1 00:00:06,815 --> 00:00:07,440 JOEL LEWIS: Hi. 2 00:00:07,440 --> 00:00:08,950 Welcome back to recitation. 3 00:00:08,950 --> 00:00:10,390 In lecture, you've begun learning 4 00:00:10,390 --> 00:00:12,940 about various different ways to describe planes 5 00:00:12,940 --> 00:00:14,380 in three-dimensional space. 6 00:00:14,380 --> 00:00:16,310 In particular, there are equations and ways 7 00:00:16,310 --> 00:00:18,830 you can translate between other characterizations. 8 00:00:18,830 --> 00:00:20,830 So I have here four different planes for you, 9 00:00:20,830 --> 00:00:22,570 described in four different ways, 10 00:00:22,570 --> 00:00:24,640 and what I'd like you to do is try and figure out 11 00:00:24,640 --> 00:00:27,210 what the equations for each of these four planes are. 12 00:00:27,210 --> 00:00:28,730 So let me see what they are. 13 00:00:28,730 --> 00:00:30,570 So we've got the first one-- part 14 00:00:30,570 --> 00:00:33,510 a-- we have a plane where I'm giving you its normal vector 15 00:00:33,510 --> 00:00:36,119 N, which is the vector [1, 2, 3]. 16 00:00:36,119 --> 00:00:37,910 And I'm going to tell you that the plane is 17 00:00:37,910 --> 00:00:41,570 passing through the point 1, 0, minus 1. 18 00:00:41,570 --> 00:00:44,820 In part b, I'm telling you that the plane passes 19 00:00:44,820 --> 00:00:47,450 through the origin, and also that it's parallel 20 00:00:47,450 --> 00:00:48,210 to two vectors. 21 00:00:48,210 --> 00:00:51,900 It's parallel to the vector 1, 0, minus 1, and to the vector 22 00:00:51,900 --> 00:00:54,510 minus 1, 2, 0. 23 00:00:54,510 --> 00:00:58,200 In part c, I'm telling you that a plane 24 00:00:58,200 --> 00:01:04,370 passes through the points (1, 2, 0), (3, 1, 1), and (2, 0, 0). 25 00:01:04,370 --> 00:01:07,540 And in part d, I'm telling you that the plane is parallel 26 00:01:07,540 --> 00:01:10,550 to the plane in part a, and also that it passes 27 00:01:10,550 --> 00:01:12,770 through the point (1, 2, 3). 28 00:01:12,770 --> 00:01:15,430 So what I'd like you to do is try, for each of these four 29 00:01:15,430 --> 00:01:17,550 descriptions, figure out what the equation 30 00:01:17,550 --> 00:01:19,600 of the associated plane is. 31 00:01:19,600 --> 00:01:21,980 So why don't you pause the video, take a few minutes, 32 00:01:21,980 --> 00:01:24,840 work those all out, come back, and we can work them out 33 00:01:24,840 --> 00:01:25,340 together. 34 00:01:33,450 --> 00:01:35,830 So hopefully you had some luck working on these problems. 35 00:01:35,830 --> 00:01:37,850 Let's get started. 36 00:01:37,850 --> 00:01:40,670 So we may as well start with the first one. 37 00:01:40,670 --> 00:01:44,590 So in part a, we're given that the normal vector 38 00:01:44,590 --> 00:01:50,480 N is the vector i plus 2j plus 3k, or [1, 2, 3], 39 00:01:50,480 --> 00:01:54,240 and that it passes through the point P-- which I'm going 40 00:01:54,240 --> 00:01:58,370 to call P-- 1, 0, minus 1. 41 00:01:58,370 --> 00:02:01,180 So this is a form you learned in lecture. 42 00:02:01,180 --> 00:02:02,850 And so it's pretty straightforward 43 00:02:02,850 --> 00:02:04,530 to write down the equation here. 44 00:02:04,530 --> 00:02:06,940 The thing to remember is that if a point (x, y, z) 45 00:02:06,940 --> 00:02:10,890 is on the plane, then we have to have that the vector N-- 46 00:02:10,890 --> 00:02:14,380 the normal-- is orthogonal to the vector 47 00:02:14,380 --> 00:02:17,370 connecting the point (x, y, z) to the point we know. 48 00:02:17,370 --> 00:02:23,310 So that's the vector x minus 1, y minus 0-- 49 00:02:23,310 --> 00:02:25,950 which is just y-- z plus 1. 50 00:02:25,950 --> 00:02:28,800 So N and this vector that lies in the plane 51 00:02:28,800 --> 00:02:32,010 have to be orthogonal, so their dot product has to be 0. 52 00:02:32,010 --> 00:02:33,640 And now you just multiply this out. 53 00:02:33,640 --> 00:02:36,950 So in our case-- so N is [1, 2, 3], 54 00:02:36,950 --> 00:02:39,800 and you take the dot product with x minus 1, y, z, 55 00:02:39,800 --> 00:02:47,300 and you get 1 times x minus 1, plus 2 times y, plus 3 56 00:02:47,300 --> 00:02:50,710 times z plus 1, equals 0. 57 00:02:50,710 --> 00:02:52,390 So that's the equation of the plane. 58 00:02:52,390 --> 00:02:54,640 You could also rewrite this a bunch of different ways. 59 00:02:54,640 --> 00:02:57,050 For example, you could multiply through and collect 60 00:02:57,050 --> 00:02:58,340 all the constants together. 61 00:02:58,340 --> 00:03:04,780 So you could write this as x plus 2y plus 3z-- and then 62 00:03:04,780 --> 00:03:09,640 we've got a minus 1 plus 3, so that's plus 2-- equals 0. 63 00:03:09,640 --> 00:03:11,390 So these are two different possible forms 64 00:03:11,390 --> 00:03:12,140 for that equation. 65 00:03:12,140 --> 00:03:13,515 And you can-- you know, sometimes 66 00:03:13,515 --> 00:03:15,993 people write the constant over on this side instead 67 00:03:15,993 --> 00:03:16,993 of leaving 0 over there. 68 00:03:16,993 --> 00:03:17,450 All right. 69 00:03:17,450 --> 00:03:19,640 So several different, equivalent ways to rewrite it. 70 00:03:19,640 --> 00:03:20,430 All right. 71 00:03:20,430 --> 00:03:22,560 So there's the equation for part a. 72 00:03:22,560 --> 00:03:24,720 Now let's take a look at part b. 73 00:03:24,720 --> 00:03:28,260 So for part b we have-- let's go just back 74 00:03:28,260 --> 00:03:30,340 and remind ourselves what the question was-- so 75 00:03:30,340 --> 00:03:33,000 we have a plane that passes through the origin. 76 00:03:33,000 --> 00:03:34,840 So we know a point on the plane, and we 77 00:03:34,840 --> 00:03:38,900 know that it's parallel to the two vectors 1, 0, minus 1, 78 00:03:38,900 --> 00:03:41,611 and minus 1, 2, 0. 79 00:03:41,611 --> 00:03:42,110 OK. 80 00:03:42,110 --> 00:03:45,940 So we've got a point and we have two direction vectors. 81 00:03:45,940 --> 00:03:49,560 And so that definitely describes a plane for us, 82 00:03:49,560 --> 00:03:52,170 as long as the two directions aren't parallel, 83 00:03:52,170 --> 00:03:53,607 which they aren't in this case. 84 00:03:53,607 --> 00:03:55,440 So the question is how do we figure out what 85 00:03:55,440 --> 00:03:57,380 the equation for that plane is? 86 00:03:57,380 --> 00:03:59,300 Well, we have this nice way of figuring out 87 00:03:59,300 --> 00:04:02,800 equations for planes when we know a point and a normal. 88 00:04:02,800 --> 00:04:04,880 And we know a point, so what would be great 89 00:04:04,880 --> 00:04:08,181 is if we could come up with a normal direction to this plane. 90 00:04:08,181 --> 00:04:08,680 So OK. 91 00:04:08,680 --> 00:04:12,050 So we have two vectors in the plane, 92 00:04:12,050 --> 00:04:13,790 and we want to find a vector that's 93 00:04:13,790 --> 00:04:15,330 perpendicular to the plane. 94 00:04:15,330 --> 00:04:17,010 Well, we have a nice tool when you're 95 00:04:17,010 --> 00:04:20,080 given two vectors to figure out a vector perpendicular to both 96 00:04:20,080 --> 00:04:22,020 of them, and that's to take the cross product. 97 00:04:22,020 --> 00:04:25,590 So our normal vector should be the cross product of these two 98 00:04:25,590 --> 00:04:29,350 vectors, or, you know, any multiple of it 99 00:04:29,350 --> 00:04:32,190 would do as well. 100 00:04:32,190 --> 00:04:39,170 So for part b, the normal N should be the cross product 101 00:04:39,170 --> 00:04:40,930 of the two vectors that are in the plane, 102 00:04:40,930 --> 00:04:46,150 so it should be the cross product of 1, 0, minus 1, 103 00:04:46,150 --> 00:04:51,370 and-- what's the other one-- minus 1, 2, 0. 104 00:04:51,370 --> 00:04:53,450 So, all right, so we just have to compute 105 00:04:53,450 --> 00:04:55,360 what that cross product is. 106 00:04:55,360 --> 00:05:07,430 So this is a determinant whose first row 107 00:05:07,430 --> 00:05:10,560 is i, j, k, and whose second and third rows are 108 00:05:10,560 --> 00:05:12,160 the two vectors we're crossing. 109 00:05:12,160 --> 00:05:14,990 And OK, so we can expand this out. 110 00:05:14,990 --> 00:05:17,930 So, if you like, so this is i-- the coordinate of i 111 00:05:17,930 --> 00:05:22,795 is going to be 0 minus minus 2, so that's 2. 112 00:05:22,795 --> 00:05:24,710 The coordinate of j is going to be 113 00:05:24,710 --> 00:05:27,530 the negative of the determinant of this minor, which is 114 00:05:27,530 --> 00:05:31,015 0 minus minus 1 times minus 1. 115 00:05:31,015 --> 00:05:34,470 So the determinant of the minor is minus 1, 116 00:05:34,470 --> 00:05:38,620 so the coordinate of j is going to be plus 1. 117 00:05:38,620 --> 00:05:40,370 And the coordinate of k is the determinant 118 00:05:40,370 --> 00:05:44,060 of this minor, which is just 2. 119 00:05:44,060 --> 00:05:48,380 So the normal vector in this case is the vector [2, 1, 2]. 120 00:05:48,380 --> 00:05:49,130 So OK. 121 00:05:49,130 --> 00:05:51,589 So now we've got a normal vector and we have a point. 122 00:05:51,589 --> 00:05:53,880 We were given that the plane passes through the origin. 123 00:05:53,880 --> 00:06:00,719 So the equation-- using the same idea 124 00:06:00,719 --> 00:06:01,885 as in the previous question. 125 00:06:01,885 --> 00:06:04,050 So the origin is just [0, 0, 0]. 126 00:06:04,050 --> 00:06:06,850 That's a nice point to know it passes through. 127 00:06:06,850 --> 00:06:14,420 So the equation is just 2x plus y plus 2z is equal to 0. 128 00:06:14,420 --> 00:06:17,440 So this is the equation in part b. 129 00:06:17,440 --> 00:06:17,940 All right. 130 00:06:17,940 --> 00:06:21,630 So part c, we're given that the plane 131 00:06:21,630 --> 00:06:23,950 passes through three points. 132 00:06:23,950 --> 00:06:25,650 So once again, three points. 133 00:06:25,650 --> 00:06:27,240 So we have a point in particular. 134 00:06:27,240 --> 00:06:29,520 We have three of them. 135 00:06:29,520 --> 00:06:32,430 And so what we need then to get to the equation 136 00:06:32,430 --> 00:06:34,560 is we need a normal. 137 00:06:34,560 --> 00:06:37,440 And we saw in part b that we could 138 00:06:37,440 --> 00:06:41,120 get a normal if we knew two vectors that lay in the plane. 139 00:06:41,120 --> 00:06:43,250 So in this case, we have three points. 140 00:06:43,250 --> 00:06:47,210 So what we'd like is to find two vectors that lie in the plane, 141 00:06:47,210 --> 00:06:50,970 and then use those two vectors to come up with a normal 142 00:06:50,970 --> 00:06:51,820 to the plane. 143 00:06:51,820 --> 00:06:55,260 So in our case that's particularly-- well, 144 00:06:55,260 --> 00:06:56,970 in any case, that's not that hard. 145 00:06:56,970 --> 00:06:58,650 You have three points, right? 146 00:06:58,650 --> 00:07:01,880 So you have three points somewhere, P, Q, and R. 147 00:07:01,880 --> 00:07:05,459 And so if you want to know two vectors in the same plane 148 00:07:05,459 --> 00:07:07,250 as these three points, well, you could just 149 00:07:07,250 --> 00:07:11,460 take the vectors that connect one of the points 150 00:07:11,460 --> 00:07:13,200 to two of others, for example. 151 00:07:13,200 --> 00:07:24,460 So in our case, the plane-- since the plane 152 00:07:24,460 --> 00:07:34,150 passes through the points (1, 2, 0), and (3, 1, 1), 153 00:07:34,150 --> 00:07:36,450 and-- what's the last one-- (2, 0, 0). 154 00:07:40,320 --> 00:07:45,640 So the plane is parallel to-- well, it 155 00:07:45,640 --> 00:07:48,230 doesn't matter which one we choose, 156 00:07:48,230 --> 00:07:50,460 so for example, we can say the vector that 157 00:07:50,460 --> 00:07:53,960 goes from here to here, so we take this and subtract that 158 00:07:53,960 --> 00:07:59,890 from it, so that would give us, for example-- 2, minus 1, 1. 159 00:07:59,890 --> 00:08:02,800 And we could say the vector from here to here, so we take this 160 00:08:02,800 --> 00:08:05,890 and subtract that from it. 161 00:08:05,890 --> 00:08:12,270 And that will give us 1, minus 2, 0. 162 00:08:12,270 --> 00:08:15,130 So from three points we could get two vectors 163 00:08:15,130 --> 00:08:16,580 that are parallel to the plane. 164 00:08:16,580 --> 00:08:19,070 And we have a choice of a point to use. 165 00:08:19,070 --> 00:08:21,850 We could use, for example, the same point, (1, 2, 0), 166 00:08:21,850 --> 00:08:22,665 as our base point. 167 00:08:22,665 --> 00:08:24,040 And so then we can go back and do 168 00:08:24,040 --> 00:08:25,830 exactly what we did in part b. 169 00:08:25,830 --> 00:08:29,650 So with those two vectors, you can take their cross product, 170 00:08:29,650 --> 00:08:31,290 and find a normal vector to the plane. 171 00:08:31,290 --> 00:08:32,790 So I'm not going to do that for you. 172 00:08:32,790 --> 00:08:35,650 I'll leave that for you as an exercise. 173 00:08:35,650 --> 00:08:41,270 Finally, in part d, we have a plane 174 00:08:41,270 --> 00:08:45,330 that's parallel to the plane in part a, 175 00:08:45,330 --> 00:08:47,950 and passes through (1, 2, 3). 176 00:08:47,950 --> 00:08:49,080 The point (1, 2, 3). 177 00:08:49,080 --> 00:08:53,100 So let me just rewrite over here, parallel 178 00:08:53,100 --> 00:08:56,730 to-- so the plane in part a had equation 179 00:08:56,730 --> 00:09:07,930 x plus 2y plus 3z plus 2 equals 0, 180 00:09:07,930 --> 00:09:13,400 and passing through the point (1, 2, 3). 181 00:09:13,400 --> 00:09:13,966 All right. 182 00:09:13,966 --> 00:09:15,590 So this is the information that we know 183 00:09:15,590 --> 00:09:17,100 about our plane in this case. 184 00:09:17,100 --> 00:09:20,060 We know that it's parallel to the plane with this equation, 185 00:09:20,060 --> 00:09:23,840 and that it passes through the point (1, 2, 3). 186 00:09:23,840 --> 00:09:26,410 Well, it's parallel to this plane. 187 00:09:26,410 --> 00:09:28,290 Two planes are parallel exactly when they 188 00:09:28,290 --> 00:09:30,390 have the same normal vector. 189 00:09:30,390 --> 00:09:32,670 So remember that the normal vector here 190 00:09:32,670 --> 00:09:36,070 is always going to be encoded by these coefficients of x, y, z. 191 00:09:36,070 --> 00:09:41,110 In part a, this plane had normal vector [1, 2, 3], 192 00:09:41,110 --> 00:09:43,260 and that [1, 2, 3] shows up in the coefficient 193 00:09:43,260 --> 00:09:45,800 of x, coefficient of y, and coefficient of z, which 194 00:09:45,800 --> 00:09:48,400 are 1, 2, and 3, respectively. 195 00:09:48,400 --> 00:09:52,820 So to get parallel planes when you have the equation already, 196 00:09:52,820 --> 00:09:54,870 one thing you could do is you can just say, oh, 197 00:09:54,870 --> 00:09:57,720 so that just means I leave these coefficients the same, 198 00:09:57,720 --> 00:09:59,950 and I have to change the constant. 199 00:09:59,950 --> 00:10:02,240 Another thing you could do is you could just go back 200 00:10:02,240 --> 00:10:04,614 to our definition and say, OK, so we know that the normal 201 00:10:04,614 --> 00:10:07,880 vector is [1, 2, 3]-- the vector [1, 2, 3]-- 202 00:10:07,880 --> 00:10:11,730 and that it passes through the point (1, 2, 3). 203 00:10:11,730 --> 00:10:13,730 Either of these two methods will work. 204 00:10:13,730 --> 00:10:15,690 So let me describe, let me show you 205 00:10:15,690 --> 00:10:18,960 what this second, this new method I mentioned is. 206 00:10:18,960 --> 00:10:21,410 So we know that the equation of the plane 207 00:10:21,410 --> 00:10:29,970 has to be x plus 2y plus 3z plus something-- 208 00:10:29,970 --> 00:10:35,710 that's a big question mark there-- equal to 0. 209 00:10:35,710 --> 00:10:37,620 We know that the equation of the plane 210 00:10:37,620 --> 00:10:39,079 is going to have to look like this. 211 00:10:39,079 --> 00:10:40,661 Because it has the same normal vector, 212 00:10:40,661 --> 00:10:42,300 it has to be parallel to this plane. 213 00:10:42,300 --> 00:10:43,883 And so then we just need to figure out 214 00:10:43,883 --> 00:10:46,370 what goes into this box in order to make 215 00:10:46,370 --> 00:10:48,340 this the equation of the right plane. 216 00:10:48,340 --> 00:10:49,680 Well, what else do we know? 217 00:10:49,680 --> 00:10:52,490 We know that it passes through the point (1, 2, 3). 218 00:10:52,490 --> 00:10:56,650 So when we put in 1 for x, 2 for y, and 3 for z, 219 00:10:56,650 --> 00:11:00,190 this equation has to be true. 220 00:11:00,190 --> 00:11:03,620 This point (1, 2, 3) has to be a solution to this equation. 221 00:11:03,620 --> 00:11:06,030 So when we put in 1, 2, and 3, we 222 00:11:06,030 --> 00:11:10,750 have to have that 1, plus 2 times 2, plus 3 times 223 00:11:10,750 --> 00:11:16,280 3, plus that same question mark, is equal to 0. 224 00:11:16,280 --> 00:11:20,520 Well, this part is 1 plus 4 plus 9 is 14, 225 00:11:20,520 --> 00:11:24,230 so 14 plus whatever goes in here has to be equal to 0, 226 00:11:24,230 --> 00:11:26,090 so this better be equal to negative 14. 227 00:11:26,090 --> 00:11:26,590 Negative 14. 228 00:11:30,180 --> 00:11:31,970 So the equation for the plane in that case 229 00:11:31,970 --> 00:11:38,210 is exactly x plus 2y plus 3z minus 14 equals 0. 230 00:11:38,210 --> 00:11:41,360 Now, if you didn't like that method, the other thing you can 231 00:11:41,360 --> 00:11:43,090 do-- which I said before, let me just 232 00:11:43,090 --> 00:11:46,150 repeat it-- is that since it's parallel to this plane, 233 00:11:46,150 --> 00:11:47,890 it has the same normal vector. 234 00:11:47,890 --> 00:11:50,550 And we knew that the normal vector to this plane 235 00:11:50,550 --> 00:11:51,740 was [1, 2, 3]. 236 00:11:51,740 --> 00:11:54,500 So you have a normal vector-- the vector [1, 2, 3]-- 237 00:11:54,500 --> 00:11:56,720 and you have a point-- the point (1, 2, 3)-- 238 00:11:56,720 --> 00:11:59,190 and so you can just use the usual process given a point 239 00:11:59,190 --> 00:12:01,570 and a normal vector. 240 00:12:01,570 --> 00:12:04,800 So just to recap, we had four different characterizations 241 00:12:04,800 --> 00:12:05,380 of a plane. 242 00:12:05,380 --> 00:12:08,390 We had a plane given in terms of its normal vector and a point 243 00:12:08,390 --> 00:12:09,470 that it contains. 244 00:12:09,470 --> 00:12:11,580 We had a plane given in terms of a point 245 00:12:11,580 --> 00:12:13,300 and two vectors parallel to it. 246 00:12:13,300 --> 00:12:17,270 We had a plane given in terms of three points on it. 247 00:12:17,270 --> 00:12:19,470 And we had a plane given in terms of a point 248 00:12:19,470 --> 00:12:21,570 and of another plane parallel to it. 249 00:12:21,570 --> 00:12:23,960 So we have-- in all these different cases, 250 00:12:23,960 --> 00:12:26,760 we can apply different methods to compute 251 00:12:26,760 --> 00:12:28,150 the equation of our plane. 252 00:12:28,150 --> 00:12:31,530 So in the first case, we just do this very straightforward 253 00:12:31,530 --> 00:12:34,010 computation that you saw in lecture here. 254 00:12:34,010 --> 00:12:36,850 Where you just realize that the normal vector has 255 00:12:36,850 --> 00:12:40,010 to be orthogonal to the vector lying in the plane. 256 00:12:40,010 --> 00:12:42,010 So you take their dot product and that gives you 257 00:12:42,010 --> 00:12:43,710 the equation right away. 258 00:12:43,710 --> 00:12:45,215 In the second case, where you had 259 00:12:45,215 --> 00:12:49,380 two parallel vector-- or sorry, yeah, two parallel 260 00:12:49,380 --> 00:12:50,840 vectors to the plane, two vectors 261 00:12:50,840 --> 00:12:53,364 lying in the plane-- you need to come up with a normal. 262 00:12:53,364 --> 00:12:55,780 And you can always come up with a normal by taking a cross 263 00:12:55,780 --> 00:12:58,170 product of those two vectors, as long as you're careful. 264 00:12:58,170 --> 00:13:00,460 If you accidentally chose your two vectors parallel 265 00:13:00,460 --> 00:13:02,490 to each other, that wouldn't work. 266 00:13:02,490 --> 00:13:04,610 You'd just get 0 here, and that's no good. 267 00:13:04,610 --> 00:13:07,000 But, so you have to choose two non-parallel vectors 268 00:13:07,000 --> 00:13:09,380 in the plane in order to make this work. 269 00:13:09,380 --> 00:13:11,340 In the third case, you have three points. 270 00:13:11,340 --> 00:13:13,310 And so with three points what you can do is you 271 00:13:13,310 --> 00:13:15,440 can choose two vectors connecting 272 00:13:15,440 --> 00:13:16,510 some of those points. 273 00:13:16,510 --> 00:13:19,290 And that gives you two vectors that lie in the plane, 274 00:13:19,290 --> 00:13:21,830 and that reduces to the case of the previous part, 275 00:13:21,830 --> 00:13:23,750 and then again, you can take a cross product 276 00:13:23,750 --> 00:13:25,050 to get a normal vector. 277 00:13:25,050 --> 00:13:27,450 Finally, we did this fourth problem 278 00:13:27,450 --> 00:13:29,570 where we were given a plane parallel to it. 279 00:13:29,570 --> 00:13:32,620 And so you can read off the normal vector 280 00:13:32,620 --> 00:13:35,380 from the coefficients of x, y, and z in the equation. 281 00:13:35,380 --> 00:13:39,770 And then either use the very first method 282 00:13:39,770 --> 00:13:42,480 with a point and normal vector, or just 283 00:13:42,480 --> 00:13:45,320 realize that you just have to find the appropriate value 284 00:13:45,320 --> 00:13:47,400 of the constant so that this point actually 285 00:13:47,400 --> 00:13:48,550 lies on the plane. 286 00:13:48,550 --> 00:13:50,397 So I'll end there.