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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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 an exercise to find the vector equation of a line. We start by considering the line passing through the point <-3, 5>, with a slope ½. This problem asks me first to sketch the line, which is pretty easy. Let me start with that. <-3, 5> is right here, and it’s got a slope of ½. So I just have to maybe go down 1, to the left 2, down 1, to the left 2. That’s another point on the line. Let me draw my line here.
What next? Identify the position vector r zero for the point <-3, 5>. Remember that a position vector is a vector that starts at the origin, and ends at the point given. This point was <-3, 5>. This is what the position vector looks like. From the origin to the point <-3, 5>. That’s my r zero. Let me actually write that down. R zero in component form that would be <-3, 5>.
Identify a direction vector V for the line. Going back to the line, let’s observe first of all that there are infinitely many direction vectors you could use. You just have to use a vector that goes in the same direction as the line, and the line goes to the right 2 and up 1. You could use this vector and that vector would be <2, 1>. You can tell just because you could easily translate it down here, 2 to the right up 1, and turn it into a position vector. This point has coordinates <2, 1> and that means that the components of your vector V are going to be <2, 1>.
Let me write that down. Vector V has components <2, 1>. Write a vector equation for the line. Recall that, the vector equation for a line is r equals r zero plus the scalar t times v. R zero is the position vector for <-3, 5> and v is our direction vector so I just fill all that in. Remember that r give you the location of any point on the line. You can writer r as , the position vector equals <-3, 5> that’s r zero, plus t times direction <2, 1>. That’s the vector equation for the line.
Let’s calculate a few points of this equation. For example t equals -1. We get equals <-3, 5>, t is -1 so this is minus <2, 1>. That will be -3 minus 2, -5 and 5 minus 1, 4. <-5, 4>, where is that on our line? <-5, 4> right here. That’s the point that corresponds to t equals -1.
T equals zero. We can probably guess what’s going to happen here, but let’s just do the calculation. <-3, 5> plus zero times <2, 1>. That’s juts going to be plus zero. You’ll get <-3, 5> and that’s our original point. This point here is the point that we get when t equals zero.
When t equals 2, we get <-3, 5> plus 2 times 2, 1. That’s going to be 2 times 2, 4 and 2 times 1, 2. That’s -3 plus 4, 1 and 5 plus 2, 7. Let’s find that point <1, 7>, that’s the point where t equals 2. This was t equals -1 and that’s t equals zero. You can see that as t advances the line’s drawn in this direction. That’s how the vector equation for a line works. You always have the form r equals r zero plus t times the direction of the vector.
Unit
Vectors and Parametric Equations