 let's take questions on this ncrt question so is it i plus 2j minus 3k or plus 3k yeah let me clarify that that's plus 3k let me write it again for you this entire stuff I am writing again down here r dot i plus 2j plus 3k minus 4 equal to 0 and this one is r dot 5i plus 3j minus 6k plus 8 equal to 0 lately I've seen cam scanner they start putting their logo unnecessarily earlier they were not like this now they are unnecessarily started branding it again don't give me lambda and all just give me the final answer because that lambda will vary depending upon which equation are you keeping first in fact there will be reciprocals of each other since you're asking Shreya yes your lambda is correct I know that's not a very nice number to see but yeah that's correct okay sir thank you but can you finally give me the answer yeah I'm doing that if you're done please feel free to type the answer and just privately message it to me that's correct Shreya that's correct nobody else okay so first you'll say that the equation of the desired family member can be written as r dot i plus 2j plus 3k minus 4 plus lambda times r dot 2i plus j minus k plus 5 okay now this is not a nice way to represent it let's represent this finally as some r dot n plus d equal to 0 form that would make more sense so for that you have to take our common so our dot in fact you have to apply opposite of the distributive property so our dot this plus this will give you I'm directly writing the result to lambda plus 1 i okay lambda plus 2j and minus lambda plus 3k okay and you'll have minus 4 plus 5 lambda equal to 0 is that fine everyone okay now this vector is perpendicular to this vector so this vector is perpendicular to 5i plus 3j minus 6k right this is another way to find out the value of lambda see it's not necessary that always a point will be mentioned right you can find the lambda through various conditions which has been provided in the question so here they gave perpendicularity to a certain plane in some question they will give you that this plane is at a certain distance from the origin right so there can be so many conditions cited which will enable you to find the value of lambda so we'll have to practice all types of those questions so if they are perpendicular can I say 5 2 lambda plus 1 3 lambda plus 2 and minus 6 minus lambda plus 3 should be 0 correct so 10 13 19 lambda and you'll have 5 plus 6 11 11 minus 18 is minus 7 so lambda is going to be 7 by 19 now please note if somebody had taken lambda with the first equation he would have got the answer as exactly reciprocal of this that is 19 by 7 so don't try to match the lambda in fact if the lambda of two people are same okay and if they're reciprocal then you both are on the right track or both are on the wrong track okay so either the lambda would be same or the lambda would be reciprocals okay now having got that all you need to do was to put this in your equation of the plane back that is at this step so let me put 7 7 by 19 there so it becomes our dot 2 lambda plus 1 give you 14 plus 1 33 by 19 I lambda plus 2 will give you 38 plus 7 45 by 19 J correct hope I have run the addition correct 38 plus 7 45 yes and then if you put over here that's 3 minus 57 minus 7 is 50 K by 19 and here I will have minus 41 by 19 if you want you can drop the 19 from everywhere and you can end up writing R dot 33 I plus 45 J plus 50 K minus 41 equal to 0 or you can write this in in Cartesian form as 33 X plus 45 Y plus 50 Z equal to 41 now may I know the reason why people could not solve this other than share what was stopping you from solving this I didn't simplify it like how we did like 2 lambda plus 1 and then do it oh so how would if you don't simplify that you will not be able to know the collective normal vector for this entire family family member right so you have to consolidate like this this is an important step so yeah I have to consolidate your family of plane like this and then use it for whatever purpose is I've been used for so let's you have been given a point then you can put a point in place of our if you have been given this is perpendicular to something then you can do dot product of this with the given vector as you know just like I did it in the present example getting my point let's say I had the question mean the distance of this plane from origin is one then how do I do this let's say the second condition is this condition I drop which is perpendicular to the plane I cut this and make it it is at a unit distance from the origin then how do I do this how do I get lambda in that scenario make it as direction cosines over there and then the constant term equated to one right so or you can do one thing we already know that in case of any equation of a plane mod d by under root of a square plus b square plus c square represents the distance from the origin right so here your d is 5 lambda minus 4 isn't it so 5 lambda minus 4 is your d and these are your a b c's so all you need to do here was do 5 lambda minus 4 by under root of 2 lambda plus 1 the whole square lambda plus 2 the whole square minus lambda plus 3 the whole square equal to 1 okay take it to the right side square both the sides find the lambda values to value of lambda will come out because it will signify there are two possible planes at a unit distance from the origin one is on this side of the origin and there will be on this side of the all correct so this is a different questions that can be framed are you getting my point I'm not solving it I'm just giving you an idea how to work with it is that clear everyone family of planes is one of the most important concepts please type clear else I will not move on now I'll go back to the same old question guys so I'll just pull out that question and we'll solve the same question I think this was the question yeah question number 3.54 now you would see how easily I can solve this question if I know my concept of family of planes see all of you please see it I have to find the equation of I have to find the point of intersection of the line this with the intersection of these two right now what I'll say is that there's a line there's a line which is made up of which is made up of the two planes which are themselves the family members so what I'll do is I'll say let my line let my line be represented in asymmetric form like this so what did I do I made a family I made a family of planes with these two planes correct okay and again I'll make a family from these two so x plus y minus 2 plus some mu times x plus z minus 2 equal to 0 so the line which I'm looking for is basically intersecting this line are you getting my point so what I did I'm again repeating it these two planes I made a family of planes with this so this is my first family this is my second family okay so this plane is a family member of the first two and the second plane is a family member of the later two equations correct now what I'm naming is that my line is in asymmetric form can be written like this do you agree with me or not agreed now in order to find lambda and mu I can use this condition that this line is passing through 001 so when you put 001 over here I think this will become completely zero and if this becomes completely zero there is no point evaluating this this will also become whatever is it so lambda has to be zero okay put 001 in the second equation right because this line is passing through 001 right so both both these planes must be satisfied by this line isn't it because this line lies on both these planes and both these planes themselves are family members okay so here if you put you get 00 minus 2 plus mu times 0 1 minus 2 equal to 0 so that becomes mu as 2 I'm sorry minus 2 correct yes or no now when you put this value back when you put this value back here and back here you actually get the equation of the line in asymmetric form that is x plus 2 y plus z minus 1 equal to 0 and here when you put you get x plus y minus 2 minus 2 x plus z minus 2 equal to 0 that's nothing but minus x plus y minus 2 z plus 2 equal to 0 yes or no yes or no now in these two equations you want to know this basically this line is for passing through these two planes that is what I'm trying to say in other words my required line has an asymmetric form of this x plus 2 y plus z minus 1 equal to 0 and minus x plus y minus 2 z plus 2 equal to 0 now you want to know from where does this line where does this line meet the x y plane so put z as 0 so when you put z as 0 you get x plus 2 y is equal to 1 and minus x plus y is equal to minus 2 okay so add it so 3 y is minus 1 so y is minus 1 by 3 put it over here so x is equal to 1 minus 2 y that is 1 plus 2 by 3 that is 5 by 3 so the point of intersection is 5 by 3 minus 1 by 3 0 see we got the answer in a much faster way I don't have to write too many things is it much easier way to do it is that clear everyone yes sir yes sir please let me know if you have any questions with respect to this if you have no question we'll move on to the next concept fine so now we'll take some generic question regarding angles so let me take this question we'll take few questions and then I'll start with the concept of angle between a line in a plane yeah here goes the question find the equation of a plane containing this line and this point yeah and I don't have much concepts left early I talk about the equation between angle between a line and a plane then I will talk about the distance of a point from a plane distance between two planes and one more concept is equation of bisector of planes so I think I should be able to finish it off by 12 30 or so online class are generally faster hope you can read this point this is 0 7 minus 7 find the equation of a plane containing this line that means this line is lying on the plane something like this and is a point 0 7 minus 7 please send the answer to me privately on the chat or you can watch that bit to me so this question belongs to the interaction of a plane and a line together so we have done line and plane now to a considerable extent so now we are trying to solve questions where there's a combination of these concepts coming up any idea how to do this so one minute and a life is please do attend problem solving classes also because as I told you 60% 70% of your learning will come through problem solving I don't think like the theory is over so you need not attend the classes yeah and in physics I have been getting the complaint that people are not attending because the theory has been completed it's ultimately the problem that you have to solve so that is more important class for you your theory will become more refined once you start solving problems based on those theories okay see here once you have been given the equation of a line basically two information is given to us one is you know a point on this line as you can see minus one three minus two is a point on this line correct and you have been given a vector in the direction of the line and that vector has direction ratios of minus three two and one correct now how do I get the equation of this plane pi from here very simple if I know this direction can I also get this direction by connecting these two points okay let me call this as B1 vector and let me call this as B2 vector so who will tell me the direction ratios of B2 vector is the difference of the coordinates of these points so let's subtract so it's one four and minus five any questions with respect to getting direction ratios of B2 vector this is the direction ratios of B1 vector any doubt so far now once you know two vectors on the plane you can get the direction of the normal to the plane by taking the cross product of these two given vectors again the basic concept should not be forgotten to get the equation of a plane I need the normal and I need a point now there's no depth of the points here two points have been given to us directly or indirectly just I needed was a normal vector so for normal vector I would need the cross product of the two vectors which are already lying in the plane of the in the plane of that particular plane so here I will get I J K B1 components are one four negative five and this is minus three two one okay so let me just expand this vector so it becomes four plus 10 14 minus J again minus 14 okay and K oh my god K is also plus 14 wow so 14 I 14 J 14 game can I drop the factor of 14 I can even if you don't want to drop it's fine but I don't want to deal with these heavy numbers 14 14 14 so what I'll do I just need a vector in the direction of you know B1 cross B2 it can be any vector I can keep it as simple as this so I'll call this as my end okay now I will use the formula r minus a or you can use your directly Cartesian form since things are given in Cartesian form so a x minus x1 by minus y1 CZ minus that one is equal to zero cool so your a is one see this is your a component B and C okay so one take any one of the points you can take the point on the line also you can take the point given to you also so let me take X minus zero B is also one Y minus seven C is also one Z plus seven okay expanded it becomes X plus Y plus Z I think there will be no constant left so this is going to be your answer okay so you should know how to deal with these situations