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 | 1 | initial version | posted 2014-07-01 10:22:44 -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.
| | 2 | No.2 Revision |
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.
| | 3 | No.3 Revision |
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!
