# Finding point on sphere furthest from point in space

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• Sep 7th 2010, 12:19 PM
SyNtHeSiS
Finding point on sphere furthest from point in space
Find the point on the sphere $\displaystyle (x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).

Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
• Sep 7th 2010, 12:48 PM
Opalg
Quote:

Originally Posted by SyNtHeSiS
Find the point on the sphere $\displaystyle (x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).

Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?

Think about it geometrically. To get from C to the furthest point on the sphere, you have to go from C to the centre of the sphere and then continue in the same direction for a distance equal to the radius of the sphere, which is $\displaystyle \sqrt{99} = 3\sqrt{11}$. You have already identified the direction of travel as the vector (1,–1,3), which has length $\displaystyle \sqrt{11}$.

Those $\displaystyle \sqrt{11}$s can't be coincidental, can they?
• Sep 7th 2010, 01:29 PM
Soroban
Hello, SyNtHeSiS!

Quote:

$\displaystyle \text{Find the point on the sphere }\,(x - 2)^2 + (y - 2)^2 + (z-7)^2 \:= \:99$

$\displaystyle \text{which is furthest from the point }\,C(1,3,4).$
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              * * *        o C           *          *  *         *              ♥ P       *              *  *                     *       *          *      *       *        o        *       *      * A        *             *       *  *              *       Q ♥              *       *  *          *     *        * * *

Construct a line through $\displaystyle \,C$ and the center of the sphere $\displaystyle \,A.$

It will intersect the sphere in two points, $\displaystyle \,P$ and $\displaystyle Q.$
. . We want $\displaystyle Q$, the point further from $\displaystyle \,C.$

The line from $\displaystyle C(1,3,4)$ to $\displaystyle A(2,2,7)$ has direction $\displaystyle \langle 1,-1,3\rangle$

The line has parametric equations: .$\displaystyle \begin{Bmatrix}x &=& 1 + t \\ y &=& 3 - t \\ z &=& 4+3t \end{Bmatrix}$

Substitute into the equation of the sphere:

. . $\displaystyle \left[(1-t) - 2\right]^2 + \left[(3+t)-2\right]^2 + \left[([4-3t)-7\right]^2 \:=\:99$

. . $\displaystyle 11t^2 + 22t + 11 \:=\:99 \quad\Rightarrow\quad t^2 + 2t + 1 \:=\:9$

. . $\displaystyle t^2 + 2t - 8 \:=\:0 \quad\Rightarrow\quad (t-2)(t+4)\:=\:0 \quad\Rightarrow\quad t \:=\:2,\:-4$

$\displaystyle \begin{array}{ccccccc}t = 2\!: & (x,y,z) &=& (\text{-}1,5,\text{-}2) & \text{point J} \\ t = \text{-}4\!: & (x,y,z) &=& (5,\text{-}1,16) & \text{point K}\end{array}$

I suspect that point $\displaystyle \,K$ is further from $\displaystyle \,C$, but let's make sure.

Distance from $\displaystyle C(1,3,4)$ to $\displaystyle J(\text{-}1,5,\text{-}2)\!:$

. . $\displaystyle \overline{CJ} \;=\;\sqrt{(\text{-}1-1)^2 + (5-3)^3 + (\text{-}2-4)^2} \;=\;\sqrt{44}$

Distance from $\displaystyle C(1,3,4)$ to $\displaystyle K(5,\text{-}1,16)\!:$

. . $\displaystyle CK \;=\;\sqrt{(5-1)^2 + (\text{-}1-3)^2 + (16-4)^2} \;=\;\sqrt{176}$

Hence, $\displaystyle K$ is the point we are seeking.

Therefore: .$\displaystyle Q(5,\text{-}1,16)$
• Sep 8th 2010, 07:30 AM
SyNtHeSiS
Quote:

Originally Posted by Soroban
The line from $\displaystyle C(1,3,4)$ to $\displaystyle A(2,2,7)$ has direction $\displaystyle \langle 1,-1,3\rangle$

Thanks for the great post. I was just curious why is it that choosing a line from A(2,2,7) to C(1,3,4) with direction <-1,1,-3> equivalent to the quote above? I thought this vector be the same magnitude but opposite, since points A and C is switched around.
• Sep 8th 2010, 09:13 AM
Opalg
Just to point out that the makings of a much simpler solution were already in the original post, where the equation of the line from C to the centre of the sphere is given as $\displaystyle x = (1,3,4) + t(1,-1,3)$. If you put t=1 in that equation then you get the point (2,2,7) (the centre of the sphere). To get to the furthest point of the sphere you have to travel a further distance along that same line. The extra distance that you need to travel is $\displaystyle \sqrt{99}$ units, which is three times the length of the vector (1,–1,3).

So to get to the furthest point of the sphere you need to take t = 4 (one unit of t to take you to the centre of the sphere, and another three units of t to take you to the furthest point). The furthest point is therefore at (1,3,4) + 4(1,–1,3) = (5,–1,16).