 okay let's start with this question can you all read this yes sir yeah two forces AB and AD are acting at the vertices of a quadrilateral AB CD and two forces CB and CD at C prove that the resultant is for EF where E and F are the midpoints of AC and BD respectively let's draw it first so let me call this as A, B, C and D so AB and AD act on A similarly CB and CD act on C okay now they're saying the resultant of AB AD CB CD that means AB plus AD plus CB plus CD we have to prove that this is equal to 4 EF is the midpoint of AC AC midpoint is here okay F is here let's say F is here so it's four times EF can you prove this any idea okay the best thing here would be take one of the points as origin okay let me call this as O okay so let me change the name over here also from AB let me make it as OB and from AD let me make it as OD okay now this choice is completely yours right choosing a portion is completely upon your whims and fancies fine now why have I done that so that I can call this as position vector B and I can call this as the position vector D and I can call this as your position vector C okay remember when you say position vector of B is B that means you are indirectly referring to this vector this is your B vector this is your D vector and vector O to C that would be a C vector okay but that is not desired actually so I'm not bringing into a picture now let's say if I talk about the left hand side left hand side of this expression is OB which is B vector OD which is D vector and CB what is CB CB as I told you destination minus source destination minus source will be B minus D correct CD will be again D minus C that means your left hand side of the expression actually ends up giving you I think some these will get cancelled off why do you write B minus D it is B minus C right this is C yeah so your left hand side will be 2B plus 2D or you can say minus 2C plus 2D okay that is nothing but twice of B minus C plus D this is your left hand side of the expression now let us check out what is EF EF means position vector of E minus position vector of F correct now how do I figure out the position vector of E now here comes your better understanding of position vectors position vectors as I told you is something which relates vectors to coordinate geometry okay so before I move on to this question this question was actually a medium for me to tell you how we can use position vectors in the same way as we use coordinates so let's say if I have a point A and B whose position vectors are vector A and vector B respectively and let's say there's a point C somewhere over here whose position vector is C says that C divides the join of A and B in the ratio of M is to N right you would be surprised to know that C vector could actually be written as MB plus NA by M plus N very much similar to the does this remind you of section formula does this remind you of section formula there's a halo on top of this smiley okay section formula correct how does it come out first it's a very simple explanation so let's say this is my reference vector oh if you connect this connect this connect this okay oh let me just pull out this N from now see if you're saying position vector of A is A that means this vector is a this vector is B okay now OC vectors what is your vector C we have to figure that out how do I get the C vector very simple we all know that this entire vector will be let's say I give this as the direction of this vector so the direction is this direction AB vector completely is nothing but B minus a okay so what is AC vector AB vector is B minus a so AC vector is nothing but a vector in the same direction as AB but its length is M by M plus N times B minus a right remember the entire length if you take it as M plus N AC is just the M part of it or it's M by M plus N it times B minus a so a to C is known to you now which is M by M plus N times B minus a correct now can we now apply the addition of vectors can I say C vector is the sum of OA and AC so C vector is OA which is your vector a AC vector is the same vector which we had figured out a little while ago that is M by M plus N times B minus a take the LCM so it is M plus N times a plus M times B minus a which makes it as MA plus NA plus MB minus MA okay so these two gets cancelled by M plus N so the answer is MB plus NA by M plus N right very similar to the section formula that we used to have in our coordinate geometry getting this point okay so in a similar way if I want to get a midpoint of two vectors all you need to do is take the end position vectors add them and divide by two so position vector of E would be 0 plus C by 2 0 plus C by 2 which is nothing but C by 2 so this is a position vector of E what's the position vector of F position vector of F will be B plus D by 2 okay is that clear so position vector of E will be C by 2 while position vector of F will be B plus D by 2 okay so EF vector would be destination minus source that is B plus D by 2 minus C by 2 so take a two common so it is B minus C plus D now what I'm claiming is that for EF should actually give me the same as my left hand side by the way for EF will be two times B minus C plus D and we can clearly see that these two expressions are same so left hand side and right hand side are same and hence proved is that fine so guys let me tell you here position vector was in fact position vector along with the additional vectors was a very important tool in our hands to solve these kind of questions where normally a coordinate geometry problem has been addressed is that fine everyone clear with this yes sir yes sir okay time to take a slightly complicated problem read the question properly if O is the ortho center and O dash is the sorry O is the circum center O dash is the ortho center of triangle ABC prove that OA plus OB plus OC is O dash any idea let's make a diagram first this is your O dash it is not option you have to solve these three questions I think I tried to like figure out these are not options then prove that they prove that question I can understand exam time now okay first let me just stop perpendicular like this OD okay I'll help you with the first part how many of you remember from your last year that in any triangle the distance of the ortho center so let me just stop a separate case over a let's say this is your ortho center O dash okay how many of you know that O dash A is 2R cos A where R is the circum circle radius where R is the circum circle radius how many of you know this if you're not if you're not aware of this not a issue I can just explain you how this figure actually comes so let's say this is your 90 degree over here right and this is also 90 degree okay so we all know that this angle okay since this is angle B this angle is going to be 90 minus B correct okay and since this angle is a this angle will be 90 minus a everybody agrees to this let's say OA dash let O dash ABX so let's say this is X okay now since this is 90 minus a can I say this length would be X sine B the opposite side to B we call it as small b now since this is 90 minus a and this hypotenuse is small b can I say the same length OA dash sorry OA dash I'm saying AP let me call it as okay AP you have already shown that it is X sine B can I say the same AP length can also be written as B B cos a now from here I can say that X sine B is B cos a correct from our sign rule we know that B by sine B by the way all of you should remember sine rule as well a by sine is B by sine B is C by sine C and this ratio is actually 2R very few people know that this is 2R okay so from these two I can say that small b is 2R sine B so when I place that small b with 2R sine B sine B sine B gets cancelled so X is equal to 2R cos a this result should be remembered because it is heavily used in properties of triangles okay now why am I referring to this result all of a sudden in this question okay do you realize that OD and O let's say I call this as okay let me call it as AO dash can I say OD vector will be parallel to AO dash vector because both are perpendicular to the same line BC correct now we know that this is the or circum center so this is your rate circum circle radius okay since this is your angle a can I say this will also be your angle a now that means OD magnitude is nothing but R cos a correct now OD magnitude is R cos a AO dash magnitude is 2R cos a and both are parallel also so can I conclude from here that from these all I can can I conclude can I pass a judgment that AO vector itself will be twice of OD vector yes or no what do you say about this again if you want to look at the figure you can have a look at it this vector and this vector are parallel and this vector is double the length of this vector and hence I can and I can make this conclusion that AO vector is double of OD vector okay is that fine no problem with respect to this fine if this is okay now let us move on to the requirement the requirement is OA plus OB plus OC have to prove it as O dash okay now see it let me start with LHS LHS is OA okay OA plus OB plus OC now OB plus OC can I say it is to OD can I say this is to OD all of you agree with me on this or not now you must be wondering how it is to OD very simple again you can see that D is the midpoint of B and C isn't it D is the midpoint of B and C so if I consider O to be your origin this is position vector B and let's say this is position vector C then D would have a position vector of B plus C by 2 in other words OD vector would be OB vector plus OC vector by 2 so OB plus OC would have been to 2 OD vector okay so from here I can say OC plus OB will be 2 OD clear yes or no okay yes and 2 OD is what 2 OD is AO dash okay let me write it in a slightly different way I'll write it AO dash plus AO dash is nothing but O dash it's very obvious because if you're going from O to A and then from A to O it is as good as going from O to O dash done first part of the proof is that clear such a huge such a huge problem but the concepts involved where your concept that you had already learned in your past okay try the next one prove that O dash A plus O dash B plus O dash C is equal to 2 O dash O again shift your origin to O dash okay now let me use black pen over here okay now O dash B plus O dash C vector O dash B plus O dash C vector do I agree that it will be 2 O dash D vector yes or no yeah correct yes now O dash A O dash A you already prove that so these two are taken care of O dash A you already proved a little while ago that O dash A is 2 DO dash sorry 2 DO O dash A is 2 DO correct that means if you add them up that is O dash A plus O dash B plus O dash C is equal to 2 O dash D plus in fact 2 you can take common DO okay so O dash D O dash D plus DO will you what O dash O so it'll be 2 O dash O done this also done last one at least you try A O dash O dash B O dash C you can use the result of the second one in the third one any idea see very simple here also now this term over here let me write it separately AO dash you want to prove AO dash plus O dash B plus O dash C is equal to 2 AO AO dash you can write it as 2 AO dash minus AO dash correct O dash B O dash C okay so I'm starting from my LHS okay now if you reverse the order of AO dash you can write this as plus O dash A this result is already known to us what is that result that result is 2 O dash O so this will become 2 AO dash plus 2 O dash O take two common AO dash O dash O is nothing but AO so 2 AO done this is your RHS now this question is saying that is equal to AP where P AP is the diameter through a of the circum circle which is obvious because AO is going to be acting like the radius okay so twice of the radius will be nothing but the diameter okay that's what this question is trying to prove okay so third part also done let me take you down to the part where I proved it any question here please ask me any doubt any question any concern no sir okay