Lines in 3D
We could represent a line by two equations that define the two planes that intersect on the line. This is not the most convenient way to think about the line usually, because when you have these two equations how do you solve them? Instead there’s another representation of a line in which we visualize it as a point moving through space, with an arrow (called a vector) attached to it pointing in the direction it is moving (the velocity of the point. If you imagine a line extending into space, you can consider what happens when a point on that line moves in time. The line is the trajectory of the moving point. [Graph] A line, or trajectory, of a moving point can be represented by a parametric equation. Let’s say that we are looking at the line through. So to specify a line in space, we can do that by giving you two points on the line, or by giving you a point and a vector parallel to the line, for example. So let’s say that we have two points on the line (-1, 2, 2) and (1, 3, -1). Q_0=(-1,2,2) Q_1=(1,3,-1) It is good to have two points on a line, but there must be more points. We can find more points by using the first two points and extending them in both directions. Imagine that we have a point on the line that is moving left to right. At time zero, it is located at Q0. At t = 1, it is going to be at Q1. [Graph] The object travels at a constant speed, so at time t=1/2 it’s going to be here. At time +=2, it will be over there. In fact, the point did not start here: it started at that previous point in space. [Graph] Let’s say that a moving point, Q(t), starts at the origin and begins to move. At time t-0, Q(t) is at (0, 0). We know that it can move in any way we want, but probably the easiest way to formulate its motion is to use functions of time. In this case, our goal is to find functions that describe the position of Q(t) as a function oft. We will then be able to use these functions to determine any point on the line defined by Q(t). To solve this problem, you must plug into the value oft depending on when it is hit by the moving point. Perhaps it is easiest to do this if we make the point move at a constant speed on the line. That speed is chosen so that at time 1 it is at Q1. Q(t) = moving point at t=0 it s at Q_0, moves at constant speed on the line The question we want to answer is, what is the position at time t to the point Q oft? Well, we have an easy observation, which is that a vector from Q0 to Q oft is proportional to the vector from Q0 to Q1. And what’s the proportionality factor here? The answer is t. Q_0Q(t)->=t->Q_0Q_1-> At time t, the distance between Q0 and Q oft is equal to t times the vector from Q0 to Q1. This means that as t increases, the distance between these two points grows larger and larger. [Graph] We can determine the position of this moving point, because we can find the vector Q0Q1 from the coordinates of Q0 and Q1. To do so, we subtract the coordinates of Q0 from those of Q1. We therefore get the result that the vector Q0Q1 is (2, 1, -3). Q_0Q(t)->=t{2,1,-3} Let’s call x, y, and z the coordinates of the point that is moving on the line. Q(t) = (x t(). y t ), z t )) Then we get (x+1) of (t*2). x(t) + 1 = t * 2 In this equation, we’re writing the components of vector Q0Q1 t. And here, we’re writing t times Q0Q1. Q_0Q(t)->=t->Q_0Q_1 The quantity t is equal to 2, and the quantity z is equal to minus 3t. y(t) -2 = t z(t) – 2 =- 3t The more familiar way to write these equations is y(t) = 2 + tz(t) = 2 – 3t. Or you can express it as Q(t) = Q0 + tQ0Q1. x (t) =- 1 + 2t y(t) = 2+ t z(t) = 2 – 3t Q(t)=Q_0+tQ_0Q_1-> The first parametric equation of a line in this class is x = 0, which we saw in the previous video. It’s not extremely hard to see that the parametric equations of lines will always look like that: x, y and z are functions oft but are of the form a constant plus a constant times t. The coefficients oft tell us about a vector along the line. Here we have the vector Q0Q1, which is (2, 1, -3). And the constant terms tell us about where we are at t equals 0. If I have t equals 0, these terms go away. We get minus (1, 2, 2). That’s the starting point.
