Calculus III:
Find an equation of a sphere if one of its diameters has endpoints
(3, 5, 6) and (5, 7, 8).

- k8lyn911

- schrodinger

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- misty1212

HI!!
this is not as hard as it looks

- misty1212

find the center first, the average in each coordinate

- misty1212

you got that?

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## More answers

- anonymous

anticipation is killing me...

- k8lyn911

Sorry! Working through some other homework problems.
Which equations do we use for that?

- misty1212

just like the midpoint in two dimensions
take the average in each coordinate

- k8lyn911

So (x1 - x2, y1 - y2, z1 - z2)?

- misty1212

so, the average

- misty1212

add up and divide by two

- misty1212

"no" i meant
just like finding the midpoint of \((3,5)\) and \((7,11)\) in two dimensions
add and divide by two in each coordinate
in my example the midpoint would be \((5,8)\)

- misty1212

clear or no?

- k8lyn911

( (3+5)/2, (5+7)/2, (6+8)/2) = (4, 6, 7)?

- misty1212

right

- misty1212

before we continue, do you remember how to find the equation of a circle given two endpoints of the diameter? this is identical, except with one extra coordinate

- misty1212

the equation is going to be \[(x-4)^2+(y-6)^2+(z-7)^2=r^2\]

- k8lyn911

Not exactly...

- misty1212

because the center is \((4,6,7)\)

- misty1212

equation for a circle with radius \(r\) and center \((h,k)\) is
\[(x-h)^2+(y-k)^2=r^2\]

- misty1212

this is exactly the same, but with one more coordinate

- k8lyn911

Oh, so it's like the other problems we've had, but backwards!
So we add these values squared and take the square root to find r?

- misty1212

ok lets go slow

- misty1212

maybe i am not sure what exactly you mean
we got the center, we need the radius

- misty1212

actually we don't need the radius, we just need the square of the radius, so forget the square root business

- k8lyn911

Okay.

- misty1212

same distance formula as with two points, only now each point has three coordinates instead of two

- misty1212

square of the distance between \((3, 5, 6) \) and \( (4, 6, 7)\) is
\[((4-3)^2+(6-5)^2+(7-6)^2\] a pretty easy calculation in this case

- misty1212

for the actual distance it would be \[\sqrt{(4-3)^2+(6-5)^2+(7-6)^2}\]

- misty1212

you get 3 pretty much instantly right?

- k8lyn911

Yes.

- misty1212

final answer:\[(x-4)^2+(y-6)^2+(z-7)^2=3\]

- misty1212

a nice picture
http://www.wolframalpha.com/input/?i=%28x-4%29%5E2%2B%28y-6%29%5E2%2B%28z-7%29%5E2%3D3

- k8lyn911

Oh, I see.
Thank you so much for explaining it to me! :)

- misty1212

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