Parametric Equations
In linear algebra, parametric equations are used to express the coordinates of the points that lie on a line or a curve using a parameter. For example, the line through the points (1,0,0) and (0,1,0) can be described by the parametric equation \[ \mathbf{r_1}(t) = (1-t, t, 0) \]Here, t is the parameter that varies along the line. As t changes, the point (1-t, t, 0) will move from (1,0,0) (when t=0) to (0,1,0) (when t=1). Similarly, the parametric equation of the line where x = y = z can be written as \[ \mathbf{r_2}(s) = (s, s, s) \]Here, s is the parameter that varies, and as s changes, the point (s, s, s) moves along this line.
Distance Formula
The distance formula helps us to measure the distance between two points in space. For two points (x1, y1, z1) and (x2, y2, z2), the distance is calculated as: \[ d = \sqrt{(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2} \]In our problem, we are finding the distance between points on the two lines represented by \[ \mathbf{r_1}(t) = (1-t, t, 0) \] and \[ \mathbf{r_2}(s) = (s, s, s) \]. This gives us the distance formula: \[ d(t,s) = \sqrt{(1-t-s)^2 + (t-s)^2 + s^2} \].
Partial Derivatives
To find the minimum distance, we use the concept of partial derivatives. A partial derivative of a function of several variables is the derivative with respect to one of those variables while keeping the others constant. For the distance function \[ d^2(t,s) = (1-t-s)^2 + (t-s)^2 + s^2 \], we need to minimize it by taking partial derivatives with respect to t and s: \[ \frac{\partial d^2}{\partial t} = -2 + 2t = 0 \] Solving this gives t = 1. Similarly, \[ \frac{\partial d^2}{\partial s} = -2 + 6s = 0 \] Solving this gives s = \frac{1}{3}. These values represent the critical points that we need to analyze to find the minimum distance.
Critical Points
Critical points are the values of the variables at which a function’s derivative is zero or undefined. In the context of our problem, these points indicate potential minima or maxima of the distance function. By setting the partial derivatives to zero, we found: \[ \frac{\partial d^2}{\partial t} = -2 + 2t = 0 \rightarrow t = 1 \] \[ \frac{\partial d^2}{\partial s} = -2 + 6s = 0 \rightarrow s = \frac{1}{3} \] To confirm that these are indeed minimum points, one must check the second derivatives or substitute back to see the distance. Substituting t = 1 into \( \mathbf{r_1}(t) \) gives (0,1,0), and substituting s = \frac{1}{3} into \( \mathbf{r_2}(s) \) gives (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}). These are the points on each line closest to each other.
