The minimal distance between a point in a plane and two points in the space

Question

The question is the following:

given the plane $\pi$ with equation : $2x − y + z − 2 = 0$ find a point M in the plane such that the sum of the distances from $P = (2, 1, 0)$ and $Q = (1, −1, 2)$ be minimal.

I know that the point $M$ lies on the line $HK$ from the projection of $PQ$ on the plane

$H = Proj \ \pi(P) = (2,1,-1)$
$K = Proj \ \pi(Q) = (1,-1,-1)$

Then define a line $s$ where our desired M point lies

$s = H + t * HK = (2,1,-1) + t * (1,2,0)$

thus $x = 2 + t, y = 1 + 2 t, z =-1$

$MQ^2 = (1-(2+t))^2 +(-1-(1+2t)^2 +(2-1)^2$ and $MP^2 = (2-(2+t))^2 +(1-(1+2t))^2+(-1)^2$

if I sum $MQ^2+MP^2$, it gives a polynomial with complex roots, is where I'm stuck right now. Could someone spot the mistake, or enlight me at other kinda solution? Thanks in advance.

Answer 1

There are two mistakes -

i) Projections of points $P$ and $Q$ in the plane are not correct.

ii) Minimizing $MP^2 + MQ^2$ does not necessarily minimize $|MP| + |MQ|$.

To find projection of $P (2, 1, 0)$ in plane $2x-y+z - 2 = 0$,

If the projection is $H \ (x_1, y_1, z_1)$ then $\vec {PH}$ is parallel to the normal vector to the plane.

$(x_1 - 2, y_1 - 1, z_1) = \lambda_1 (2, -1, 1)$

$x_1 = 2 + 2 \lambda_1, y_1 = 1-\lambda_1, z_1 = \lambda_1$

Plugging into equation of plane, $\lambda_1 = -\frac{1}{6}$ and point $H$ is $\left(\frac{5}{3}, \frac{7}{6}, - \frac{1}{6}\right)$.

Similarly find point $K$, which is projection of $Q$ in the plane.

Now if $M$ is on $HK$, minimize $|MP| + |MQ|$.

But here is an easier approach -

Check whether both points are on the same side of the plane or opposite sides. For that, plug in coordinates of $P$ and $Q$ in the equation of the plane $2x-y+z -2 = $ and see if both values have the same sign, which is indeed true.

As they are on the same side, we now find reflection of point $P$ about the plane, which is on the other side of the plane from $Q$.

We earlier found equation of $PH$ as $(2, 1, 0) + \lambda_1 (2, - 1, 1)$.

As $\lambda_1 = - \frac{1}{6}$ gave us point $H$, $\lambda_1 = - \frac{1}{3}$ will give us $P'$ which is reflection of $P$ about the plane.

So we get $P'$ as $ \ \left(\frac{4}{3}, \frac{4}{3}, \frac{1}{3}\right)$.

Now observe that for any point $M$ on the plane, $|PM| = |P'M|$.

Hence $ \ |PM| + |MQ| = |P'M| + |MQ|$

But $|P'M| + |MQ|$ is minimum if $P', M, Q$ are on a straight line.

Equation of line $P'Q$ is $(1, -1, 2) + \left(\frac{1}{3}, \frac{7}{3}, -\frac{7}{3} \right) t$

Plugging in coordinates of point on the line in the equation of the plane, we get $t = \frac{3}{4}$.

That gives us coordinates of $M$ as $\left(\frac{5}{4}, \frac{3}{4}, \frac{1}{4} \right)$.

Answer 2

If $P$ and $Q$ are on opposite sides of the plane, then $M$ is just the point of intersection of $\pi$ with the line $PQ$.

Otherwise, let $Q'$ be the reflection of the point $Q$ in the plane $\pi$. then $M$ is the point of intersection of $\pi$ with the line $PQ'$.

Answer 3

Another way to calculate the minimum distance.

Given a plane $\Pi\to (p-p_0)\cdot \vec n = 0$ and two points $(p_1,p_2)$ such that $\{p_1,p_2\}\notin \Pi$, first we determine if both $(p_1,p_2)$ are located into the same semi-space defined by $\Pi$ or if they are located each at separated semi-spaces. This is done verifying the sign for $(p_i-p_0)\cdot\vec n,\ i = \{1,2\}$. If they have the same sign, they are located at the same semi-space otherwise are located at complementary semi-spaces. Now if they are located at complementary semi-spaces, the minimum distance point is calculated as the intersection of the segment $s(\lambda)$ and $\Pi$ or as the solution for

$$ \cases{ (p-p_0)\cdot\vec n = 0\\ s(\lambda)= \lambda p_1+(1-\lambda)p_2 }\Rightarrow (\lambda p_1+(1-\lambda)p_2-p_0)\cdot\vec n = 0\Rightarrow \lambda_0 = \frac{(p_1-p_0)\cdot\vec n}{(p_2-p_0)\cdot\vec n} $$

given by $p_m = p_1+\left(\frac{(p_1-p_0)\cdot\vec n}{(p_2-p_0)\cdot\vec n}\right)(p_1-p_2)$ and the minimum distance distance is $\|p_1-p_m\|+\|p_2-p_m\|$.

Now is $\{p_1,p_2\}$ are located into the same semi-space we take a point, $p_2$ for instance, and after determining their reflected image regarding $\Pi$ called $p_2'$ we proceed as in the previous case.

NOTE

The reflected point is determined as follows.

First we calculate the intersection point $p_i$, $(p_2+\mu\vec n-p_0)\cdot \vec n = 0$ in $p_i = p_2 - \frac{(p_2-p_0)\vec n}{\|\vec n\|}\vec n$ and then we have $p_2'= p_2 - 2\frac{(p_2-p_0)\vec n}{\|\vec n\|}\vec n$