1.31 | Review Of Vector Algebra | Reciprocal System Of Vectors

Transcript

in today's lecture we will discuss reciprocal system of vectors to begin with let us consider three known coplanar vectors A B and C C the these are known Co planar vectors now if they are non coplanar this would mean that the scalar product the scalar triple product of the three vectors is going to be not equals to zero it cannot be equals to zero if the three vectors are non coplanar now we are going to define a set of three new vectors we'll call it a vector prime vector B Prime and Vector C Prime then these three vectors are reciprocal system of vectors to a b and c if a vector a prime as given by the cross product between Vector B and C divided by the scalar triple product of the vector a b n c for Vector B Prime we defined it as the cross product between Vector c and a divided by the scalar triple product between the vector a b n c and for the third vector Vector C Prime that will be equals to a vector cross B vector divided by a B and C now there are a couple of things that we can pick out from this particular idea the first one would be if we take the dot product between the vector a with its reciprocal that is a vector Prime then this will be equals to 1. it also follows that this will be true for Vector B dotted with Vector B Prime and vectors A dotted with Vector C Prime let's try to prove this particular idea I will begin with Vector a dotted to Vector a prime that does means we have a vector a dotted bit so we'll use the definition of R reciprocal Vector a prime so this will be Vector B crossed with Vector C divided by e b and c then the numerator we have Vector a dotted to Vector B Cross C which is divided by Vector a b c now here by definition a vector dotted to B Cross C this is the scalar triple product between the vector a b and c so we have the scalar triple product of vector a b and c divided by the scalar triple product of vector a b n c now there's been this since both the numerator and denominator comprises of a scalar number so that means they cancel out to give us 1. where you can prove this for Vector B dotted with B Prime as well and for Vector C dotted with C Prime Point number two so if we have the vector a dotted with a vector from another system that means we have dot product between Vector A and B Prime or let's say we take the vector a and dotted with Vector C prime or similarly Vector B dotted mid Vector a prime or vector B dotted to Vector C Prime or vector C dotted to Vector a prime and Vector C DOTA to Vector B Prime in all the three all these cases the dot product will be zero so let's try out any one of them let's try out the last one that means Vector C dotted with Vector B Prime so we have C Vector dotted to the PE Vector Prime so that is given by Vector C cross Vector a divided by the scalar triple product between the vector a b in C that means we have Vector C dotted to Vector C cross a divided by the scalar triple product between the vector a b and c this will be equals to now this is by definition is the scalar triple product of the vector C and a divided by the scalar triple product of the vector a b n c now if we consider the new the numerator since we have two vectors that exactly equals which is the vector C so that means the scalar triple product of this Vector is going to be equals to 0 so we have zero value so this is what we wanted to show that the scalar product of any other pair vectors one from each system is going to be zero let's go to point number three so we have already discussed that the vector a prime B Prime and C Prime these are the these are three non coplanar vectors and are reciprocal to the vector a b and c then in that case we can say that the scalar triple product of vector a prime B Prime and C Prime this is also going to be the reciprocal of the vectors of the scalar triple product formed by these three non coplinear vectors in another sense we can say this will be equals to 1 divided by the scalar triple product intervector is a b and c let's prove that this is indeed true so we'll start from the left hand side so we have the scalar triple product between the vectors A Prime B Prime and C Prime when we use the definition to expand it we have Vector a dotted with Vector B cross Vector C and then we'll have to use the definition of vector B Prime and Vector C Prime we will use for a prime a little bit later so we have a prime dot is with so the definition of b Prime will be C cross a divided by the scalar triple product between the vector a b c this is crossed with C Prime so C Prime will be in Vector a crossed with Vector B divided by Vector a b c this is the scalar triple product so we have so what we now have is this K and this is the vector product of four vectors C cross a cross bit A cross B we'll have to use the definition so here there's a particular way to expand this Vector cross product of four vectors so we'll begin with the third Vector so we have the vector a so this will be multiplied to the scalar triple product of the remaining vectors and to note down the remaining vectors the scalar triple product will start with the next Vector so this is p vector then we'll come back to the origin beginning and again read from left to right so we are left with Vector C followed by Vector a minus then we'll consider the fourth Vector reading from left to right so we have Vector B and this will be multiplied to the scalar triple product between the remaining Vector so after B we'll come back to the beginning so we have Vector C followed by Vector a followed by Vector a whole thing divided by Vector a the scalar triple beta of the vector a b n c Square by definition this vectors the scalar triple product between the vector C A and A this will go to zero because we have two vectors that are exactly equal so we are only left with one term so that means we have a Vector Prime this will be dotted to this particular Vector there's a reason for that because this is just a scalar number we cannot multiply take the dot protographer scalar number with a vector so the only way we can do that is take the dot product of two vectors so here we are taking the dot product between a prime and a so this is Multiplied to Vector b c a so b c a so this is cyclical in nature so we can also write it as a b and c and it would mean the same thing divided by e B c squared so this is a scalar number this cancels out with one of the scalar value in the denominator and a prime dot a we have already talked about this this value is 1 so we have left it 1 divided by the scalar triple product of the vector a b and c so we started out with the scalar triple product of the vector a prime B Prime and C Prime so this is equals to the reciprocal of the scalar triple product the vector a b in C now we have already mentioned that the vector a B and C are non-clope non coplanar meaning that the scalar triple product of the vector a b and c that's a non-zero quantity if this is a non-zero quantity then it follows that the scalar triple product of the reciprocal set are the reciprocal vectors A Prime B Prime and C Prime that is also not equals to zero that means the three reciprocal vectors are also non-co planar one is more evident that if a Prime B Prime C Prime these are the reciprocal Vector set to our Vector a B and C then a b c will be the reciprocal Vector set to a prime P Prime n c Prime as well also if we look at this particular concept we can say that a the scalar triple product within the vector a b and c multiplied to the scalar triple product between the vector a prime B Prime and C Prime that's equals to one so this would mean that both of them has to be a positive or both of them has to be negative at the same time in other words what we are trying to say here that if a b c is a right-handed system then a prime B Prime says C Prime must also be a right-handed system and if one is a left-handed the other automatically becomes a left-handed system and finally from all this result we can say that our autonomial Vector Triads which is defined by I cap J cap and keycap the corresponding reciprocals will be I cap Prime J cap Prime and K cap Prime that means we can say that I cap Prime this will be defined as J cap Cross Key cap divided by the scalar triple product between the vectors I cap J cap and K cap well the scalar triple product of I cap JK pen keycap that's always equals to one we can check that out so by definition I cap J cap k-cap it's the scalar triple product will be defined as I cap dotted to the cross product between Vector J and K Vector J Dot and across with K this gives us the vector I cap and the dot product between the unit Vector I with itself is equals to 1. so we have J cap Cross Key cap divided by 1 and J cap Cross Key cap is I cap so that means the reciprocal Vector set so we can see that icap Prime will be equals to I cap similarly J K Prime will be equals to G cap and k k Prime will be equals to K cap fine so now that we have understood the idea of the reciprocal set of vectors let's work on few numerical problems problem number one we have three vectors let's name them as a b and c so we have Vector a which is given by 2 I cap plus 3 J cap minus K cap where Vector B given by I cap minus J cap minus 2 K cap and we have Vector C given by negative I cap plus 2 G cap plus 2 K cap we have to find a set of vectors that is reciprocal to the given set sometimes we need to find out Vector a prime B Prime n c Prime so we'll use the relation to obtain the value of a prime B Prime Min C Prime so let us Begin by first obtaining the value of the scalar triple product between the vector a B and C the defined of the scalar triple product all we have to do is obtain the three by three determinant where in the first row we'll have components of vector a that will be 2 3 negative one in the second row we have components of B so that will be one negative one negative two and in the third row we have components of vector C that it will be negative one two and two expanding along row number one so we have two times so negative one times two so that will be negative two minus two times negative two so that's negative four but we'll have positive four minus three times that we have one times two is a little bit two minus negative two times negative one so that is 2. minus one multiplied to one times two so that's two minus one so this will be equals to two times two minus three times zero minus one times one so we have two times two so that's four four minus one which is equals to three in the next step we will find out the cross product between Vector B and C cross product between Vector C and A and cross product between Vector a n b so let's begin so we have actually a b and c let's determine the cross product between Vector p and C so this will be given by a two by two determinant in the first row we have the unit vectors I cap J cap K cap and the second row we have the components of vector B that will be 1 negative 1 negative two and in the third row we have components of vector C that it will be negative 1 2 2.

so let's expand it along the first row we have icap multiplied to negative 1 times 2 so that's negative two then we have positive 4 minus J cap times we have 2 minus 2 plus K cap so K cap will be 2 minus one so solving this we get 2 I cap so this will be 0 minus 0 g cap plus K cap so this is the value of vector B Cross C we'll then find out the value with or the cross product written Vector C and A so we have I cap J cap K cap into first row in the second row we have components of vector C so that's negative one two two in the third row we have components Vector a so that's 2 3 negative one so this will be when we expand it along the first row we should get negative eight I cap plus 3 J cap minus seven K cap and similarly when we find out the cross product between Vector A and B that was going to give us negative 7 I cap plus 3 J cap minus 5 K cap so we have these three cross products uh now what we'll do we will try to evaluate the value of each of the reciprocal vectors e Prime B Prime and C Prime so let's do that so the reciprocal of vector a which is a prime so that will be equals to will be Cross C that's 2i cap Plus key cap that this will be divided by 3. then the reciprocal Vector for B which is B Prime so this will be equals to with C cross e so that's negative 8 I cap plus 3 J cap minus seven K cap divided by three and finally the reciprocal for Vector C that will be C Prime will be given by negative 7 I cap plus 3 J cap minus five k k all divided by three so these are the three reciprocal vectors of a b and c now let's go to the next problem so we have problem number two which is quite similar to the previous problem so let us begin we will consider the vector a to be given by negative I cap plus J cap plus K cap the vector B to be represented with I cap minus J cap plus K cap Vector C represented by I cap plus J cap plus K cap we'll have to find out the reciprocals of each of the given Vector assessments we have to find out a prime B Prime and C Prime now let's begin by determining the scalar triple product of the vector a b and c that means we have for in the first row this is a three by three determinant in the first row we have components of vector a so that would be negative one one second row we have components of vector B so that will be one negative one positive one in the third row we have components of vector C and when we expand it along the first row this is going to give us a value now I'll note down the value you can check them out so we will also need to find out the value of B Cross C cross A and A cross B so this will be equals to so B Cross C so this will be negative two I cap Plus 2 keycap the value of C cross a that will be negative 2 J cap plus 2 K cap and the value of A cross B so that will be positive 2 I cap plus 2 J cap well that meant a value of a reciprocal Vector a a prime will be equals to so by definition we have Vector B cross with Vector C divided by the scalar triple product of vector a b m c that means this will be equals to Vector C negative 2 I cap plus 2 K cap divided by 4 or simply negative I cap plus K cap divided by 2. let's find out the reciprocal Vector B the B Vector Prime so that will be given by C cross a divided by the scalar triple product between a b and c so C cross a is negative 2 J cap plus 2 K cap so we will have negative J cap plus K cap divided by two and C Vector Prime this will be Vector a crossed with Vector B divided by the scalar triple product of vector a b and c that means we will get I cap plus J cap divided by 2. so these three are the reciprocal vectors of a b n c respectively let's go to the next problem number three we have three vectors A Prime B Prime C Prime these are the reciprocal system of vectors for three non coplanar vectors APN see and what we have to do we have to prove that any Vector in the position Vector R can be expressed in this particular format so let's begin what we'll do here we will write our Vector r as a linear combination of these three vectors A B and C such that we have x times a vector plus y times V vector and c times C Vector where X Y and Z are some scalars let's call it equation number one now what we'll do we'll take equation number one and take the dot product with the vector B Cross C on each side so we have R Vector dotted with B Cross C similarly we will dot each of the terms on the right hand side with B cross attachment we have X for the first time x multiplied to a vector dotted to P Cross C Plus y times B Vector Dota 2 p Cross C Plus Z multiplied to C Vector dotted with B Cross C so we have x times so this is the definition of the scalar triple product of a vector a b n c plus y times so this will be the scalar triple product of vector B Vector p and Vector C plus Z times will use the definition so this will become Vector C Vector B and Vector C now the scalar triple part of vector B B and C this will go to 0 and also scale of triple product of vector c b and c will also go to zero that's because we have two vectors that are exactly the same that means we are only left with the first term so that is x times Vector a the scalar triple product of vector a b and c so this is Vector R dotted with B Cross C since a the scalar triple product is just the scalar number we can divide both sides with the scalar triple product of a b and c and solve for x then this was going to give us X is equals to R Vector dotted with B Cross C holding divided by a b c the scalar triple product now here if we look carefully this is the foreign reciprocal vector a prime so that means what we have done we have Express DX DX as X was actually the X component of vector r along a direction or the component of vector R along a vector Direction which is X so this is given by Vector R dotted with the reciprocal Vector a let's call that equation number two if we use the same method that it means next time what we'll do we'll take Vector R in equation number one and dotted with C cross a reminder C cross a then this process is going to give us the value for Y which will be Vector R dotted with p and B Prime D reciprocal of vector B so this will be equation number three and similarly we'll take equation number one take the dot productive Vector r with a cross B and this is going to give us the value for Z which will be Vector R dotted with d reciprocal for Vector C so this will be 4. so what we'll do we will replace the value of x y and z obtained in 2 3 and 4 and equation number one so this is our equation number one so that means we have R Vector equals to x times a vector so X here will be R Vector dotted with the reciprocal of vector a so this is X multiplied to a vector plus we have y the Y will be R Vector Dota 2 B Prime so this is Multiplied to B Vector plus r Vector dotted with C Prime so this is Multiplied to C Vector so we have successfully represented the position Vector R in terms of this given expression root let's move on to the next problem then problem number four we have three vectors A B and C which constitutes a set of three non coplanar vectors such that the reciprocal vectors A Prime B Prime and C Prime are given as followed we have already talked about this this is the definition here a b a prime B Prime C Prime divides D reciprocal system of vectors to a b and c so what we have to show that that the converse is also true that means if we have the vector a this is given by the cross product between Vector B Prime and C Prime divided by the scalar triple product of the reciprocal vectors of a b and c so let's do that so we'll begin with B Prime crossed with Vector C Prime let's begin with this so here we will use the definition of b Prime and C Prime that means we have Vector C cross with Vector a divided by the scalar triple product between the vector a b and c and we're going to cross it with the vector C Prime and C Prime is defined as a vector crossed with P Vector divided by the scalar triple product of vector a b and c fine so that means in the numerator we have the vector product of four vectors which is C cross a crossed with a cross B and in the denominator we have just to scale the numbers that is the scalar triple product of a b and c squared now the task will be to find out the value of the vector product of four vectors so here we are going to use the expansion so first what we'll do we will begin with the third Vector as we read from left to right so that means we have Vector a this will be multiplied to the scalar triple product of the remaining vectors so right next to Vector a we have Vector p then we come back to the beginning and again read from left to right so we are left to Vector C followed by Vector a minus now we will consider the fourth Vector as we read from left to right so this will be Vector B will be multiplied to the scalar triple product of the remaining Vector so since we are at V the vector next to P for that we'll have to come back to the beginning so we are left with Vector C followed by a and Then followed by a divided by to scalar triple product a vector a b and c squared now the scalar triple product of vector c e and a this will go to zero because two of the entries the two vectors A are exactly the same because that means in the numerator we adapted Vector a times BCA but PCA is equivalent to a b c that is because the scalar triple product is a cyclical in nature divide that means what I mean to say here if we consider Vector a vector B Vector C equals to b b goes to C and C comes back to a so we have initially the vector B so we start at B it goes to C and comes back to e this is the same machine we start out from Vector a go to B and then reach out to C so that means why we can write b c a the scalar triple product is equals to A B C and also in the denominator we have a b c Square the scalar triple product of vector a b and c and we can cancel out one term from the numerator and the denominator so this leaves us with Vector a divided by a b and c so this is the value for the cross product between Vector B Prime and C Prime so let's cross multiply so cross multiplying will give us Vector a this will be equals to B Prime crossed with C Prime so this is Multiplied to Vector a Vector B and Vector C and we have already proved that the scalar triple product of a b and c this is the reciprocal of this scalar triple product of a prime B Prime n c Prime we've already defined as we are going to replace this value with B Cross C divided by a prime P Prime C Prime and D gets killer triple product so this is the expression for our Vector a similarly using the same logic we can also show that for Vector B this will be given by C Prime cross a prime divided by the scalar triple product of the vector a prime B Prime n c Prime and Perfecto C we will have a Vector Prime cross B Vector Prime divided by a prime B Prime and C Prime so we have successfully proved the following problem let's move on to the final problem so here we have to do the three vectors A B and C with the reciprocal vectors given by a prime B Prime Min C Prime and we have to prove the following three so let's begin with the first one so we'll start from the left hand side we have Vector a crossed with it reciprocal Vector a prime plus Vector B crossed with its reciprocal vector B prime plus Vector C cross versus reciprocal Vector C Prime that will be equals to a vector we will be Crossing with a prime so by definition a prime is given by B Cross C divided by the scalar triple product of a b n C Plus Vector B crossed with Vector B Prime so B Prime will be given by Vector C crossed with Vector a divided by Vector a Vector B Vector C plus C Vector this will be crossed with C Prime so C Prime is defined as Vector a crossed with Vector P divided by Vector a vector B and Vector C now let's combine them so in the this will give us so we have a big denominator so we have Vector a the scalar triple product between Vector a b and c and in the numerator we have a vector crossed with B Cross C plus b Vector crossed with C cross a plus C vector crossed with a cross B our next task will be to expand each of the vector triple products so let us begin all we have to do is remember the niponic so when we have a vector a and this is crossed with B Cross C then their expansion is given by the mnemonic back cap well B Vector is simply multiplied to the dot product between Vector A and C and we have Vector C multiplied to the dot product between Vector A and B so we'll use this to expand each of the three Vector triple production for the first one we have Vector P this will be multiplied to a DOT C minus C multiplied to a dot b further second term in the numerator the second Vector triple product we have C Vector multiply to B dot e minus a vector multiplied to P dot DC all you have to do is remember the position for back cap then this becomes fairly easy plus we have C Vector cross A cross B so we have a vector multiplied to C dotted with B minus B Vector multiplied to C dotted with Vector a all divided by the scalar triple product of a b n c now if we consider the first term we have B Vector multiplied to a DOT C and we also have uh look at the third term we have negative B multiplied to C dot a so this will cancel out the second term negative C Vector multiplied to a dot b so second terminal third term cancels out and we have the fourth term and the fifth terms cancel though that means in the numerator we only have zero so that means this entire fraction goes to zero so we have successfully proved that if we take the vector a cross it with it reciprocal plus take the vector B cross head position reciprocal plus take the vector C cross it versus reciprocal then the sum of them will be equals to zero let's go to the second part let's begin with the second part so we'll start from the left hand side so in the left hand side we have Vector a prime crossed with Vector B Prime this is added to Vector P Prime crossed with Vector C Prime plus Vector C Prime crossed with Vector e Prime so Vector a prime this is given by Vector B crossed with Vector C divided by a b c to scalar triple product this is crossed with Vector B Prime so B Prime is given by Vector C cross Vector a divided by a b c plus Vector B Prime so RB Prime is C cross a divided by a b c this is crossed with Vector C Prime so C Prime is Vector a cross Vector B divided by a b c plus C Prime so C Prime will be Vector a crossed with Vector B divided by Vector a vector B Vector C this is cross to Vector a prime so a prime is Vector B cross to Vector C divided by e b n c that will be equals to so that means in the numerator we have B crossed with C this is crossed with C cross a divided by the scalar triple product of a b n c Square so let's write the remaining two terms as well so we have the vector product of of four vectors that means we have a vector quadruple product in each of the three numerator so that means we'll have to expand it so for the first one we have we'll start with Vector C so C Vector will be multiplied to the scalar triple product of the remaining Vector so that will be Vector a vector B Vector C minus the way start with the fourth vector rho vector a this is Multiplied to the scalar triple product of the remaining Vector so that will be p c c divided by e b c squared similarly let's compute the vector quadruple product for the numerator we will have a vector multiplied to the scalar triple product of vector p c a minus B Vector multiplied to vector c a a whole divided by a b c squared plus the third term so in the third term we have Vector B multiplied to the scalar triple product of c e b minus C Vector multiplied to a b B holding divided by e b c squared now if we use the definition of a scalar triple product so this term will go to zero because we have C vector and C Vector repeating here this will also go to zero a vector a vector is repeating and we have B Vector B Vector repeating so this scale of triple product goes to zero now what we'll do we'll combine all the three terms to get from the first time we have C Vector multiplied to a b c Plus from the second term we have a vector multiplied to BCA BCA is actually the same as a b c plus the third Vector which is B Vector multiplied to cab so cab also the same as a b c whole thing divided by the common denominator from each of the three terms which is a the scalar triple product of a b and c squared now we can take the scalar triple product a b and c command from both the numerator and the denominator and rearrange the terms till the first word Vector a plus Vector B plus Vector C all divided by vector the scalar triple product of vector a b n c and this is the right hand side this is what we were asked to prove let's do the third and the final part the third part is very simple we have already talked about that of when you take a vector a and Dot it with its reciprocal then this value is always equals to one same goes for the vector B dotted with it reciprocal that's also equals to 1 and Vector C dotted to its reciprocal gives us one when we add the three terms then we have Vector a dotted with a prime plus Vector B dotted with Vector B prime plus Vector C8 Vector C Prime so this will be equals to three and so this was all about reciprocal vectors [Music]