 Hello and welcome to the session, let us work out the following problem. It says for any two vectors a and b show that 1 plus square of magnitude of vector a plus 1 plus square of magnitude of vector b is equal to 1 minus vector a dot vector b whole square plus square of magnitude of vector a plus b plus vector a cross vector b. So, let us now move on to the solution. We will start with RHS and we will first simplify RHS and we will show that it is equal to the LHS. RHS is 1 minus vector a dot vector b whole square plus square of magnitude of vector a plus vector b plus vector a cross vector b. Now this becomes 1 minus vector a dot vector b square it should be plus minus 2 into 1 into vector a dot vector b applying the formula of a minus b whole square plus. Now again here we will apply the formula of a plus b whole square or x plus y whole square where x is vector a plus vector b and y is vector a cross vector b. So, we have vector a plus vector b whole square plus vector a cross vector b whole square plus 2 into vector a plus vector b into vector a cross vector b. Now again this is equal to 1 plus vector a dot vector b square minus 2 into vector a dot vector b. Now here we will apply the formula of x plus y whole square. So, this becomes vector a square plus vector b square plus 2 into vector a into vector b into a cross b whole square. Here we have plus vector a cross vector b whole square plus 2 into vector a plus vector b into vector a cross vector b is 2 into vector a dot a cross b plus 2 into vector b dot vector a cross vector b. Now again this is equal to 1 plus vector a dot vector b is given by magnitude of vector b a into magnitude of vector b into cos theta square 2 into vector a dot vector b. Let us cancel with 2 into vector a dot vector b and here we have vector a square is square of magnitude of vector a plus square of magnitude of vector b and a cross b is given by magnitude of vector a into magnitude of vector b into sin theta into vector n where n is a unit vector perpendicular to both a and b and here we have the square of this and here we have a dot a cross b which is 0. Here also we have b dot a cross b which is also 0 as we know that a dot a cross b is 0 and we have used the identity vector a dot vector b is given by magnitude of vector a into magnitude of vector b into cos theta and vector a cross vector b is given by magnitude of vector a into magnitude of vector b into sin theta into vector n. Now again this is equal to 1 plus square of magnitude of vector a plus square of magnitude of vector b. Now taking square of magnitude of vector a into square of magnitude of vector b common, we have cos square theta into sin square theta and square of vector n dot square of vector n is 1 as we know that vector a dot vector a is 1 where a is the unit vector. So again this is equal to 1 plus square of magnitude of vector a taking square of magnitude of vector b common we have square of magnitude of vector b into 1 plus square of magnitude of vector a. Now taking 1 plus square of magnitude of vector a common we have 1 plus square of magnitude of vector a into 1 plus square of magnitude of vector b and that is what we have to prove right? So this is the LHS, hence the result is proved. So this completes the question and the session. Why for now take care, have a good day.