 ###### 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 2

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 another example. Consider the line passing through <-5,8> and <3,-4>. We’re first asked to sketch the line, so let’s take a look at the graph, <-5,8> would be, that’s -5; 1,2,3,4,5,6,7,8 is here. And then <3, -4> is here, 1, 2, 3, 4, that’s <3,-4> right there. Let me connect those two points and draw my line.

Second step: find a vector equation of the line. Now we need two things, we need a position vector r zero, for one of the two points the line passes through, and a direction vector, v. Let’s pick the left hand point, <-5, 8>, this point. And construct a position vector to it. Let me just draw that really quickly. It’s going to be strange looking because the line itself passes so close to the origin, but our position vector goes from the origin to a point. That’s the position vector for this point; let’s call this r sub zero. r sub zero would be <-5, 8>.

What about v; the direction vector? We could just make the direction vector, the vector from -5, 8> all the way down to the point <3, -4>. If we do that, then the components are going to be 3 minus -5 and -4 minus 8. That’s 8, -12. One vector equation for the line would be equals the initial point,<-5, 8> plus t times the direction vector <8, -12>.

Find the points corresponding to t equals zero, 1 and ½. Because two people doing the same problem may come up with different initial points and different direction vectors, you’re going to get different points corresponding to these. These are not going to be unique answers. For my vector equation let’s calculate these. T equals zero, we get equals <-5, 8> plus zero times, well zero times any vector is going to be the zero vector. We have the zero vector to <-5, 8>. We get <-5, 8>. We get that initial point when t equals zero.

What about t equals 1? We get <-5, 8> plus 1 times <8, -12> plus ½ of <8, -12>. This is going to be -5 plus 8, which is 3 and then 8 plus -12 , -4. <3, -4> is the other point that we use. This makes sense because, if you add the vector that connects this point, the first point with the second point, you’re going to get the second point. Let’s see what happens when you get t equals ½. is <-5, 8> plus ½ of <8, -12>. ½ of <8, -12> is <4, -6>. So we’re adding that to <-5, 8>. We add component-wise, we get -5 plus 4, -1. 8 plus -6, 2. When t equals ½, we get <-1, 2>.

Let’s just take a look our graph.<-1, 2> is right here. It turns out that that’s going to be exactly the midpoint between these two guys here. That was our t equals zero point, that was our t equals 1, and that’s our t equals ½.

There’re many ways to find a vector equation for a line, but one way is when you’re give two points, use one of the points as your initial point, get a position vector for it and then use the two points to create your direction vector. This is the form, r zero plus t times v.