Finding point on sphere furthest from point in space

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September 7th 2010, 12:19 PM
SyNtHeSiS

Finding point on sphere furthest from point in space

Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ 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?
September 7th 2010, 12:48 PM
Opalg

Quote:

Originally Posted by SyNtHeSiS

Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ 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 $\sqrt{99} = 3\sqrt{11}$ . You have already identified the direction of travel as the vector (1,–1,3), which has length $\sqrt{11}$ .

Those $\sqrt{11}$ s can't be coincidental, can they?
September 7th 2010, 01:29 PM
Soroban

Hello, SyNtHeSiS!

Quote:

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

$\text{which is furthest from the point }\,C(1,3,4).$

Code:

* * * o C * * * * ♥ P * * * * * * * * o * * * A * * * * * Q ♥ * * * * * * * *

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

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

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

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

Substitute into the equation of the sphere:

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

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

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

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

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

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

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

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

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

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

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

Quote:

Originally Posted by Soroban

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

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.
September 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 $x = (1,3,4) + t(1,-1,3)$ . 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 $\sqrt{99}$ 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).