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Find a vector equation and parametric equations for the line segment that joins $P$ to $Q$. $P(a, b, c), \quad Q(u, v, w)$

   Find a vector equation and parametric equations for the line segment that joins $P$ to $Q$.
$P(a, b, c), \quad Q(u, v, w)$
Single Variable Calculus: Early Transcendentals
Single Variable Calculus: Early Transcendentals
James Stewart,… 9th Edition
Chapter 13, Problem 24 ↓
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Find a vector equation and parametric equations for the line segment that joins $P$ to $Q$. $P(a, b, c), \quad Q(u, v, w)$
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Transcript

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00:01 So they want us to find a vector equation and the parametric equation that joins this line segment between the point abc and the point uvw.
00:11 So this follows pretty much like what we did in the previous questions.
00:16 So first, what we want to do is find our vector going from p to q.
00:22 So the vector p to q is going to be equal to it's q minus p.
00:29 Because remember what we're really doing is if we start from p and we add this vector to it we should land on q so like if we do plus p the p's consulate we're just left with q so that's the way i always remember it so let's go ahead and plug these in really quickly uh so q so i'm going to write this as a vector as opposed to a point it's going to be u v w and then p is going to be a b let me write that a little bit better a, b, c.
01:03 Then to add or subtract a vector, we do it component -wise.
01:08 So it's going to be u minus a.
01:12 So u -minus a.
01:17 Then it's going to be v minus b.
01:22 And then lastly, it's going to be w minus c.
01:27 So this is going to be our vector going from p to q.
01:30 Now, if we go back and think, well, what was a line in algebra? that was y is equal to mx plus b well now instead we're working with a three -dimensional thing so let's just go ahead and call this r over here so we can say it's r of t and this is going to be equal to well our slope is really our vector p to q so this right here is really p to q this x is really our t and then b is just going to be some point we have so just a point on the graph but we'll need a specific point so let's fill these two things in first so we can figure out what b should actually be all right so we said our slope was p to q so that's going to be u minus a v minus b w minus c and then our x is now just going to be t and now for this point here let's think about this we want when we plug in zero essentially or whatever we're starting from to then land on q or to start on p and then after that go towards q so if we want to start from p we can just go ahead and write p here so it doesn't really matter if you just want to point in general or a a line in general, you could use any point on it.
03:15 But since we want it starting from p, it's a good idea to just go ahead and put p here.
03:21 Because notice, if t was equal to zero, we'd have zero times p to q, which would still be the zero vector.
03:29 And then zero vector plus p is just going to give us p back.
03:33 So we already have it where it's starting from zero with this.
03:37 But let's go ahead and actually write what p was.
03:41 Let me erase this.
03:44 P is so a b c all right now let's go ahead and combine these into one expression so first we could distribute the t so it's going to give us u minus a times t b minus b times t w minus c times t and then plus a b b b minus c times t and then again we go ahead and add these component wise so it's going to to be u minus a t plus a so u minus a times t plus a then it'll be v minus b times t plus b so v minus b times t plus b and then lastly we'll get w minus c times t plus c and so now this is our vector expression for this.
05:00 And if we want it as the parametric, we just say like x of t, or you can just say x is equal to u minus a, t plus a, y of t is equal to v minus b, t plus b, and then z of t is equal to w minus c, t plus c.
05:27 And you can see how this looks more like that...
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