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Lines in 3D - Problem 1
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Let’s find the vector and the parametric equations of a line through two given points. The points a; which is 4, -1, 2, and b; which is <6, 2, 3>. Now recall that the vector equation is of the form r equals r0 plus t times v. And recall that r0 is a position vector for one of the points that you know, is on the line, and v is the direction vector of the line.
Let’s start by defining r0 as the position vector of this point a. So that will be <4, -1, 2>. And the direction vector, you can always define if you’re given two points you can always define as the vector from one of the two points to the other. I’ll define it as vector AB. That’s going to be 6 minus 4, 2. 2 minus -1, 3, and 3 minus 2, 1. That means that the vector equation is going to be <x, y, z>, these are the components of vector r, equals r0 <4, -1, 2> plus t times my direction vector <2, 3, 1>. This is the vector equation.
How do we get the parametric equation? We just look at the components, so we get x equals 4 plus 2t. y equals -1 plus 3t and z equals 2 plus t. These three are the parametric equations for my line. Now it turns out that there is one more form for the equation of a line in space.
You get it by eliminating the parameter. So I take my parametric equations; x equals 4 plus 2t, y equals -1 plus 3t, and z equals 2 plus t. I eliminate the parameter. Starting with the x equation. I subtract 4, I get x minus 4 equals 2t, and then I divide by 2. I get t equals x minus 4 over 2. In this equation I can do the same thing. I add 1 to both sides, y plus 1 equals 3t and then divide by 3. I get t equals y plus 1 over 3.
Finally all I have to do to this one is just subtract 2. I get t equals z minus 2. So you get this new form of the equation by observing that all of these expressions equal t. So you can set them all equal to each other. X minus 4 over 2 equals y plus 1 over 3, equals z minus 2, which can also be written as z minus 2 over 1.
This is the symmetric equation of the line, notice it doesn’t involve any parameters, or vectors. So this is a third way to express the equation of a line, and you’ll also notice that it’s very easy to derive this from the parametric equations. I would always come up with the vector equation of a line first then parametric and then the symmetric equation.
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