1 00:00:06,785 --> 00:00:07,410 JOEL LEWIS: Hi. 2 00:00:07,410 --> 00:00:08,720 Welcome to recitation. 3 00:00:08,720 --> 00:00:11,200 In lecture, you started learning about vectors. 4 00:00:11,200 --> 00:00:13,050 Now vectors are going to be really important 5 00:00:13,050 --> 00:00:15,360 throughout the whole of this course. 6 00:00:15,360 --> 00:00:16,860 And I wanted to give you one problem 7 00:00:16,860 --> 00:00:19,211 just to work with them in a slightly different context 8 00:00:19,211 --> 00:00:20,960 than what we're going to do in the future. 9 00:00:20,960 --> 00:00:23,010 So this is the context of Euclidean geometry. 10 00:00:23,010 --> 00:00:27,359 So some of you have probably seen this problem 11 00:00:27,359 --> 00:00:29,150 that we're going to solve, but you probably 12 00:00:29,150 --> 00:00:30,910 haven't seen it solved with vectors. 13 00:00:30,910 --> 00:00:32,470 So let's take a look at it. 14 00:00:32,470 --> 00:00:34,260 So what I'd like you to do is show 15 00:00:34,260 --> 00:00:37,590 that the three medians of a triangle intersect at a point, 16 00:00:37,590 --> 00:00:40,370 and the point is 2/3 of the way from each vertex. 17 00:00:40,370 --> 00:00:42,780 So let me just remind you of some terminology. 18 00:00:42,780 --> 00:00:45,870 So in a triangle, a median is the segment 19 00:00:45,870 --> 00:00:49,890 that connects one vertex to the midpoint of the opposite side. 20 00:00:49,890 --> 00:00:55,761 So here, this point M is exactly halfway between B and C. 21 00:00:55,761 --> 00:00:56,260 So OK. 22 00:00:56,260 --> 00:00:58,360 So every triangle has three medians-- 23 00:00:58,360 --> 00:01:01,100 one from each vertex connected to the midpoint 24 00:01:01,100 --> 00:01:03,660 of the opposite side-- and what I'm asking you to show 25 00:01:03,660 --> 00:01:07,850 is that these three medians all intersect in the same point. 26 00:01:07,850 --> 00:01:10,950 And also, that this point divides the median 27 00:01:10,950 --> 00:01:14,390 into two pieces, and the big piece is twice 28 00:01:14,390 --> 00:01:15,970 as large as the small piece. 29 00:01:15,970 --> 00:01:20,530 So this is 2/3 of the median, and this is 1/3 of the median. 30 00:01:20,530 --> 00:01:23,520 So, why don't you take a few minutes, work that out-- try 31 00:01:23,520 --> 00:01:28,215 and do it using vectors as much as possible-- pause the video, 32 00:01:28,215 --> 00:01:29,965 come back, and we can work on it together. 33 00:01:39,080 --> 00:01:42,320 So hopefully you had some luck working on this problem. 34 00:01:42,320 --> 00:01:44,460 Let's get started on it. 35 00:01:44,460 --> 00:01:48,020 So to start, I actually want to rephrase 36 00:01:48,020 --> 00:01:50,110 the question a little bit. 37 00:01:50,110 --> 00:01:53,680 And I'll rephrase it to an equivalent question 38 00:01:53,680 --> 00:01:56,210 that's a little bit more clear about how 39 00:01:56,210 --> 00:01:57,250 we want to get started. 40 00:01:57,250 --> 00:02:00,140 So another way to say this problem is 41 00:02:00,140 --> 00:02:03,430 that it's asking us to show-- so for each median, 42 00:02:03,430 --> 00:02:08,110 say this median AM here, where M is the midpoint of side BC, 43 00:02:08,110 --> 00:02:11,220 there exists a point on the median that divides it 44 00:02:11,220 --> 00:02:15,120 into a 2:1 ratio, so the point that's 2/3 from the vertex 45 00:02:15,120 --> 00:02:16,960 to the midpoint of the opposite side. 46 00:02:16,960 --> 00:02:20,160 So for example, you know, there's a point-- so, 47 00:02:20,160 --> 00:02:25,330 let's call it P, say, at first-- so there's a point P 48 00:02:25,330 --> 00:02:27,930 such that AP is twice PM. 49 00:02:27,930 --> 00:02:28,430 OK? 50 00:02:28,430 --> 00:02:30,520 And similarly, there's some point-- maybe called 51 00:02:30,520 --> 00:02:36,110 Q-- that's 2/3 of the way from B to the midpoint of this side. 52 00:02:36,110 --> 00:02:39,030 And there's some point that's 2/3 of the way from C 53 00:02:39,030 --> 00:02:40,450 to the midpoint of this side. 54 00:02:40,450 --> 00:02:43,240 And so an equivalent formulation of the question 55 00:02:43,240 --> 00:02:45,440 is to show that these three points are really 56 00:02:45,440 --> 00:02:46,890 the same point. 57 00:02:46,890 --> 00:02:48,880 That they're all in the same place. 58 00:02:48,880 --> 00:02:52,360 So one way we can do that is that we 59 00:02:52,360 --> 00:02:55,590 can compare the position vectors of those three points. 60 00:02:55,590 --> 00:02:58,490 And if those three points all have the same position vector, 61 00:02:58,490 --> 00:03:00,800 then they're all in exactly the same position. 62 00:03:00,800 --> 00:03:03,689 So in order to do that we need some origin. 63 00:03:03,689 --> 00:03:05,230 And it happens that for this problem, 64 00:03:05,230 --> 00:03:07,405 it doesn't matter where the origin is, 65 00:03:07,405 --> 00:03:09,030 and so I'm not going to draw an origin, 66 00:03:09,030 --> 00:03:10,770 but I'm going to call it O. 67 00:03:10,770 --> 00:03:14,490 So we're going to set up a vector coordinate system 68 00:03:14,490 --> 00:03:20,240 with origin O. And now I want to look 69 00:03:20,240 --> 00:03:26,970 at what the vector from O to P is in terms of the vectors 70 00:03:26,970 --> 00:03:29,420 connecting O to A, B, and C. Right? 71 00:03:29,420 --> 00:03:32,160 Those are the vectors that determine 72 00:03:32,160 --> 00:03:33,530 the vertices of the triangle. 73 00:03:33,530 --> 00:03:36,190 And so I want to relate the location of P 74 00:03:36,190 --> 00:03:38,620 to the locations of A, B, and C. 75 00:03:38,620 --> 00:03:42,330 So the first thing to do is that-- well, in order 76 00:03:42,330 --> 00:03:46,000 to talk about where P is, I know how P is related to A and M 77 00:03:46,000 --> 00:03:48,770 and I know how M is related to B and C. 78 00:03:48,770 --> 00:03:51,580 So let's first figure out what the position vector of M 79 00:03:51,580 --> 00:03:54,337 is in terms of the position vectors of A, B, and C, 80 00:03:54,337 --> 00:03:55,920 and then we can use that to figure out 81 00:03:55,920 --> 00:04:03,240 the position vector of P. 82 00:04:03,240 --> 00:04:10,260 So M is the midpoint of the segment BC. 83 00:04:13,110 --> 00:04:15,670 So I think we saw this in lecture. 84 00:04:15,670 --> 00:04:22,232 What this means is that the position vector OM is exactly 85 00:04:22,232 --> 00:04:25,770 the average of the position vectors of B and C. 86 00:04:25,770 --> 00:04:31,430 It's 1/2 of the quantity OB plus OC. 87 00:04:31,430 --> 00:04:31,930 All right? 88 00:04:31,930 --> 00:04:35,470 So it's easy to express the position 89 00:04:35,470 --> 00:04:38,300 vector of the midpoint of a segment in terms 90 00:04:38,300 --> 00:04:40,280 of the position vectors of the endpoints. 91 00:04:40,280 --> 00:04:42,450 You just add the position vectors of the endpoints 92 00:04:42,450 --> 00:04:43,970 and divide by 2. 93 00:04:43,970 --> 00:04:46,920 So if you like, this is equivalent 94 00:04:46,920 --> 00:04:49,910 to the geometric fact that the diagonals of a parallelogram 95 00:04:49,910 --> 00:04:51,470 bisect each other. 96 00:04:51,470 --> 00:04:55,340 So that's the position vector of M. 97 00:04:55,340 --> 00:04:59,490 Now we have to figure out what the position vector of P is. 98 00:04:59,490 --> 00:05:03,930 So in order to do this we can note, 99 00:05:03,930 --> 00:05:06,560 that in order to get from the origin to point P, 100 00:05:06,560 --> 00:05:09,700 well, what we have to do is we have to go from the origin-- 101 00:05:09,700 --> 00:05:11,460 wherever it is-- to A, and then we 102 00:05:11,460 --> 00:05:15,720 have to go from A 2/3 of the way to M. All right? 103 00:05:15,720 --> 00:05:29,440 So the vector OP is equal to OA plus 2/3 of the vector AM. 104 00:05:29,440 --> 00:05:29,940 Right? 105 00:05:29,940 --> 00:05:31,530 Because we go 2/3 of the way from A 106 00:05:31,530 --> 00:05:33,730 to M in order to get from A to P. 107 00:05:33,730 --> 00:05:36,570 This is because we've chosen P to be the point that 108 00:05:36,570 --> 00:05:42,390 divides segment AM into a 2:1 ratio so that AP is 2/3 of AM. 109 00:05:42,390 --> 00:05:43,000 OK. 110 00:05:43,000 --> 00:05:43,590 So good. 111 00:05:43,590 --> 00:05:46,030 So now we need the vector AM. 112 00:05:46,030 --> 00:05:48,970 Well, we know what the position vector of A is. 113 00:05:48,970 --> 00:05:50,190 It's just OA. 114 00:05:50,190 --> 00:05:53,500 And we also know what the position vector of M is. 115 00:05:53,500 --> 00:05:54,500 It's OM. 116 00:05:54,500 --> 00:06:01,880 So that means that AM is just the difference of those two 117 00:06:01,880 --> 00:06:02,870 vectors. 118 00:06:02,870 --> 00:06:09,360 It's going to be OM minus OA. 119 00:06:09,360 --> 00:06:12,520 Another way to say this is that if you add OA to both sides, 120 00:06:12,520 --> 00:06:16,120 you have that OA plus AM equals OM. 121 00:06:16,120 --> 00:06:18,160 In other words, to go from O to M, 122 00:06:18,160 --> 00:06:20,510 first you can go from O to A, and then go from A to M. 123 00:06:20,510 --> 00:06:21,340 All right. 124 00:06:21,340 --> 00:06:24,410 And I've just subtracted OA onto the other side here. 125 00:06:24,410 --> 00:06:28,640 So we can write AM in terms of OM and OA. 126 00:06:28,640 --> 00:06:30,440 And we also-- we have an expression for OM 127 00:06:30,440 --> 00:06:33,290 here in terms of OB and OC. 128 00:06:33,290 --> 00:06:36,190 So that means we can get an expression for AM 129 00:06:36,190 --> 00:06:39,120 in terms of OA, OB, and OC. 130 00:06:39,120 --> 00:06:39,830 So let's do that. 131 00:06:39,830 --> 00:06:43,000 So that's just by substituting from here into here. 132 00:06:43,000 --> 00:06:47,960 So if I do that, I get that AM is 133 00:06:47,960 --> 00:06:58,550 equal to-- so OM is 1/2 OB plus 1/2 OC, 134 00:06:58,550 --> 00:07:01,090 and now I just subtract OA. 135 00:07:03,685 --> 00:07:04,510 All right. 136 00:07:04,510 --> 00:07:08,790 So that's what AM is, putting these two equations together. 137 00:07:08,790 --> 00:07:10,830 I get that that's AM. 138 00:07:10,830 --> 00:07:13,100 And so now I need to figure out what OP is. 139 00:07:13,100 --> 00:07:14,980 So for OP, I just need to substitute 140 00:07:14,980 --> 00:07:18,170 in this new expression that I've got for AM. 141 00:07:18,170 --> 00:07:21,444 So I have OP is equal to, well it's 142 00:07:21,444 --> 00:07:28,542 equal to OA plus 2/3 of what I've written 143 00:07:28,542 --> 00:07:30,860 just right above-- 2/3 of AM. 144 00:07:30,860 --> 00:07:40,500 So that's 1/2 OB plus 1/2 OC minus OA. 145 00:07:40,500 --> 00:07:41,020 OK. 146 00:07:41,020 --> 00:07:43,760 And so now you can multiply this 2/3 in-- you know, 147 00:07:43,760 --> 00:07:46,710 just distribute the scalar multiplication 148 00:07:46,710 --> 00:07:49,300 across the addition there-- and then we can rearrange. 149 00:07:49,300 --> 00:07:52,890 We'll have two terms involving OA and we can combine them. 150 00:07:52,890 --> 00:07:58,280 So we'll see we have a plus OA minus 2/3 OA. 151 00:07:58,280 --> 00:08:03,450 So that's going to be equal to 1/3 OA. 152 00:08:03,450 --> 00:08:07,280 And then we have, OK so 2/3 times 1/2 OB. 153 00:08:07,280 --> 00:08:16,730 So that's plus 1/3 OB plus 1/3 OC. 154 00:08:16,730 --> 00:08:19,160 So this gives us a simple formula 155 00:08:19,160 --> 00:08:24,050 for the position vector of P-- that vector OP-- in terms 156 00:08:24,050 --> 00:08:25,980 of the position vectors of A, B, and C. 157 00:08:25,980 --> 00:08:28,040 So in particular, it's actually-- 158 00:08:28,040 --> 00:08:32,920 because P is the special point-- it's 1/3 of their sum. 159 00:08:32,920 --> 00:08:35,830 Of the sum OA plus OB plus OC. 160 00:08:35,830 --> 00:08:38,010 OK, so that's where P is. 161 00:08:38,010 --> 00:08:39,680 Now to finish the problem, I just 162 00:08:39,680 --> 00:08:43,460 have to show that this is the same location as the point that 163 00:08:43,460 --> 00:08:45,470 trisects the other medians. 164 00:08:45,470 --> 00:08:46,940 So how would I do that? 165 00:08:46,940 --> 00:08:49,380 Well, I could go back to my triangle 166 00:08:49,380 --> 00:08:51,912 and I could do exactly the same thing. 167 00:08:51,912 --> 00:08:54,120 So I could-- maybe I'll give this point a name, also. 168 00:08:54,120 --> 00:08:56,770 I'll call this midpoint N, say. 169 00:08:56,770 --> 00:09:01,890 So I could let Q be the point that lies 2/3 of the way 170 00:09:01,890 --> 00:09:04,540 from B to N. And then I could write down 171 00:09:04,540 --> 00:09:08,277 the position vector of N in terms of OA, OB, and OC. 172 00:09:08,277 --> 00:09:10,860 And then I can use that to write down the position vector of Q 173 00:09:10,860 --> 00:09:14,580 in terms of OA, OB, and OC, and I'll get some expression. 174 00:09:14,580 --> 00:09:16,390 And what will happen at the end-- 175 00:09:16,390 --> 00:09:18,720 I hope if I'm lucky-- that expression 176 00:09:18,720 --> 00:09:22,220 will be equal to this expression that I found over here. 177 00:09:22,220 --> 00:09:22,760 OK? 178 00:09:22,760 --> 00:09:24,950 So you can go through and do that, 179 00:09:24,950 --> 00:09:28,210 and if you do that, what you'll find is that in fact it works. 180 00:09:28,210 --> 00:09:31,140 So there's actually a sort of clever, shorter way 181 00:09:31,140 --> 00:09:31,960 of seeing that. 182 00:09:31,960 --> 00:09:36,730 Which is that this formula is symmetric in A, B, and C. 183 00:09:36,730 --> 00:09:42,860 So that means if I just relabel the points A, B, and C, 184 00:09:42,860 --> 00:09:45,570 this expression for the position vector doesn't change. 185 00:09:45,570 --> 00:09:48,010 So rather than going through that process 186 00:09:48,010 --> 00:09:49,900 that I just described, you can also 187 00:09:49,900 --> 00:09:55,150 say, well, in order to look at, say Q, 188 00:09:55,150 --> 00:09:57,210 instead of P, what I need to do is I just 189 00:09:57,210 --> 00:10:00,150 need to switch B and A. I need to do exactly the same thing 190 00:10:00,150 --> 00:10:02,184 but the roles of A and B are interchanged. 191 00:10:02,184 --> 00:10:04,350 Well, if the roles of A and B are interchanged, then 192 00:10:04,350 --> 00:10:05,891 in the resulting formula, I just have 193 00:10:05,891 --> 00:10:07,570 to interchange the roles of A and B, 194 00:10:07,570 --> 00:10:09,960 but that won't change the value of this expression. 195 00:10:09,960 --> 00:10:14,170 So by symmetry, the point I get really is going to be the same. 196 00:10:14,170 --> 00:10:16,240 If you don't like that argument, I 197 00:10:16,240 --> 00:10:20,360 invite you to go through this computation 198 00:10:20,360 --> 00:10:24,600 again in the case of the other medians. 199 00:10:24,600 --> 00:10:27,860 In either case, what you'll find is that the points that trisect 200 00:10:27,860 --> 00:10:31,100 the three medians all have position vector 1/3 201 00:10:31,100 --> 00:10:34,700 OA plus 1/3 OB plus 1/3 OC, but that means 202 00:10:34,700 --> 00:10:36,230 they're the same point. 203 00:10:36,230 --> 00:10:41,260 So what we've shown then, is that the points that 204 00:10:41,260 --> 00:10:45,290 trisect the three medians-- that trisects, that divide them 205 00:10:45,290 --> 00:10:48,809 into 2:1 ratios from the vertex to the midpoint 206 00:10:48,809 --> 00:10:51,100 of the opposite side-- that those three points all have 207 00:10:51,100 --> 00:10:52,390 the same position vector. 208 00:10:52,390 --> 00:10:54,210 So in fact, they're the same point, 209 00:10:54,210 --> 00:10:56,390 and that's what we wanted to show, right? 210 00:10:56,390 --> 00:10:59,660 We wanted to show that there's one point that trisects 211 00:10:59,660 --> 00:11:01,380 all three of those medians. 212 00:11:01,380 --> 00:11:04,080 So we've shown that the three points that trisect them 213 00:11:04,080 --> 00:11:05,480 are actually the same. 214 00:11:05,480 --> 00:11:09,530 So that's the same conclusion, phrased differently. 215 00:11:09,530 --> 00:11:11,906 So I think I'll end there.