1 00:00:01,439 --> 00:00:02,439 Hi everyone 2 00:00:02,439 --> 00:00:05,680 In this video, we will be talking about two topics of 3D chapter for 3 00:00:05,680 --> 00:00:10,210 mathematics of JEE preparation. 4 00:00:10,210 --> 00:00:15,349 The topics are: doing problems on finding perpendicular distance of a point from line 5 00:00:15,349 --> 00:00:18,880 and we will be finding equations of a line perpendicular to two lines given. 6 00:00:18,880 --> 00:00:27,870 These are small topics but as in other maths topics, you should learn the approach to solve 7 00:00:27,870 --> 00:00:28,870 problems like these. 8 00:00:28,870 --> 00:00:33,910 Let us start by finding perpendicular distance of a point from a line. 9 00:00:33,910 --> 00:00:35,870 We have a line given. 10 00:00:35,870 --> 00:00:43,010 Let us saw we have point A, and that is a_vector, and we have been given b_vector, which is 11 00:00:43,010 --> 00:00:44,470 the parallel vector. 12 00:00:44,470 --> 00:00:49,670 And we have been given a point P. Let us call it p_vector. 13 00:00:49,670 --> 00:00:54,479 And we have been asked to find the perpendicular distance of the point from the line. 14 00:00:54,479 --> 00:00:58,999 You may recall these kind of problems in 2D in straight lines. 15 00:00:58,999 --> 00:01:02,430 But now its 3D or 3-Dimensinal Geometry. 16 00:01:02,430 --> 00:01:05,030 So how will we solve this problem? 17 00:01:05,030 --> 00:01:13,270 One of the approaches to start this to think that there is a point C here, let us call 18 00:01:13,270 --> 00:01:14,270 it c_vector. 19 00:01:14,270 --> 00:01:15,550 And the idea is that we know, 20 00:01:15,550 --> 00:01:18,680 c_vector=a_vector + \lambda b_vector 21 00:01:18,680 --> 00:01:24,680 because this point is on the line. 22 00:01:24,680 --> 00:01:30,760 And, PC_vector is perpendicular to b_vector. 23 00:01:30,760 --> 00:01:31,800 This is what we know, right? 24 00:01:31,800 --> 00:01:37,590 These are the two conditions we know for the PC_vector to be perpendicular to the parallel 25 00:01:37,590 --> 00:01:41,590 vector because this is perpendicular to the line, so it has to be perpendicular to the 26 00:01:41,590 --> 00:01:43,270 parallel vector. 27 00:01:43,270 --> 00:01:48,310 It should flash to you that the condition of perpendicularity it PC_vector \dot b_vector 28 00:01:48,310 --> 00:01:49,310 = 0. 29 00:01:49,310 --> 00:01:51,620 Again, the knowledge of vectors is super critical in this chapter. 30 00:01:51,620 --> 00:01:54,650 I have emphasized that many times in the last two videos also. 31 00:01:54,650 --> 00:01:58,630 PC_vector is nothing but c_vector - p_vector. 32 00:01:58,630 --> 00:02:04,810 Thus, c_vector-p_vector \dot b_vector = 0. 33 00:02:04,810 --> 00:02:19,150 And c_vector is nothing but (a_vector + \lambda b_vector - p_vector) \dot b =0. 34 00:02:19,150 --> 00:02:31,020 If you were to find the value of \lambda, you will get, \lambda = ( (p_vector - a_vector) 35 00:02:31,020 --> 00:02:35,000 \dot b_vector, divided by |b|^2. 36 00:02:37,000 --> 00:02:44,000 Now you have got the value of \lambda, so you can calculate the point, and then you can calculate the distance. 37 00:02:45,000 --> 00:02:52,900 I hope you are able to understand and follow what we are doing here. You can solve this 38 00:02:53,000 --> 00:02:57,000 equation using the associative property by opening the brackets, and rearranging, you get this. 39 00:02:57,100 --> 00:03:02,000 Let us do a problem on this so that you can quickly understand. 40 00:03:02,100 --> 00:03:33,000 You have been given P=(1,2,3). And the line is (x-6)/3 = (y-7)/2 = (z-7)/(-2). 41 00:03:33,000 --> 00:03:44,000 How will we find \lambda? I will suggest you to remember this but you may want to derive it in the paper. 42 00:03:44,000 --> 00:03:47,800 It is up to you. I will recommend you to remember this, it is 43 00:03:47,880 --> 00:03:52,050 very easy to remember ( (p_vector - a_vector) \dot b_vector ) / |b_vector|^2. 44 00:03:52,050 --> 00:04:02,000 Here, p_vector = i_cap + 2j_cap+ 3k_cap 45 00:04:02,000 --> 00:04:24,000 a_vector = 6i_cap + 7j_cap+ 7k_cap b_vector = 3i_cap + 2j_cap - 2k_cap 46 00:04:24,000 --> 00:04:28,660 If I were to find \lambda here, this would become 47 00:04:28,660 --> 00:04:46,150 p_vector - a_vector is -5i - 5j - 4k \dot with 3i + 2j -2k 48 00:04:46,150 --> 00:04:50,460 ( (p_vector - a_vector) \dot b_vector ) / |b_vector|^2. 49 00:04:50,460 --> 00:04:59,090 |b_vector|^2 = 3^2 + 2^2 + 2^2 = 8 + 9 = 17. 50 00:04:59,090 --> 00:05:10,070 If you will do this, this will become: -15-10+8 = -17. 51 00:05:10,070 --> 00:05:16,120 Thus \lambda = -1. 52 00:05:16,120 --> 00:05:35,169 The point c_vector = a_vector - b_vector, and it comes out to be, 3i_cap + 5j_cap + 53 00:05:35,169 --> 00:05:36,600 9k_cap. 54 00:05:36,600 --> 00:05:52,300 And the distance |PC| = (2^2+3^2+6^2)^(1/2) = 7. 55 00:05:52,300 --> 00:06:01,790 You know the vector PC now, so you can find the distance by taking the sum of x_component^2, 56 00:06:01,790 --> 00:06:07,310 y_component^2 and z_component^2, and that comes out to be 7. 57 00:06:07,310 --> 00:06:12,180 And I hope that is very obvious for you to solve these types of problems. If you are 58 00:06:12,180 --> 00:06:16,850 still not very comfortable, please practice. Just make some problems yourself, make an 59 00:06:16,850 --> 00:06:22,290 equation of a line, find a point, and then do the problem. It is very easy. If you do 60 00:06:22,290 --> 00:06:25,750 this several times, it should be quick for you to remember. 61 00:06:25,750 --> 00:06:38,919 Let me give you one more HW problem so that you can do it on your own. And the HW problem 62 00:06:38,919 --> 00:06:41,620 is: 63 00:06:41,620 --> 00:06:58,880 Find a point on y_axis such that 64 00:06:58,880 --> 00:07:16,290 perpendicular distance from the point to line x=y=z is unity. 65 00:07:16,290 --> 00:07:24,800 So this is an opposite question. You have been given the task to find the point on y-axis 66 00:07:24,800 --> 00:07:28,550 such that the perpendicular distance from the line to the point is unity. So, how will 67 00:07:28,550 --> 00:07:29,550 you do this problem? 68 00:07:29,550 --> 00:07:38,120 You can assume a point: (0, k, 0), and proceed now, completely the same way. You can very 69 00:07:38,120 --> 00:07:43,259 quickly find the value of this thing (\lambda), and then you can calculate PC, and you can 70 00:07:43,259 --> 00:07:45,139 equate it to 1, to find the value of k. 71 00:07:45,139 --> 00:07:51,460 Please do this problem. This will give you a nice practice to solve the problems of perpendicular 72 00:07:51,460 --> 00:07:57,310 distance of a point from a line. 73 00:07:57,310 --> 00:08:05,910 Let us do another problem which I wanted to do. I will not be giving you any method, I 74 00:08:05,910 --> 00:08:12,699 just want to do an example problem since that example will be enough to understand the topic. 75 00:08:12,699 --> 00:08:14,770 The question says we have been given two lines: 76 00:08:14,770 --> 00:08:23,590 (x-1)/1 = (y-2)/2 = (z-3)/3, L1 is this 77 00:08:23,590 --> 00:08:40,729 L2 has been given as (x-2)/4 = (y-3)/5 = (z-1)/6. 78 00:08:40,729 --> 00:08:56,990 The 79 00:08:56,990 --> 00:09:09,630 question says find the line perpendicular to L1 and L2 and passing through 80 00:09:09,630 --> 00:09:21,350 1,1,1. How to solve this problem? 81 00:09:21,350 --> 00:09:26,410 So you have been given line L1. Let me write answer here: 82 00:09:26,410 --> 00:09:33,839 Let us say this is L1 and this is L2. You have been given a1_vector, b1_vector, a2_vector, 83 00:09:33,839 --> 00:09:42,720 b2_vector. You have been asked to find line 3 for which you know that it is perpendicular 84 00:09:42,720 --> 00:09:44,930 to both, and it passes through the point. 85 00:09:44,930 --> 00:09:48,830 To find the equation of line, you need a point, which you already have, and you need a parallel 86 00:09:48,830 --> 00:09:53,990 vector. What is speciality of parallel vector? That the line is perpendicular to line L1 87 00:09:53,990 --> 00:10:02,310 and L2. Hence, the parallel vector will be perpendicular to both b1 and b2. 88 00:10:02,310 --> 00:10:06,519 And as soon as I say this, it should flash into your mind, that this is basically asking 89 00:10:06,519 --> 00:10:10,560 you to take a cross product. Thats it. 90 00:10:10,560 --> 00:10:15,810 If you cannot recall this, please become more thorough with understanding the physical meaning 91 00:10:15,810 --> 00:10:20,600 of dot product and cross product. We have discussed here dot product: perpendicularity 92 00:10:20,600 --> 00:10:23,820 condition of dot product already. We have done in the last video, that dot product is 93 00:10:23,820 --> 00:10:31,519 used for finding angles. So as soon as I say, it should come to your mind, the vector perpendicular 94 00:10:31,519 --> 00:10:35,220 to two vectors is the cross product of two vectors. 95 00:10:35,220 --> 00:10:42,320 So what you have to do is that you have to find b3_vector = b1 \cross b2, and if you 96 00:10:42,320 --> 00:10:51,640 do, you should remember that for cross product you have to do this i, j, k — and opening 97 00:10:51,640 --> 00:10:52,730 up the determinant. 98 00:10:52,730 --> 00:11:06,890 So if you do that, this will be: i(-3) + j(6) + k(-3). Is this perpendicular to b1 and b2? 99 00:11:06,890 --> 00:11:14,770 You should remember that condition of perpendicularity is dot product is zero. 100 00:11:14,770 --> 00:11:21,800 So if you take -3 + 12 - 9, that is dotted with b1, it comes out to be 0. 101 00:11:21,800 --> 00:11:27,029 If you dot with b2, -12 + 30 -18, so 30 - 30 = 0. So it is perpendicular to both b1 and 102 00:11:27,029 --> 00:11:28,029 b2. 103 00:11:28,029 --> 00:11:29,700 And you already have point. 104 00:11:29,700 --> 00:11:44,529 So the equation of line L3 could be easily written as: (x-1)/3 = (y-1)/6 = (z-1)/(-3). 105 00:11:44,529 --> 00:11:52,029 This line is perpendicular to both the lines and passes through (1,1,1). 106 00:11:52,029 --> 00:11:56,630 I hope that you are able to understand this, understand how important the understanding 107 00:11:56,630 --> 00:12:02,220 of vectors is, and how to solve the problems. We discussed two things here: how to find 108 00:12:02,220 --> 00:12:07,910 the perpendicular distance of a point from a line. For that, we can remember this formula 109 00:12:07,910 --> 00:12:15,220 of \lambda. Calculate \lambda, calculate the point, and when you get the coordinates of 110 00:12:15,220 --> 00:12:19,670 the point, you subtract from the original point to get the distance. I have given you 111 00:12:19,670 --> 00:12:22,390 a homework problem, so please do that at home. 112 00:12:22,390 --> 00:12:27,380 And then we did a problem of finding a line perpendicular to two lines. For that you are 113 00:12:27,380 --> 00:12:31,580 taking the cross product of the parallel vectors, and then you will get the equation of the 114 00:12:31,580 --> 00:12:32,580 line. 115 00:12:32,580 --> 00:12:35,970 I hope you enjoyed this video. There are a lot more interesting videos and topics in 116 00:12:35,970 --> 00:12:39,600 this chapter. So I hope to see you in the next video. Thank you.