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Lines in 3D - Problem 2

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s do another problem that involves the vector equation of a line. Here, lineLis given by <x, y, z> equals <-2, 1, 0> plus t, times the direction vector <4, 1, -1>. The first question is, find parametric equations for l. That’s pretty easy. You just look at the components. You get x equals -2 plus 4t. Y equals 1 plus t, and z equals 0 plus –t. z equals –t.

Part b says, identify two points on L. So I’m going to go the easy route. I’ll identify the point where t equals 0 and t equals 1. That’s probably the easiest two points to plug in. When t equals 0, I get, these are all going to be 0, so I'm going to get <-2, 1, 0>. When t equals 1, I get -2 plus 4, 2. I get 1 plus 1, 2 and I get -1. Here are two points I know. You would want to find two points, if for example you wanted to graph the line. That would help you graph the line.

Part C; which of these points are on L? If you need to check to see if a point is on the line, one way to do it is to look at the very first coordinate of the point and see if the value of t that gets you that x coordinate will get you the y and z coordinate you need as well. Looking at -10 I could get -10 out of this. If t equals -2, I’d have -2, plus -8. Let’s try that. So I plug in t equals -2. I get -2 plus -8, -10. 1 plus -2, -1, and z equals the opposite of -2, +2. That’s exactly point A. So that one is on the line.

Now what about point B? To get 18, I have to add 20 to -2. So t would have to be 5. So I'll get 20 for x. What about for y? 1 plus 5, 6, and for z -5. This should be 18. This actually is point B so that’s on the line.

What about point C? It starts with 14. How do I get 14? I’ve got to add 16 to -2, so t would be 4. And so I’ll get 14. When t equals 4, I get 1 plus 4, 5, and -4. Unfortunately that’s not the same, not the same point. Not equal to c, so c is not on the line.

This is an interesting part. It says write a vector equation of the line through <2, 1, 0> that is parallel to L. All you have to do to find a line parallel to L is use the same direction vector. The direction vector was <4, 1, -1>. That’s all we need to do is use this direction vector. So I’ll have <x, y, z> equals and I’ll use this as my initial point, <2, 1, 0> plus t times <4, 1, -1>. Same exact direction vector and there you go. Really easy. This will be a line parallel to L that passes through <2, 1, 0>.

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