 Hello and welcome to the session. In this session we are going to discuss a few details on vector product. We know that vector A cross vector B is equal to A into B into sin of theta into n-tap. A is equal to modulus of vector A. B is equal to modulus of vector B and n-tap is a unit vector dependent to both vector A and vector B. First we have vector product of two parmen or collinear vector is 0. If the vectors A and B, parmen or collinear then the value of theta is equal to 0 of 180 degrees. So in each case the value of sin of theta will be equal to 0 which implies that vector A cross vector B which is equal to A into B into sin of theta into n-tap will be equal to 0 into n-tap since the value of sin of theta is 0 which is equal to 0 vector. Second result is vector A cross vector A is equal to 0 for any vector A. Therefore if we have vector A cross vector A then it is equal to modulus of vector A into modulus of vector A into sin of 0 into n-tap which is equal to 0 into n-tap since sin of 0 is equal to 0 that is equal to 0 vector. Third result is vector A cross vector B is equal to 0 vector then vector A is equal to 0 or vector B is equal to 0 or vector A and vector B are parmen or collinear. Let theta be the angle between vector A and vector B then vector A cross vector B is equal to 0 vector implies that A into B into sin of angle theta into n-tap is equal to 0 vector which implies that A into B into sin of angle theta is equal to 0 which implies that A is equal to 0 or B is equal to 0, sin of angle theta is equal to 0, A is equal to 0 implies that vector A is equal to 0 vector and B is equal to 0 implies that vector B is equal to 0 vector and sin of angle theta is equal to 0 implies that theta is equal to 0 as time of theta is equal to 0 for theta is equal to 0 or theta is equal to pi which implies that vector A is equal to 0 vector or vector B is equal to 0 vector A theta is equal to 0 or theta is equal to pi implies that vector A and vector B are collinear or parallel. Next we are going to discuss vector product of two perpendicular vectors. We know that vector A cross vector B is equal to A into B into sin of angle theta into n-tap then the value of theta is given as pi by 2 then the value of sin of theta that is sin of pi by 2 is equal to 1. Therefore vector A cross vector B is equal to A into B into sin of pi by 2 that is 1 into n-tap or we can say modulus of vector A cross vector B is equal to modulus of A into B into n-tap which is equal to A into B into modulus of n-tap and modulus of n-tap is equal to 1 therefore we have A into B. Now we shall discuss vector product of two unit vectors if A and B are two vectors such that A is equal to modulus of vector A which is equal to 1 and B is equal to modulus of vector B which is equal to 1 that is A and B are unit vectors therefore vector A cross vector B which is equal to A into B into sin of angle theta into n-tap is equal to sin of angle theta into n-tap since the value of A is equal to 1 and the value of B is also equal to 1. In other words we can say that the cross product of two unit vectors is a vector whose magnitude is equal to the sin of the angle between the direction of the given vectors. Now we are going to discuss vector product of unit vectors i-tap, j-tap and k-tap. Let i-tap, j-tap, k-tap be the also normal child of the unit vectors then i-tap cross j-tap which is equal to 1 into 1 into sin of 90 degrees into n-tap since i-tap and j-tap are the unit vectors and i-tap and j-tap are perpendicular to each other therefore theta is equal to 90 degrees and n-tap is the unit vector perpendicular to both i-tap and j-tap and is therefore equal to j-tap therefore by definition of also normal child of unit vectors therefore i-tap cross j-tap which is equal to sin of 90 degrees into n-tap is equal to 1 into k-tap since sin of 90 degrees is equal to 1 and n-tap is equal to k-tap which is equal to k-tap this implies that i-tap cross j-tap is equal to k-tap similarly j-tap cross k-tap will be equal to i-tap and k-tap cross i-tap is equal to j-tap therefore we have i-tap cross j-tap is equal to k-tap j-tap cross k-tap is equal to i-tap and k-tap cross i-tap is equal to j-tap also as we know vector A cross vector B is equal to minus of vector B cross vector A i-tap cross j-tap is equal to minus of j-tap cross i-tap here we have i-tap cross j-tap which is equal to j-tap is equal to minus of j-tap cross i-tap also we have j-tap cross k-tap which is equal to i-tap will be equal to minus of k-tap cross j-tap and k-tap cross i-tap which is equal to j-tap will be equal to minus i-tab cross k-tab. Therefore we have i-tab into j-tab is equal to k-tab, j-tab into k-tab is equal to i-tab, k-tab into i-tab is equal to j-tab and similarly i-tab into k-tab is equal to minus of j-tab, k-tab into j-tab is equal to minus of i-tab, j-tab into k-tab is equal to minus of k-tab. Now we shall learn how vector product can be expressed as a determinant. Let vector a be equal to a-1 i-tab plus a-2 j-tab plus a-3 k-tab. Vector b be equal to b-1 i-tab plus b-2 j-tab plus b-3 k-tab. Now vector a cross vector b is equal to a-1 i-tab plus a-2 j-tab plus a-3 k-tab cross b-1 i-tab plus b-2 j-tab plus b-3 k-tab which is equal to a-1 b-1 into i-tab cross i-tab plus a-1 b-2 into i-tab cross j-tab plus a-1 b-3 into i-tab cross j-tab plus a-2 b-1 into j-tab cross i-tab plus a-2 b-2 into j-tab cross j-tab plus a-2 b-3 into j-tab cross j-tab plus a-3 b-1 into j-tab cross i-tab plus a-3 b-2 into j-tab cross j-tab plus a-3 b-3 into j-tab cross j-tab and from our previous results we know that i-tab cross j-tab is equal to j-tab, j-tab cross j-tab is equal to i-tab and j-tab cross i-tab is equal to j-tab. Also we have i-tab cross j-tab which is equal to j-tab is also equal to minus of j-tab cross i-tab. j-tab cross j-tab is equal to i-tab and is also equal to minus of j-tab cross j-tab. j-tab cross i-tab which is equal to j-tab is also equal to minus of i-tab cross j-tab and from the second results we know that vector a cross vector a is equal to 0 vector. Using the above two results we get a-2 into b-3 minus of a-3 into b-2 into i-tab plus a-3 into b-1 minus of a-1 into b-3 into j-tab plus a-1 into b-2 minus of a-2 into b-1 into k-tab which can be written in the determinant form as vector a cross vector b is equal to the determinant containing elements i-tab j-tab k-tab a-1 a-2 a-3 b-1 b-2 b-3. Next is angle between two non-zero non-parallel vectors we have vector a cross vector b which is equal to modulus of vector a into modulus of vector b into sin of angle theta into n cap or we can write modulus of vector a cross vector b is equal to modulus of modulus of vector a into modulus of vector b into sin of angle theta into n cap which can be written as modulus of modulus of vector a into modulus of vector b into sin of angle theta into modulus of unit vector n cap which again can be written as modulus of vector a into modulus of vector b into sin of angle theta into 1 since modulus of unit vector n cap is equal to 1 which is equal to modulus of vector a into modulus of vector b into sin of angle theta so we have modulus of vector a cross vector b is equal to modulus of vector a into modulus of vector b into sin of angle theta which implies that sin of angle theta is equal to modulus of vector a cross vector b upon modulus of vector a into modulus of vector b next we are going to discuss unit vector perpendicular to two given vectors let n cap be a unit vector perpendicular to two non zero and non parallel vectors vector a and vector b are non zero and non parallel the cross product vector a cross vector b is non zero now we know that vector a cross vector b is equal to modulus of vector a into modulus of vector b into sin of angle theta into n cap and we can write it as modulus of vector a cross vector b is equal to modulus of modulus of vector a into modulus of vector b into modulus of sin of angle theta into modulus of n cap since theta lies in between zero and pi therefore the value of sin of theta will always be greater than zero and n cap is the unit vector and modulus of unit vector n cap will be equal to 1 therefore we get modulus of vector a cross vector b is equal to modulus of vector a into modulus of vector b into sin of angle theta into 1 vector a cross vector b is equal to modulus of vector a into modulus of vector b into sin of angle theta into n cap we know this equation as 1 and equation 1 implies that vector a cross vector b is equal to modulus of vector a into modulus of vector b into sin of angle theta which is equal to modulus of vector a cross vector b into n cap which implies that unit vector n cap is equal to vector a cross vector b upon modulus of vector a cross vector b sin theta vector minus of vector a cross vector b is also equivalent to the plane of vector a and vector b therefore minus of vector a cross vector b upon modulus of vector a cross vector b is also a unit vector the perpendicular to the plane of vector a and vector b therefore unit vectors the perpendicular to the plane of vector a and vector b plus minus of vector a cross vector b upon modulus of vector a cross vector b which would also note that unit vector of length lambda where lambda is not equal to 0 normally to the plane of vector a and vector b are given by plus minus of lambda into vector a cross vector b upon modulus of vector a cross vector b which is equal to plus minus of lambda upon modulus of vector a cross vector b into vector a cross vector b this completes my question hope you enjoyed this question