1 00:00:00,000 --> 00:00:07,000 2 00:00:07,000 --> 00:00:07,410 JOEL LEWIS: Hi. 3 00:00:07,410 --> 00:00:08,720 Welcome to recitation. 4 00:00:08,720 --> 00:00:11,200 In lecture, you started learning about vectors. 5 00:00:11,200 --> 00:00:13,620 Now vectors are going to be really important throughout 6 00:00:13,620 --> 00:00:15,370 the whole of this course. 7 00:00:15,370 --> 00:00:18,200 And I wanted to give you one problem just to work with them 8 00:00:18,200 --> 00:00:20,070 in a slightly different context than what we're going 9 00:00:20,070 --> 00:00:20,960 to do in the future. 10 00:00:20,960 --> 00:00:23,010 So this is the context of Euclidean geometry. 11 00:00:23,010 --> 00:00:27,890 So some of you have probably seen this problem that we're 12 00:00:27,890 --> 00:00:29,730 going to solve, but you probably haven't seen it 13 00:00:29,730 --> 00:00:30,910 solved with vectors. 14 00:00:30,910 --> 00:00:32,470 So let's take a look at it. 15 00:00:32,470 --> 00:00:35,330 So what I'd like you to do is show that the three medians of 16 00:00:35,330 --> 00:00:39,080 a triangle intersect at a point, and the point is 2/3 of 17 00:00:39,080 --> 00:00:40,370 the way from each vertex. 18 00:00:40,370 --> 00:00:42,780 So let me just remind you of some terminology. 19 00:00:42,780 --> 00:00:46,830 So in a triangle, a median is the segment that connects one 20 00:00:46,830 --> 00:00:49,890 vertex to the midpoint of the opposite side. 21 00:00:49,890 --> 00:00:53,630 So here, this point M is exactly halfway between B and 22 00:00:53,630 --> 00:00:55,810 C. 23 00:00:55,810 --> 00:00:56,260 So OK. 24 00:00:56,260 --> 00:01:00,070 So every triangle has three medians-- one from each vertex 25 00:01:00,070 --> 00:01:02,500 connected to the midpoint of the opposite side-- and what 26 00:01:02,500 --> 00:01:05,590 I'm asking you to show is that these three medians all 27 00:01:05,590 --> 00:01:07,850 intersect in the same point. 28 00:01:07,850 --> 00:01:11,380 And also, that this point divides the median into two 29 00:01:11,380 --> 00:01:15,200 pieces, and the big piece is twice as large 30 00:01:15,200 --> 00:01:15,970 as the small piece. 31 00:01:15,970 --> 00:01:20,530 So this is 2/3 of the median, and this is 1/3 of the median. 32 00:01:20,530 --> 00:01:23,520 So, why don't you take a few minutes, work that out-- try 33 00:01:23,520 --> 00:01:27,610 and do it using vectors as much as possible-- 34 00:01:27,610 --> 00:01:29,200 pause the video, come back, and we 35 00:01:29,200 --> 00:01:30,450 can work on it together. 36 00:01:30,450 --> 00:01:39,080 37 00:01:39,080 --> 00:01:42,320 So hopefully you had some luck working on this problem. 38 00:01:42,320 --> 00:01:44,460 Let's get started on it. 39 00:01:44,460 --> 00:01:48,710 So to start, I actually want to rephrase the question a 40 00:01:48,710 --> 00:01:50,110 little bit. 41 00:01:50,110 --> 00:01:53,910 And I'll rephrase it to an equivalent question that's a 42 00:01:53,910 --> 00:01:57,250 little bit more clear about how we want to get started. 43 00:01:57,250 --> 00:02:00,450 So another way to say this problem is that it's 44 00:02:00,450 --> 00:02:01,850 asking us to show-- 45 00:02:01,850 --> 00:02:05,820 so for each median, say this median AM here, where M is the 46 00:02:05,820 --> 00:02:08,110 midpoint of side BC-- 47 00:02:08,110 --> 00:02:11,560 there exists a point on the median that divides it into a 48 00:02:11,560 --> 00:02:15,430 2:1 ratio, so the point that's 2/3 from the vertex to the 49 00:02:15,430 --> 00:02:16,960 midpoint of the opposite side. 50 00:02:16,960 --> 00:02:19,910 So for example, you know, there's a point-- 51 00:02:19,910 --> 00:02:24,070 so, let's call it P, say, at first-- 52 00:02:24,070 --> 00:02:28,020 so there's a point P such that AP is twice PM. 53 00:02:28,020 --> 00:02:28,430 OK? 54 00:02:28,430 --> 00:02:32,490 And similarly, there's some point-- maybe called Q-- 55 00:02:32,490 --> 00:02:36,110 that's 2/3 of the way from B to the midpoint of this side. 56 00:02:36,110 --> 00:02:39,220 And there's some point that's 2/3 of the way from C to the 57 00:02:39,220 --> 00:02:40,450 midpoint of this side. 58 00:02:40,450 --> 00:02:43,400 And so an equivalent formulation of the question is 59 00:02:43,400 --> 00:02:45,180 to show that these three points are 60 00:02:45,180 --> 00:02:46,890 really the same point. 61 00:02:46,890 --> 00:02:48,880 That they're all in the same place. 62 00:02:48,880 --> 00:02:53,060 So one way we can do that is that we can compare the 63 00:02:53,060 --> 00:02:55,590 position vectors of those three points. 64 00:02:55,590 --> 00:02:58,490 And if those three points all have the same position vector, 65 00:02:58,490 --> 00:03:00,800 then they're all in exactly the same position. 66 00:03:00,800 --> 00:03:03,820 So in order to do that we need some origin. 67 00:03:03,820 --> 00:03:06,220 And it happens that for this problem, it doesn't matter 68 00:03:06,220 --> 00:03:08,590 where the origin is, and so I'm not going to draw an 69 00:03:08,590 --> 00:03:10,770 origin, but I'm going to call it O. 70 00:03:10,770 --> 00:03:16,300 So we're going to set up a vector coordinate system with 71 00:03:16,300 --> 00:03:24,180 origin O. And now I want to look at what the vector from O 72 00:03:24,180 --> 00:03:28,600 to P is in terms of the vectors connecting O to A, B, 73 00:03:28,600 --> 00:03:29,420 and C. Right? 74 00:03:29,420 --> 00:03:32,360 Those are the vectors that determine the 75 00:03:32,360 --> 00:03:33,530 vertices of the triangle. 76 00:03:33,530 --> 00:03:36,970 And so I want to relate the location of P to the locations 77 00:03:36,970 --> 00:03:38,620 of A, B, and C. 78 00:03:38,620 --> 00:03:41,340 So the first thing to do is that-- 79 00:03:41,340 --> 00:03:44,220 well, in order to talk about where P is, I know how P is 80 00:03:44,220 --> 00:03:48,770 related to A and M and I know how M is related to B and C. 81 00:03:48,770 --> 00:03:52,290 So let's first figure out what the position vector of M is in 82 00:03:52,290 --> 00:03:54,920 terms of the position vectors of A, B, and C, and then we 83 00:03:54,920 --> 00:04:03,240 can use that to figure out the position vector of P. 84 00:04:03,240 --> 00:04:10,260 So M is the midpoint of the segment BC. 85 00:04:10,260 --> 00:04:13,110 86 00:04:13,110 --> 00:04:15,670 So I think we saw this in lecture. 87 00:04:15,670 --> 00:04:22,232 What this means is that the position vector OM is exactly 88 00:04:22,232 --> 00:04:27,490 the average of the position vectors of B and C. It's 1/2 89 00:04:27,490 --> 00:04:31,450 of the quantity OB plus OC. 90 00:04:31,450 --> 00:04:31,930 All right? 91 00:04:31,930 --> 00:04:36,510 So it's easy to express the position vector of the 92 00:04:36,510 --> 00:04:38,930 midpoint of a segment in terms of the position 93 00:04:38,930 --> 00:04:40,280 vectors of the endpoints. 94 00:04:40,280 --> 00:04:42,720 You just add the position vectors of the endpoints and 95 00:04:42,720 --> 00:04:43,970 divide by 2. 96 00:04:43,970 --> 00:04:47,700 So if you like, this is equivalent to the geometric 97 00:04:47,700 --> 00:04:49,100 fact that the diagonals of a 98 00:04:49,100 --> 00:04:51,470 parallelogram bisect each other. 99 00:04:51,470 --> 00:04:55,340 So that's the position vector of M. 100 00:04:55,340 --> 00:04:59,490 Now we have to figure out what the position vector of P is. 101 00:04:59,490 --> 00:05:04,780 So in order to do this we can note, that in order to get 102 00:05:04,780 --> 00:05:09,250 from the origin to point P, well, what we have to go from 103 00:05:09,250 --> 00:05:11,750 the origin-- wherever it is-- to A, and then we have to go 104 00:05:11,750 --> 00:05:15,720 from A 2/3 of the way to M. All right? 105 00:05:15,720 --> 00:05:29,720 So the vector OP is equal to OA plus 2/3 of the vector AM. 106 00:05:29,720 --> 00:05:29,940 Right? 107 00:05:29,940 --> 00:05:32,790 Because we go 2/3 of the way from A to M in order to get 108 00:05:32,790 --> 00:05:36,430 from A to P. This is because we've chosen P to be the point 109 00:05:36,430 --> 00:05:40,140 that divides segment AM into a 2:1 ratio so that 110 00:05:40,140 --> 00:05:42,390 AP is 2/3 of AM. 111 00:05:42,390 --> 00:05:43,000 OK. 112 00:05:43,000 --> 00:05:43,590 So good. 113 00:05:43,590 --> 00:05:46,030 So now we need the vector AM. 114 00:05:46,030 --> 00:05:48,970 Well, we know what the position vector of A is. 115 00:05:48,970 --> 00:05:50,190 It's just OA. 116 00:05:50,190 --> 00:05:53,500 And we also know what the position vector of M is. 117 00:05:53,500 --> 00:05:54,500 It's OM. 118 00:05:54,500 --> 00:06:01,310 So that means that AM is just the difference 119 00:06:01,310 --> 00:06:02,870 of those two vectors. 120 00:06:02,870 --> 00:06:09,360 It's going to be OM minus OA. 121 00:06:09,360 --> 00:06:11,950 Another way to say this is that if you add OA to both 122 00:06:11,950 --> 00:06:16,120 sides, you have that OA plus AM equals OM. 123 00:06:16,120 --> 00:06:19,173 In other words, to go from O to M, first you can go from O 124 00:06:19,173 --> 00:06:21,340 to A, and then go from A to M. All right. 125 00:06:21,340 --> 00:06:24,410 And I've just subtracted OA onto the other side here. 126 00:06:24,410 --> 00:06:28,640 So we can write AM in terms of OM and OA. 127 00:06:28,640 --> 00:06:31,120 And we also, we have an expression for OM here in 128 00:06:31,120 --> 00:06:33,290 terms of OB and OC. 129 00:06:33,290 --> 00:06:36,930 So that means we can get an expression for AM in terms of 130 00:06:36,930 --> 00:06:39,120 OA, OB, and OC. 131 00:06:39,120 --> 00:06:39,830 So let's do that. 132 00:06:39,830 --> 00:06:43,000 So that's just by substituting from here into here. 133 00:06:43,000 --> 00:06:55,210 So if I do that, I get that AM is equal to-- so OM is 1/2 OB 134 00:06:55,210 --> 00:07:01,090 plus 1/2 OC, and now I just subtract OA. 135 00:07:01,090 --> 00:07:03,685 136 00:07:03,685 --> 00:07:04,510 All right. 137 00:07:04,510 --> 00:07:08,790 So that's what AM is, putting these two equations together. 138 00:07:08,790 --> 00:07:10,830 I get that that's AM. 139 00:07:10,830 --> 00:07:13,100 And so now I need to figure out what OP is. 140 00:07:13,100 --> 00:07:15,800 So for OP, I just need to substitute in this new 141 00:07:15,800 --> 00:07:18,170 expression that I've got for AM. 142 00:07:18,170 --> 00:07:28,000 So I have OP is equal to, well it's equal to OA plus 2/3 of 143 00:07:28,000 --> 00:07:29,850 what I've written just right above-- 144 00:07:29,850 --> 00:07:30,860 2/3 of AM. 145 00:07:30,860 --> 00:07:40,500 So that's 1/2 OB plus 1/2 OC minus OA. 146 00:07:40,500 --> 00:07:41,020 OK. 147 00:07:41,020 --> 00:07:43,980 And so now you can multiply this 2/3 in-- you know, just 148 00:07:43,980 --> 00:07:47,170 distribute the scalar multiplication across the 149 00:07:47,170 --> 00:07:49,300 addition there-- and then we can rearrange. 150 00:07:49,300 --> 00:07:52,890 We'll have two terms involving OA and we can combine them. 151 00:07:52,890 --> 00:07:58,280 So we'll see we have a plus OA minus 2/3 OA. 152 00:07:58,280 --> 00:08:03,450 So that's going to be equal to 1/3 OA. 153 00:08:03,450 --> 00:08:07,280 And then we have, OK so 2/3 times 1/2 OB. 154 00:08:07,280 --> 00:08:16,730 So that's plus 1/3 OB plus 1/3 OC. 155 00:08:16,730 --> 00:08:20,680 So this gives us a simple formula for the position 156 00:08:20,680 --> 00:08:22,160 vector of P-- 157 00:08:22,160 --> 00:08:23,500 that vector OP-- 158 00:08:23,500 --> 00:08:26,290 in terms of the position vectors of A, B, and C. So in 159 00:08:26,290 --> 00:08:29,370 particular, it's actually because P is the special 160 00:08:29,370 --> 00:08:32,920 point, it's 1/3 of their sum. 161 00:08:32,920 --> 00:08:35,830 Of the sum OA plus OB plus OC. 162 00:08:35,830 --> 00:08:38,010 OK, so that's where P is. 163 00:08:38,010 --> 00:08:41,340 Now to finish the problem, I just have to show that this is 164 00:08:41,340 --> 00:08:43,460 the same location as the point that 165 00:08:43,460 --> 00:08:45,470 trisects the other medians. 166 00:08:45,470 --> 00:08:46,940 So how would I do that? 167 00:08:46,940 --> 00:08:49,940 Well, I could go back to my triangle and I could do 168 00:08:49,940 --> 00:08:52,180 exactly the same thing. 169 00:08:52,180 --> 00:08:54,120 So I could-- maybe I'll give this point a name, also. 170 00:08:54,120 --> 00:08:56,770 I'll call this midpoint N, say. 171 00:08:56,770 --> 00:09:02,130 So I could let Q be the point that lies 2/3 of the way from 172 00:09:02,130 --> 00:09:06,290 B to N. And then I could write down the position vector of N 173 00:09:06,290 --> 00:09:08,320 in terms of OA, OB, and OC. 174 00:09:08,320 --> 00:09:10,460 And then I can use that to write down the position vector 175 00:09:10,460 --> 00:09:13,400 of Q in terms of OA, OB, and OC, and I'll get some 176 00:09:13,400 --> 00:09:14,580 expression. 177 00:09:14,580 --> 00:09:16,390 And what will happen at the end-- 178 00:09:16,390 --> 00:09:17,990 I hope if I'm lucky-- 179 00:09:17,990 --> 00:09:20,540 that expression will be equal to this expression that I 180 00:09:20,540 --> 00:09:22,220 found over here. 181 00:09:22,220 --> 00:09:22,760 OK? 182 00:09:22,760 --> 00:09:26,540 So you can go through and do that, and if you do that, what 183 00:09:26,540 --> 00:09:28,210 you'll find is that in fact it works. 184 00:09:28,210 --> 00:09:30,890 So there's actually a sort of clever, shorter 185 00:09:30,890 --> 00:09:31,960 way of seeing that. 186 00:09:31,960 --> 00:09:36,730 Which is that this formula is symmetric in A, B, and C. 187 00:09:36,730 --> 00:09:40,830 So that means if I just relabel the points A, B, and 188 00:09:40,830 --> 00:09:44,210 C, this expression for the position 189 00:09:44,210 --> 00:09:45,570 vector doesn't change. 190 00:09:45,570 --> 00:09:48,460 So rather than going through that process that I just 191 00:09:48,460 --> 00:09:54,730 described, you can also say, well, in order to look at, say 192 00:09:54,730 --> 00:09:57,830 Q, instead of P, what I need to do is I just need to switch 193 00:09:57,830 --> 00:10:00,770 B and A. I need to do exactly the same thing but the roles 194 00:10:00,770 --> 00:10:02,280 of A and B are interchanged. 195 00:10:02,280 --> 00:10:04,470 Well, if the roles of A and B are interchanged, then in the 196 00:10:04,470 --> 00:10:07,080 resulting formula, I just have to interchange the roles of A 197 00:10:07,080 --> 00:10:09,960 and B, but that won't change the value of this expression. 198 00:10:09,960 --> 00:10:13,020 So by symmetry, the point I get really is 199 00:10:13,020 --> 00:10:14,170 going to be the same. 200 00:10:14,170 --> 00:10:17,530 If you don't like that argument, I invite you to go 201 00:10:17,530 --> 00:10:21,840 through this computation again in the case 202 00:10:21,840 --> 00:10:24,600 of the other medians. 203 00:10:24,600 --> 00:10:27,340 In either case, what you'll find is that the points that 204 00:10:27,340 --> 00:10:31,390 trisect the three medians all have position vector 1/3 OA 205 00:10:31,390 --> 00:10:34,700 plus 1/3 OB plus 1/3 OC, but that means 206 00:10:34,700 --> 00:10:36,230 they're the same point. 207 00:10:36,230 --> 00:10:44,680 So what we've shown then, is that the points that trisect 208 00:10:44,680 --> 00:10:45,290 the three medians-- that trisect-- that divide them 209 00:10:45,290 --> 00:10:49,000 into 2:1 ratios from the vertex to the midpoint of the 210 00:10:49,000 --> 00:10:51,440 opposite side, that those three points all have the same 211 00:10:51,440 --> 00:10:52,390 position vector. 212 00:10:52,390 --> 00:10:54,850 So in fact, they're the same point, and that's what we 213 00:10:54,850 --> 00:10:56,390 wanted to show, right? 214 00:10:56,390 --> 00:10:59,900 We wanted to show that there's one point that trisects all 215 00:10:59,900 --> 00:11:01,380 three of those medians. 216 00:11:01,380 --> 00:11:04,170 So we've shown that the three points that trisect them are 217 00:11:04,170 --> 00:11:05,480 actually the same. 218 00:11:05,480 --> 00:11:09,530 So that's the same conclusion, phrased differently. 219 00:11:09,530 --> 00:11:12,120 So I think I'll end there. 220 00:11:12,120 --> 00:11:12,406