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Problem 21 Easy Difficulty

(a) Prove that the midpoint of the line segment from $ P_1 (x_1, y_1, z_1) $ to $ P_2 (x_2, y_2, z_2) $ is
$$ \left(\frac{x_1 + x_2}{2} , \frac{y_1+ y_2}{2} , \frac{z_1+ z_2}{2} \right) $$
(b) Find the lengths of the medians of the triangle with vertices $ A (1, 2, 3) $, $ B (-2, 0, 5) $, $ C (4, 1, 5) $. (A median of a triangle is a line segment that joins a vertex to the midpoint of the opposite side.)

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Calculus 3

Calculus: Early Transcendentals

Chapter 12

Vectors and the Geometry of Space

Section 1

Three-Dimensional Coordinate Systems

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Video Transcript

Hello and welcome. We're looking at chapter eight, Section one, problem 17. Basically asked to approve the midpoint, um, the room And then use that to help a self part Be something about triangles. So let's start with party here. Um, problem. One way we can prove it. There's more than one way we think about it in this way. What the midpoint actually is. We have a line segment from if you want to pee, too. The midpoint is defined. It's the point where the distance between the midpoint and both points is the same. We're gonna use distance formula to prove this. If we can show that the distance between these two the same as the distance between those two than it is the truth midpoint. So let's get started with that. Um, let's find the distance between P one and em. It's gonna be the square root of just using our distance formula. That'll be x one minus x one plus x two over to. So just the difference of the two X values squared. And I just repeat that for each variable. So big square root here. I'm not going to do too much more of this. Um, but I am going to simplify inside of each of those pregnancies because we have some, like, terms and their recon simplify. So, uh, x one minus x one plus X two over too. That's the same as two x one over to. So just turning X one common denominator. And I'm subtracting this whole thing, so it should be to x one minus x one over to If I split the fraction up. Plus, uh, that being minus there. That was my point there, minus x two over to some subtracting the whole fraction. So if I split it apart like that, um, you can simplify it that way. So that would be X one minus x two squared. Go over to Let's just put it is one fraction here, keep it simple. So the same logics gonna apply to each of these. So it's a little bit simpler. Probably be easier to compare. Uh, when we do the second stuff, the second step, we do the distance from P, too. So it's gonna look almost exactly the same. Except we're doing X two minus that midpoint. All right. And we're gonna simplify it the same way as well. This is where the question happens. Um, I do two X to over two minus x one plus X to over two. It's gonna work. Similarly. Um, I can just distribute the negative kind of and combining that way. And I would get x two minus x one over two square. So it's the same as up here, except the two, uh, variables inside each set of princes are SWAT. So the question is, are these two the sinks? If they're the same, then we're done. We've proved it. So what we can show is that since everything inside the princes is being squared like a simple example here, if you do, uh, two minus three and then three minus two they're very different answers, negative one and positive one. But when you square them, it doesn't matter. All right, Uh, it comes out to the positive answer. So what we're doing, because we're doing the difference here of two different things and we're swapping the order, so they're gonna be the same value. Just one is gonna be opposite the negative. So when you square it, that disappears and there's a couple different ways you could show that you could show that, um, you could foil out each of these princes and, uh, and kind of show the expanded form that way. Or you could say, uh, since x to bias x one. Or actually, let's make it even more general. Since a, uh, doing subtraction here, a minus B, uh, equals C and be minus a equals. Uh, get out of myself. It was negative, C. Then if you square them space here, but my point will come across regardless. When you square them, these two values are equal. So you could combine this idea here, maybe write it out. Um, complete sentences combined this idea here with the fact that these two are then equal. So if the distances between the two are equal than it has to be the mid right, so that's a pretty full and complete proof. Um, hopefully you can put the finishing touches on. All right, let's move on to part B. So a median joins a vertex to the midpoint of the opposite side. We want to find the lengths of the medians of the triangle with specific courtesies. So, um, you want to use the distance formula so Let's start with, uh, a first. We want to find the distance from point A, and I'll make up my own little notation here to the midpoint of B and C's used used sub script there. So if I want to find the distance to that, it will be the square root. Take a eight. X value is one. Subtract that from the average of the two other ex values. So So just to show you the X value of B is negative, too. X value of C is four. So I'm gonna average those two together. That gives me one. So I'm going to one minus one, I swear, do the same process two. That's gonna be subtracted by the average of the Y values ups. The lie value is zero plus one over to side B one, right? And then Z. So my ex value for point A is three, and I subtract it by the average. I'm just the midpoint, essentially of being C. So I would be five plus five over to which would be five. The average of 55 is so just what I did. Maybe it was a little bit of a shortcut methods will explain real quick. So this right here is the midpoint of being C. So I kind of calculated it as I needed it for my distance for me. So you could have found the midpoint and then plugged it in. You would get the same thing. Oh, so we'll have for the distance between a in the mid point of being, C uh, this would be zero be 9/4 two minus 1/2 is three halves square that you get nine for us. Three miles. Five is negative, too. Negative. Two squared is for. So this forced the Samos 16/4 eso We'd get the square root of 25/4. That is five halves. All right, so, um, this is the distance from a to the midpoint of B C. All right, same process. Let's go to the distance between be in the mid point of a C so square root. Um, so just distance formula here. So, uh, negative to minus in the mid point of a and sees before, plus, one is five five over, too. Then for the y values be zero minus. Um, and then that would be two plus one over to side B three halves for the Z value, we have five minus and then a is three CS five. That adds up to eight. Divided by two is four to simplify this negative too. Negative too is the same as negative or over to that becomes Uh huh. Negative 9/2 or 81/4. Hey. And then, uh, next one becomes 9/4 3 have squared, and then the last one is just one. So if we add these together, one, of course, is 4/4. We get, uh, 90 four over for so that simplifies the denominator Would obviously be to see 94 divided by two would be 47 which doesn't simplify force. So I guess we just leave that as Route 94. So that is the distance between point B and the midpoint of a C. All right, so one more to go. You want the midpoint of the distance between sea and the midpoint of a B user distance formula one more time. So this would be four, which is the X value of C minus the average of A and B the X coordinates of a B, so that'd be one plus negative to side B. Negative 1/2 weird. Why? Value of C is one. The average of the Y values of A and B is one post zero about. And then Z value of C is five. And the average of the values of A and B are Bible stories eight divided by two is four, thanks to four plus 1/2. That would be nine halves squared would be 81/4, um, plus zero. You know, square to zero plus one five minutes for is one squared is what? So that gives us the square root of 85. Over four. Squared of a five doesn't simplify for us. I believe that's five times 17 17 seconds. Factor any further force. Very nicely. At least this simplifies to route 85 over, too. And that is the distance from point C to the midpoint of the two other sides. All right, so this one had a lot of working, right? Um that is everything. That part am part B

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James Stewart

Calculus: Early Transcendentals

Lily An

Catherine Ross

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Lily An

Catherine Ross

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Lily An

Catherine Ross

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals