For #2
(x-3)^2 + (y-8)^2 + (z-1)^2 = r^2 is the equation for all points lying on the sphere including(4,3,-1)
substitute (4,3,-1) for x, y, and z to find r
just got into calc3, the professor taught us one class and gave us this homework which I don't feel he prepared us for.. I'm trying here, but I'm stuck on two problems
1) determine whether the points lie on a straight line
A (2,4,2); B (3,7,-2); C(1,3,3)
don't even know where to begin.
2) Find the equation of a sphere that passes through the point (4,3,-1) and has center (3,8,1)
now for this one, obviously the left side of the equation is going to look something like
(x-3)^2 + (y-8)^2 + (z-1)^2 = something
the back of the book says that it equals 30, and I don't know at all how to get this answer.
any helps or tips appreciated
nope, we didn't. my "trig" class was basically pre-calc, then went straight to calc1 and 2 without ever doing a question like this.
If you wouldn't mind, a little more explanation would help me get this, so tell me if I'm understanding this correctly:Consider the vector from (1,3,3) to (2,4,2) i + j -k
however from (2,4,2) to (3,7,-2) is i + 3j -4k
since they are not parallel the pts don't lie on the same line
you take the first two sets of points, (1,3,3) and (2,4,2)
and you take the difference of each of the points, so the difference between the Xs is 1, the Ys is 1, the Zs is -1
then you compare two different sets, so (2,4,2) and (3,7,-2)
the difference between the Xs is 1, Ys is 3, and Zs is -4
so since the numbers don't match (1,1,-1) versus (1,3,-4), then the points don't lie on the same line?
You have the right idea
Two vectors are parallel if one is a mutiple of the other
for eg the first is (1,1,-1) so if they other is a multiple,not necessarily
the same for eg (3,3,-3) they would still be parallel.
If you haven't learned it yet you will soon learn the vector
from (x1,y1,z1) to (x2,y2,z2) is v = (x2-x1) i + (y2-y1) j +(z2-z1)k
you'll use this a half a million times before calc 3 is over