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

1. misty1212

HI!! this is not as hard as it looks

2. misty1212

find the center first, the average in each coordinate

3. misty1212

you got that?

4. anonymous

anticipation is killing me...

5. anonymous

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

6. misty1212

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

7. anonymous

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

8. misty1212

so, the average

9. misty1212

add up and divide by two

10. 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 $$(5,8)$$

11. misty1212

clear or no?

12. anonymous

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

13. misty1212

right

14. 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

15. misty1212

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

16. anonymous

Not exactly...

17. misty1212

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

18. misty1212

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

19. misty1212

this is exactly the same, but with one more coordinate

20. anonymous

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?

21. misty1212

ok lets go slow

22. misty1212

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

23. misty1212

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

24. anonymous

Okay.

25. misty1212

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

26. 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

27. misty1212

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

28. misty1212

you get 3 pretty much instantly right?

29. anonymous

Yes.

30. misty1212

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

31. misty1212
32. anonymous

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

33. misty1212

$\huge \color\magenta\heartsuit$