$

I'm not exactly sure if that is right, since I still don't exactly understand the $\vec u $ and $\vec v$. So hopefully someone else can help clarify a bit more on that.

** Awesome on explaining the U and V, I now understand where he got those from when he worked an equation like this on the board! Thanks Justin!