Lines in 3D - Problem 2

Let’s do another problem that involves the vector equation of a line. Here, lineLis given by 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 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>.