distance between two points $A$ and $B$, one point is set parametrically

Question

$A$ is a point set parametrically with $x = 2u+4v$, $y = 2u+2v$, $z = 3u+4v$. Point $B$ is given by $(-5, -1, -3)$.

I need to find the smallest distance between points $A$ and $B$.

Answer 1

In general, the squared-distance between points $(a,b,c)$ and $(d,e,f)$ is

$$d^2 = (a - d)^2 + (b - e)^2 + (c - f)^2.$$

We want to minimize the squared-distance (which is the same as minimizing the square-root distance, since it's a monotonic transformation, since the square root is positive). Between $A$ and $B$ we have the squared-distance is

$$ \begin{align*} d^2 &= (x - (-5))^2 + (y - (-1))^2 + (z - (-3))^2 \\ \\ &= (2u + 4v + 5)^2 + (2u + 2v + 1)^2 + (3u + 4v + 3)^2. \end{align*}$$

To minimize a multivariable function (of $u$ and $v$ here), we take the gradient and set it to $0$ to find critical points. We use the chain rule and get

$$ \begin{align*} \frac{\partial d^2}{\partial u} &= 4(2u + 4v + 5) + 4(2u + 2v + 1) + 6(3u + 4v + 3) \\ \\ &= 34u + 48v + 42 \\ \\ \frac{\partial d^2}{\partial v} &= 8(2u + 4v + 5) + 4(2u + 2v + 1) + 8(3u + 4v + 3) \\ \\ &= 48u + 72v + 68 \end{align*} $$

Thus we set them to $0$ to get the system of equations $$\begin{cases} 34u + 48v + 42 = 0 \\ 48u + 72v + 68 = 0 \end{cases}$$

Whatever method you use to solve the system, we get the only critical point as $u = \frac{5}{3}$ and $-v = \frac{37}{18}$. To verify this is indeed a minimum, we need to do the second derivative test. We take the second partials (i.e., the Hessian) to get $$ \begin{align*} \frac{\partial^2 d^2}{\partial u^2} &= 34 \\ \\ \frac{\partial^2 d^2}{\partial v^2} &= 72 \\ \\ \frac{\partial^2 d^2}{\partial u \partial v} = \frac{\partial^2 d^2}{\partial v \partial u} &= 48 \\ \\ \end{align*} $$

Notice that $\frac{\partial^2 d^2}{\partial u^2} > 0$. The determinant is $$ D = \frac{\partial^2 d^2}{\partial u^2} \frac{\partial^2 d^2}{\partial v^2} - \left(\frac{\partial^2 d^2}{\partial u \partial v}\right)^2 = 144 > 0.$$

Since these two conditions hold, the second derivative test does indeed confirm this point minimizes the squared-distance, and therefore also the distance, between the points.

To finally get the distance, we just plug in $u$ and $v$ into the distance formula and get

$$d^2 = (2u + 4v + 5)^2 + (2u + 2v + 1)^2 + (3u + 4v + 3)^2 = \frac{1}{9} \implies d = \frac{1}{3}.$$

Answer 2

Point $A$ can be expressed as

$ A = u (2,2,3) + v (4,2,4) $

And this a parametrization of the plane passing through the origin, with a normal vector given by

$ N = (2, 2, 3) \times ( 4,2,4) = ( 2, 4 , -4 ) = 2 ( 1, 2, -2) $

Since $ N$ is a scalar multiple of $(1, 2, -2)$ we can take $N$ to be $(1, 2, -2)$

Hence, point $A$ is on the plane $ x + 2 y - 2z = 0 $

Now the minimum distance between $A$ and $B$ is the distance between $B = (-5, -1, -3)$ and the plane containing $A$, and this is given by the distance formula between a point and a plane which is

$ d = \dfrac{ |(-5) + 2 (-1) - 2( - 3)|}{\sqrt{ 1^2 + 2^2 + (-2)^2 }} = \dfrac{ 1 }{3 } $

And this is the desired minimum distance between $A$ and $B$.