# Math Help - Shortest distance plane, sphere [difficult]

1. ## Shortest distance plane, sphere [difficult]

Given the sphere
$(x-3)^2 + (y-4)^2 + z^2 = 36$

And the plane
$3x-2y+4z=8$

Find the shorterst distance from the center of the sphere to the plane.
___________
This is what I've tried:

Center of the sphere $C(3, 4, 0)$

The normal vector of the plane $\vec n =[3, -2, 4]$

I chose an arbitrary point on the plane $P(x, y, z)$

Which gives us;
$[x, y, z]=[3, 4, 0]+[3, -2, 4]t$

So the distance should be
$\rm{D} = \sqrt{(3+3t)^2 + (4-2t)^2 + (4t)^2}$

Where;
$f(t) = 29t^2 + 2t + 25$

Which gives;
$f\prime(t) = 58t+2$

And
$f\prime(t) = 0$

$t = -\frac{2}{58}=\underline{-\frac{1}{29}}$

So
$f(-\frac{1}{29}) = 29(-\frac{1}{29})^2 + 2(-\frac{1}{29}) + 25 = \underline{\frac{724}{29}}$

$\rm{D} = \sqrt{\frac{724}{29}}$

But this is not correct according to my book. What am I doing wrong?

2. Originally Posted by MatteNoob
Given the sphere
$(x-3)^2 + (y-4)^2 + z^2 = 36$

And the plane
$3x-2y+4z=8$

Find the shorterst distance from the center of the sphere to the plane.
___________
This is what I've tried:

Center of the sphere $C(3, 4, 0)$

The normal vector of the plane $\vec n =[3, -2, 4]$

I chose an arbitrary point on the plane $P(x, y, z)$

Which gives us;
$[x, y, z]=[3, 4, 0]+[3, -2, 4]t$
- Write the line as a vector function in parametric form.
- Find the limits of t so the line will be limited between (3,4,0) and the plane.
- The length is $\int_a^b |V|~dt$ where a and b are the limits of t, and V is the velocity of the line (derivative of the line function).

3. Find the shorterst distance from the center of the sphere to the plane.
Don't make it hard.
If $A$ is a point not on the plane $\Pi$ and $P \in \Pi$ then the distance from $A$ to $\Pi$ is $D(\Pi ,A) = \frac{{\left| {\overrightarrow {PA} \cdot n} \right|}}{{\left\| n \right\|}}$.

You can find a point on the plane. You already know the center and normal.

4. Before there was calculus or vector notation, there was geometry and algebra.

Plane: 3x - 2y + 4z - 8 = 0
Point: (3,4,0)

$\frac{|3(3) - 2(4) + 4(0) - 8|}{\sqrt{3^{2}+(-2)^{2}+4^{2}}}$

You will pardon me if I updated the notation from the ancient days.

5. Hello, MatteNoob!

You started off great . . . I'd take a different direction.

Given the sphere: . $(x-3)^2 + (y-4)^2 + z^2 \:= \:36\;\;{\color{blue}[1]}$

and the plane: . $3x-2y+4z\:=\:8\;\;{\color{blue}[2]}$

Find the shortest distance from the center of the sphere to the plane.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This is what I've tried:

Center of the sphere: . $C(3, 4, 0)$

The normal vector of the plane: . $\vec n \:=\:[3, -2, 4]$ . . . . Good!
The equation of the line through $C$, perpendicular to the plane: . $\begin{array}{ccc}x &=& 3+3t \\ y &=& 4 - 2t \\ z &=&0 + 4t \end{array}\;\;{\color{blue}[3]}$

Now where does this line intersect the plane?

. . Substitute [3] into [2]: . $3(3+3t) - 2(4-2t) + 4(4t) \:=\:0$

. . Solve for $t\!:\;\;t \:=\:-\frac{1}{29}$

Substitute into [3]: . $\begin{array}{ccccc}x &=& 3 + 3\left(\text{-}\frac{1}{29}\right) &=& \frac{84}{29} \\ \\[-3mm]
y &=& 4 - 2\left(\text{-}\frac{1}{29}\right) &=& \frac{118}{29} \\ \\[-3mm]
z &=& 4\left(\text{-}\frac{1}{29}\right) &=& \text{-}\frac{4}{29} \end{array}$

Hence, $P\left(\frac{84}{29},\:\frac{118}{20},\:\text{-}\frac{4}{29}\right)$ is the nearest point on the plane to center $C.$

Now, find the distance: . $\overline{CP}.$

6. I chose an arbitrary point on the plane $P(x, y, z)$

Which gives us;
$[x, y, z]=[3, 4, 0]+[3, -2, 4]t$
Is this really a description of points in the plane? It appears to me that only one point in the plane fits this description.