minimal distance between two points and point on a plane

Question

I'm stuck on this problem for about an hour and on the edge of giving up. The problem goes as follow:

On the $YZ$ plane, find a point $P$ such that the sum of the distances from the points $A(2,4,6)$ and $B(6,2,2)$ to $P$ would be minimal.

I've tried to define a point to be $P(0,y,z)$, and just take the derivative of the sum of $|AP|+|AB|$ but I'm getting a function with two variables, and I'm kind of stuck from that point. Any ideas on how to continue will be great!

Answer 1

Let's consider the plane $\mathcal P$ orthogonal to $YZ$ and containing $(AB)$.

$\mathcal P$ intersects $YZ$ along a line $(HK)$ where $H=proj(A)=(0,4,6)$ and $K=proj(B)=(0,2,2)$.

Then I claim that $P\in(HK)$

For $M\in YZ$ and $N$ the orthogonal projection of $M$ onto $(HK)$ in the plane $YZ$ then $\vec{MA}=\vec{MN}+\vec{NA}$ so $MA^2=MN^2+NA^2+\underbrace{2\vec{MN}\cdot\vec{NA}}_{=0}$

So $MA^2\ge NA^2$ which is equivalent to $MA\ge NA$ since distances are positive.

Same property for point $B$, thus the minimum can only be realized on $(HK)$.

The equation of this line is $K+t\,\vec{KH}=(0,2+2t,2+4t)$

Now $P=(0,y,z)$ with $\begin{cases}y=2+2t\\z=2+4t\end{cases}$ depend of only one parameter.

$\vec{PA}=(2,2-2t,4-4t)$ and $\vec{PB}=(6,-2t,-4t)$

$f(t)=PA+PB=...=2\sqrt{6-10t+5t^2}+2\sqrt{9+5t^2}$

Then I find a minimum for $t=\frac 34$ whose value is $2\sqrt{21}$.

The point $P=(0,\frac 72,5)$.

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Edit

In fact I was thinking about something related to amd's method this afternoon, which is elliptic billard.

Remember also that if $F_1,F_2$ are focii of an ellipse then for any point $M$ on the ellipse, we have $MF_1+MF_2=\text{constant}$

I represented in the figure amd's idea, with the notations we have used above. The plan $(YZ)$ is only represented by a transversal section (thus just a line).

So if we consider the nested ellipsoids $AP+BP=\alpha$ we are searching an $\alpha$ such that the ellipsoid touches the plane YZ in one point $P$, so that it is tangent.

The tangency property makes it behave like an elliptic billard thus the construction of $B'$ such that the ray $\vec{BP}$ reflects into $\vec{PA}$ direction.

Thanks, amd for reminding me of this pretty idea.

Answer 2

This is a three-dimensional version of a 2-D problem that was posed quite recently and can be solved without resorting to calculus.

Consider the reflection $B'$ of the point $B$ in the $y$-$z$ plane. Obviously, $B'P=BP$, so the same point that minimizes $|AP|+|BP|$ also minimizes $|AP|+|B'P|$. The shortest distance between two points is along the straight line segment joining them, therefore the point $P$ that you seek is the intersection of $AB'$ with the $y$-$z$ plane.

We have $B'=(-6,2,2)$ so $AB'$ can be parameterized as $tA+(1-t)B'=(8t-6,2t+2,4t+2)$. Solving $8t-6=0$ for $t$ gives $t=\frac34$, so $P=\left(0,\frac72,5\right)$.