# maximum distance from point to sphere

• Jan 31st 2011, 03:24 PM
harriette
maximum distance from point to sphere
I have to calculate the maximum distance from the point $\displaystyle T(1, -1, 3)$ to the sphere $\displaystyle x^2+y^2+z^2-6x+4y-10z-62=0$.

Here's what I did.
The equation for the sphere is $\displaystyle (x-3)^2+(y+2)^2+(z-5)^2=100$.
That means the center of the sphere is $\displaystyle C(3, -2, 5)$, and the radius is $\displaystyle r=10$.
I think the maximum distance $\displaystyle d_m$ is the distance from the point $\displaystyle T$ to the center of the sphere plus the radius:
$\displaystyle d_m=d(T, C)+r$.

Is this okay?
Thank you!
• Jan 31st 2011, 03:34 PM
Ackbeet
Everything looks good so far. What do you get?
• Jan 31st 2011, 03:39 PM
harriette
I get
$\displaystyle d((1, -1, 3), (3, -2, 5))=\sqrt((3-1)^2+(-2+1)^2+(5-2)^2 )=\sqrt(9)=3$ so $\displaystyle d_m=3+10=13$.

It seems too simple, are you sure I'm not thinking something terribly wrong?

• Jan 31st 2011, 03:49 PM
Ackbeet
You're not computing the distance formula correctly. It should be more like

$\displaystyle \sqrt{(3-1)^2+(-2+1)^2+(5-2)^2}=\sqrt{4+1+9}=\sqrt{14}.$

[EDIT]: Ignore this post. See below for correction.
• Jan 31st 2011, 03:52 PM
harriette
Why do we have (5-2)^2 instead of (5-3)^2? I'm afraid I don't understand.
• Jan 31st 2011, 03:56 PM
Ackbeet
I'm sorry. You're correct. It should be 3, like you got.
• Jan 31st 2011, 03:57 PM
harriette
Thank you so much! I'm quite insecure when I'm doing the exercises on my own, it really helps when someone checks your work.
• Jan 31st 2011, 04:25 PM
Ackbeet
If you feel insecure when you're doing exercises, I would come up with a system of checks you can run against your answers to justify them to yourself. It's good problem-solving strategy anyway.

You're very welcome. Have a good one!