hello welcome to simplified Minds numer one Mark is very very thinks out of I mean it be very very easy coming to 3D Direction cosin so very simple concept Direction Co dire raos and it's very important for even exams equation of line in space actually equation of line given point and given Direction two points two point okay angle between two lines very very simple and the distance between the SK distance between the lines that is Q lines simple concept very very simple very simple chapter okay first what is Direction ratios and Direction cosines so very simple Direction ratios very simple for example a vector equal to 2 I 2 J and 2 K maybe two 3 3 J and 4 K and vector Vector coefficient 2 3 4 Direction ratios this 2A what is it component in X Y vector vector coefficient Direction ratios in so what are Direction coin important water Direction cosines as a word says Direction cosin wishen some last something oh take so the made by that line line is it makes Alpha Beta gamma with the xaxis y axis Z axis so angle Alpha with x-axis it is towards you it is away from you chemistry organicist you have some kind of notation it's towards you it's behind you obviously it will make some angles Alpha Beta GMA with the axis x axis Y axis Z axis so cost of the alpha C of the beta C of the gamma is itself called as Direction cosin and Direction cosin you know we call it as Dr and the direction cosin sorry Direction DCS Direction DCS and the direction cosin so it is nothing but C of the angle made that is cos Alpha cos beta and cos gamma it is cosine of the angle made by the direction Vector with respect to the x axis Y axis Z axis this is called as curve D is Direction cos so this we we give a notation L MN L MN simple so we can for example point y j and z l will be X divid modulus of x² + y² + z s y / root of x² + y² + z s n z / root of x² + y² + z relation so cos beta cos G is but X of that's very simple /un of 2 s + 3 S + 4 s m 3 / root of 2 s + 3 S + 4 S 4 / n root of 2 s + 3 4 this is a form this will be what happens root of x² + y sare root of x sare + y Square py the horizontal projection now previous previous chapter find projections physics component X component is cos Alpha towards the angle cos Alpha X+ Y squ what is this length will becomes so this is root of this length x will become root of x² + y² cos Alpha cos Alpha x / x + y sare same thing for 3D add one Z Square One yare xare some I don't feel anything they this Square this Square equal one anyway so this is about Direction ratios and direction it tells about what is a ratio word ratio total Vector component y component line makes 60 90 60 30 with positive direction xais direction so what is cos 90 so cos 90 and cos 60 and then afterwards cos 30 so this is a direction signs cos 90 0 cos 60 half cos 30 so a line has direction cosin as 0 1 by 2 < tk3 by2 then the angle made by the line with Y AIS istion okay 0 180 something what is angle made by that particular line with respect to the Y AIS half so they given cos beta as half and what is beta ask that is 60° understand given ratios 2 1us one as in one markers because you don't have one markers 2id by square plus square plus Square 4 so it is again- 1 /un 9 one more - 2 /un 9 so this is the direction cosine that is 2 by 3 - 1 by 3 and - 2 by that's answer they can to M write the direction cosin of xaxis not xaxis we'll write all y axis and then Z AIS x-axis direction cosin direction coin know angle made by a particular line with respect to xaxis Y axis Z axis now they're asking direction cosin of x- axis what is the angle made by xais with respect to xaxis angle xais XIs both are same only xaxis the axis will make they asking Direction Co of angle x axis itself xaxis will make with with xaxis itself 0° no in xaxis xaxis with respect to Y axis 90° xaxis with respect to Z axis 90° so what is the direction what are the angles made by x-axis so it makes 0° 90° and 90° so what is 0 one cos 9 0 cos 90 0 xais so xais Direction cos 1A 0 0 y yis 0 1 0 0 0 0 1 0 0 1is will make itself so the xaxis with respect to xaxis xaxis x x AIS x axxis x axis Y axis Z axis z x with respect to x x with respect to Y 90 y with respect to X 90° y with respect to Y 0° y with respect to Z 90° so cos apply cos 90 0 cos 0 1 cos 90 0 same thing same concept next one of the most easy dations know standed easy simple okay Dimension how do you find it you know uniquely unique unique there should not be any Liney for example question simple find the find the equation passing through this point simp how many equation how many lines are passing through this point there are 100 crores of there are million hundreds of infinite number of lines passing through given points so is there any if I say find equation Point passing through that point point you should give one more definition find equation of a line passing through a point passing through a point and it is parall to this line passing through a point and parall to this line so there's only one line which is passing through this point at the same time it's par to this one more line so it became unique now Ma you can't directly find you have to there's a one only one line which is passes through a given point and parall to a given direction if I to you find equation of line passing through these two points there's only one line there's only one line that can pass through the these two points Al can you draw more than one line no not possible so it is unique in terms of vors you can learn anything you want 35 you cross that very anyway very there can be no line equation of a line passing through a given point and given Direction find equation of line which is passing through this point and parall to a given Direction a consider a with Direction point a with you can take this as X1 y1 Z1 where so X1 y1 Z1 point a with coordinates X1 y1 Z1 with Direction Vector as position Vector as a vector b b is a direction of a given line consider arbitary point arbitary point is the random point arbitary point p with Direction the position Vector R Vector I show in the figure problem one L line I gave you an example for last lecture can I say this AP this AP Vector is Lambda * of B Vector why because it's given they have given that they have given that the B is a direction B Vector is is a Direction they given that it's and they also given is parall to the given you have to find a line that's parallel to the given Direction they already given a parallel lines par one is Lambda time 2 3 4 last parar some show that these two points are you to show one L Lambda other line simple a v b Vector what op minus o op minus o last vect so what is op Vector op Vector is R Vector a vector is what a vector o a vector is a vector Lambda * B Vector so what is your R Vector will be equal to a vector shift a vector plus very expor form vector vect vector form form cian formally we should not have vectors we just have the points we going to let let us take let us take R Vector y j random point which satisfy the given condition it can be anywhere random point and the given point so that mean the given it as X1 i y1 j Z1 k b vector uh Direction I I will take it as a i Cap B J cap c c already all they given it as a I show you so sub okay so what I can do is basically I'll substitute here I'll sub in this equation so r x i x i y j z k must be equal to e that is X1 I cap y1 J cap Z 1 K cap plus Lambda * of a i x- X1 already y- y1 JC Z minus Z 1 K cap is equal Lambda * of Lambda Lambda a i Lambda BJ L Cal Lambda a x - X1 by a equal sorry x - X1 is equal to Lambda a y - y1 byal Lambda B Z - Z 1 equal Lambda C what is my Lambda Lambda Comm Lambda x - X1 by a in this case Al it is y - y1 by B which is equal to Z - Z 1 by simple deration not requ okay L Lambda X Lambda y y by a Lambda 1 by C obviously Lambda we can equate this is called asan own words a is a point P isit point B is a given Direction R are position vectors position locates the particular position1 very you should know this let's go next one and as I told you last so you can check for my last year videos of find the vector andan equation of the line through the point simple b b b Das so what is my a vector is have 5 I + 2 J cap - 4 K Al what is B Vector will be uh 3 I + 2 J cap - 8 K cap Vector B Vector next one is I2 uh then afterwards uh a I2 B I2 dire what they ask vector Vector r vectoral a vector plus Lambda time of B Vector simple Lal 3 IUS 2 minus 8 K simple plus Lambda * okay you can write it as this is direct you can write it x - X1 by a by y - y1 by b z minus Z 1 by C what is it now x - X1 that is point point so xus 5 by a that is 3 would be equal to y - y1 y - y1 that is 2 y - 2 / B that is 2 = to zus Z 1 that isus of-1 + 4/ - caran form and know you write this as like X1 X1 I cap y1 J cap you're taking this as like this at 1K cap this has a I cap this as BJ cap then we are writing this but you can write this simple okay very very simple next question simple topic next angle between two lines very simple direction for example direction of this line is this let's imagine direction of don't don't you think that between the lines don't you think that that is same as this angle like thatle angle between the lines is same as angle between the directions because they both are parall lines right yes or no Direction no direction for example let's imagine the direction of this line is B1 L1 line to B1 L2 L2 B2 don't you think this angle is same as this angle yes it same because they both are parall lines so the direction between the lines is nothing but Direction between the vectors how to findle between them you to use a DOT product two lines suppose if I take first line as R1 equal A1 Vector plus Lambda * of B1 second R2 Vector equal to A2 Vector plus Lambda * of how if an angle dot product so cos Theta is equal to B1 CR B2 / modulus of V1 and modul of B2 there was student actually last so dot angle find it is not a vector it's angle so you can't findle between these two vectors using that formula Vector so we always use a DOT product so B1 B2 by mod of simp B1 B you just have to find the angle like how we did for the previous chapter2 a Lambda B A1 B1 A2 + Lambda * of B2 ble I hope you found the angle by yourself by I'm saying yourself is sometimes B1 basically and B2 multiply 1 into 3 1 into 3 3 2 into 2 that is is 4 2 into 6 that is 12 divided by so this modulus modulus of that that is modulus of 1 sare + 2 square + 2 square 2 square + 2 square is 8 8 4 + 1 9 9 into root 3 square + 2 sare + 3 square is 6 6 + 6 sare 36 36 + 4 40 40 + 9 49 so 49 equal to 4 5 6 19 19 / 3 into 7 so cos inverse of 19 by 21 21 19 by 21 so cos inverse of 19id 21 next next show that these angles are perpendicular to each other actually question find the value of K for which perpendicular simple example you won't get dis the vector chap I example also so x - 5 7 and this one so question first anyway first we'll solve this question okay how do you show this two vectors are you know perpendicular dot product zero dot product between the direction should be zero direction direction form form x - X1 by a y - y1 by b z - Z 1 by C you get General format x - 5 by 7 actually [Music] x- x - 5 by 7 actually Yus of-2 so the point is X1 y1 it won't affect the denominator but sometime it affect the denominator I'll show you by changing the format so anywhere this is the denominator for example chinat but change because general form already there same format you do x- 0 by 1 why I'm p x - 1 by y - 0 by 2 Z - 0 by 3 it isn't form there's no change 7 into 1 7- 5 into 2 that's basically - 10 3 into 1 that is 3 so what answer now basically 7 + 3 10 - 10 0 answer is z they are perpendicular so if we find the B1 Vector the direction of the first line dot B2 Vector should be equal to Z you have to find first B 7 i- 5 J + 1 k b B2 what is B2 1 I 2 into 1us 5 into 2 1 into 3 total if B1 into B2 is z therefore you write steps only two show both the steps and then and show this Z let's go for next question question one more I want to tell you one more thing in model paper same question find the value of inste of S inste of seven they given K so you'll get like K into 1 K and - 5 into 2 that is - 10 and you have 1 into 3 that is 3 n that should be equal to Zer find the value of K now k - 10 + 3 is - 10 + 3 - 7 - 7al 0 K = 7 so they ask this kind of questions also for example type question look find the value of P for which this two are perpendicular you to get the format of xus we have to convert that and know negative common instead of wrting negative here I can write negative denominator y - 2 / 2p 7 sameus a X11 equ not no problem no issues x because- x + - 7 so if I take - 7 out then I'll get x - multiply - 7 - 7us is next so but again I want this minus 7 to be down x -1 / 3 p by - 7 X - 1 by 3 p by 7us that must be equal what's the value for this multiply corresponding this is this this a direction now - 3 into this- 3 into again minus - 3 Pi by 7+ 2 p by 7 into 1 + -2 into 5 is - 10 that must be equal to Z - minus plus 9 P by 7 2 p by 7 - 10 = 0 11 P 11 P by 7 = 10 and what is my P will be P will be 70 / 11 simple again if you wanted to write steps you can write the B1 Vector as then for perpendicular condition b1. B2 is Z on the simple let's go for next one next up so distance between the lines so distance between lines and the lines in the space two there can be two par lines if they not par 100% they will meet somewhere 100% they will meet so in case of 2D two dimension lines can be either par or intersecting of course we have coining L again only very nice you to enjoy your AG very nice nice okay they non parall and non intersecting so non parall non intersecting lines are called as SK lines L lines so non par non intersecting lines shortest distance is2 one you get a shortest distance the perpendicular distance the shortest distance I think I last year so two lines which are not parall non intersecting in the space in case of 2D if they are not par they will intersect but space intersect L if this is one line this is one line the shortest distance which is perpendicular distance between them is D is called as this is the value B1 cross B2 first new B1 find B2 find cross product find A2 mod of B1 B22 here not on the plane specially SP if you want to find the angle if you find the you know distance between them simple dou exactly ra let's do this question now find the shortest distance between the lines what isal that is I and JC okay CR 2 I minus J plus K what is A2 A2 I cap will be A2 2 I + J cap minus K cap what is your B2 B2 will be 3 i - 5 J + 2 K1 2 2- 1 A2 2 1 - 1 3 - 5 2 okay now form B1 B2 do A2 - A1 by modulus of B let's find 1 by one first A2 - A1 find what is a 2 - A1 Vector 2 - 1 that is 1 that is I cap 1 - 1 0 good this 0us one k so basically okay what is B1 cross B2 how do you find how find cross product very simple i j k i j k 2 - 1 1 3 - 5 2 find find very simple minus 2 so this into this minus this and I into this into this - 2us of - 5 that's + 5 that's 3 I cap minus J into this into this leave this for this column 2 into - 5 10 minus of - 3 + 3 okay minus plus K cap that is into column aor this this column 2 is 2 is 4 4 - 3 okay yeah K into leave this this column 2 5- 10-- + 3 something I'll get like uh uh 3 I cap minus J cap - 7 this is the your A1 CR A2 modulus at the same time find modulus also what is modulus of B1 CR B2 root of 3 square is 9 plus 1 square is 1+ 7 square is 49 49 so 49 50 root 59 simp dot A2 - A1 into model of B1 into B2 so what is B1 CR B2 3 i - J cap - 7 K cap dot one 1 i 0 JC minus K divid by modulus B root 59 so what is answer for this 3 into 13 7 into 1 7un 59 7 + 3 10 10 byun simple question change values now for 4 IUS 2 J + 2 K two twice of that minus one twice of that tce of this so I can this Vector for example1 B1 is Lambda * of B2 or you can say B2 is lamb you can do anything you want B2 is Lambda * of B1 can do that's all done this I love not talk too much and see you in the next lectures bye-bye take care