 ###### 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.

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# The Vector Equation of a Line - Problem 3

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.

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We’re talking about finding the vector equation of a line. Here’s a problem, consider the points A, <-17, 8> and B, <-2, 17>. First, let’s find the vector equation for this segment AB. I first want to graph these two points, <-17, 8>, will be right, here is -17, up 4, 8. That’s <-17, 8>. That’s A and point B is <-2, 17>, that’s B.

Unlike in previous problems, I want to find an equation for just this segment. I can do it the same way I’ve done it before. I first want to find a position vector for one of the two points and I’ll choose point A. let me do that now. Let me first draw that vector, vector r zero, I’ll call it.

What are the components of that vector? They’re precisely the coordinates of point QA. Let me write that down. R zero is going to be <-17, 8>. The only other thing I need is a direction vector. What I’m going to do is I’m going to create a direction vector from A to B. I’m going to use the whole length of it. Remember how to get a vector from A to B. I’ll make vector v equal AB. All you have to do is subtract components. I take -2, minus -17. That’s -2 plus 17, 15. And then 17 minus 8, is 9. This is my direction vector, and my equation is going to be r equals r zero plus t times v, that’s the general equation. R in this case is the components of any point on the line r zero, <-17, 8> plus t times <15, 9>. This will be my equation.

What makes this the equation of a segment rather than the entire line? I have to make some kind of restriction on t. Let’s make the observation that when t equals zero, this whole term is going to go away. And I’m going to get <-17, 8> which is point A. How do I get point B? Remember that this is exactly the displacement between point A and point B. So if I added this to this, I would get point B, and I’m adding this <15, 9> to this, when t equals 1. I get point B when t equals 1, point A when t equals zero. I need t to go between zero and 1. This domain restriction makes it so that this vector equation gives me the line segment AB and not the whole line.

Let’s deal with the next part. Find the coordinates of the point 2/3 from A to B. 2/3 form A to B is one where around here, let’s say. Since we just finished finding the equation, the vector equation for this segment why don’t we use this? And I say that the point 2/3 the way form A to B happens when t equals 2/3. We get B when t equals 1 and we get A when t equals zero.

Let me plug 2/3 in and I get equals; my fist point<-17, 8> plus 2/3 of <15, 9>. That’s <-17, 8> plus, 2/3 of 15 is 10, 2/3 of 9 is 6. And so I get -17 plus 10, -7, 8 plus 6, 14. Let’s see how close I get. <-7, 14>; -7 is here, pretty close. That’s the point right there, <-7, 14>.

Now it says, find an equation for ray AB. Remember, this arrow is different from the vector arrow. The vector uses the half arrow. When you have a full arrow, that means the ray AB. We want the ray that starts at point A and goes through B and continues forever. All we have to do is make a little, tiny adjustment on the equation for this guy. I need the equation to start of at point A, which is <-17, 8>, and I need it to go through B which I get when t equals 1 and keep going. It makes sense that I want t greater than or equal to zero, but the same equation will work. equals -17, 8> plus t times <15, 9>, only now t is greater than or equal to zero. This restriction means I’ll get all the points from A onward and none of the points from the line to the left. Here is my equation for the ray AB.