 Hello and welcome to this session. In this session we are going to discuss properties of vector product. Property 1 states that the vector product is not commutative rather vector A cross vector B is equal to minus of vector B cross vector A. Let vector A and vector B be to non-zero, non-parallel vectors. Let theta be the angle between them where theta lies in between 0 and pi and let n cap be a unit vector. Dependicular to both vector A and vector B, vector A and vector B are to non-zero and non-parallel vectors and theta is the angle between vector A and vector B and n cap is a unit vector which is perpendicular to both vector A and vector B. Then by definition we have vector A cross vector B is equal to A into B into sin of angle theta into n cap. Or we can also write it as modulus of vector A cross vector B is equal to A into B into sin of angle theta and mark this equation as 1 where the orientation is anticlockwise and is taken as pivotal. Now similarly we have modulus of vector B cross vector A is equal to B into A into sin of angle theta. Here orientation is clockwise and we can write modulus of vector B cross vector A as A into B into sin of angle theta and as we have A B is equal to B A. Hence we can say vector A cross vector B and vector B cross vector A are equal in magnitude but opposite in direction therefore vector A cross vector B is equal to minus of vector B cross vector A. Next we have the vector product of our vector say vector A with itself is a null vector that is vector A cross vector A is equal to zero vector as angle between the vectors is equal to zero degrees that is the value of theta is equal to zero. Therefore sin of theta is equal to sin of zero degrees that is zero which implies that the value of vector A cross vector A that is vector A into vector A into sin of zero degrees into n cap will be equal to zero since sin of zero degrees is equal to zero. Note that vector product is associated with respect to us thus if vector A and vector B are any two vectors and n is scalar then n into vector A cross vector B is equal to vector A cross vector B is equal to vector A cross vector B let theta be the angle between vector A and vector B and let n be our scalar then the following cases arise case one when n is greater than zero here theta is the angle between vector A and vector B and n is our scalar then we have n into vector A and vector A are in the same direction and vector B and n into vector B are also in the same direction therefore n into vector A cross vector B is equal to modulus of n into vector A into modulus of vector B into sin of angle theta into n cap which is equal to n into A into B into sin of angle theta into n cap which can be written as n into A into B into sin of angle theta into n cap which is equal to n into vector A cross vector B also we have vector A cross n into vector B is equal to modulus of vector A into modulus of n into vector B into sin of angle theta into n cap which can be written as A into n into B into sin of angle theta into n cap that is we have n into A into B into sin of angle theta into n cap which is equal to n into vector A cross vector B as vector A cross vector B is equal to AB sin of angle theta into n cap now we have n into vector A cross vector B is equal to n into vector A cross vector B and vector A cross n into vector B is equal to n into vector A cross vector B Therefore we can say N2 vector A cross vector B is equal to N into vector A cross vector B and is equal to vector A cross N into vector B. Second case is when N is less than 0, where theta is the angle between vector A and vector B. In this case N into vector A and vector A are N-like vectors and so the angle between N into vector A and vector B is pi minus theta and here N into vector A vector B and minus N cap form a right-handed system. Therefore N into vector A cross vector B is equal to modulus of N into vector A into modulus of vector B into sin of angle pi minus theta into minus N cap which is equal to modulus of N into modulus of vector A into modulus of vector B into sin of pi minus theta which is equal to pi minus theta into minus N cap since N is less than 0, therefore modulus of N is equal to minus of N into into B into sin of angle theta into minus of N cap which is equal to N into A into B into sin of angle theta into N cap which can be written as N into vector A cross vector B where theta is the angle between vector A and vector B and again N into vector B and vector B are N-like vectors and so the angle between vector A and N into vector B is pi minus theta in this case vector A N into vector B and minus of N cap form a right-handed system therefore vector A cross N into vector B is equal to modulus of vector A into modulus of N into vector B into sin of pi minus theta into minus of N cap which can be written as modulus of vector A into modulus of N into modulus of vector B into sin of pi minus theta which is equal to sin of theta into minus of N cap. Now modulus of N is equal to minus N since the value of N is less than 0 so we get minus N into A into B into sin of angle theta into minus of N cap which is equal to N into A into B into sin of angle theta into N cap which can be written as N into vector A cross vector B so we have got N into vector A cross vector B is equal to N into vector A cross vector B and vector A cross N into vector B is equal to N into vector A cross vector B thus in this case we get N into vector A cross vector B is equal to N into vector A cross vector B is equal to vector A cross N into vector B. Now case 3 is when the value of N is equal to 0 in this case we have N into vector A cross vector B which is equal to 0 vector cross vector B that is 0 vector since N into vector A is equal to 0 into vector A which is equal to 0 vector also we have vector A cross N into vector B is equal to vector A cross 0 vector which is equal to 0 vector since N into vector B is equal to 0 into vector B which is equal to 0 vector and N into vector A cross vector B is equal to 0 into vector A cross vector B which is equal to 0 vector so in this case also we get N into vector A cross vector B is equal to N into vector A cross vector B and is equal to vector A cross N into vector B as the case is N into vector A cross vector B is equal to N into vector A cross vector vector b and is equal to vector a cross m into vector b, property 4 is distributive law that is for any three vectors vector a, vector b and vector c, vector a cross vector b plus vector c is equal to vector a cross vector b plus vector a cross vector c. Let vector a be equal to a1 i cap plus a2 j cap plus a3 k cap, vector b be equal to b1 i cap plus b2 j cap plus b3 k cap and vector c be equal to c1 i cap plus c2 j cap plus c3 k cap therefore vector a cross vector b plus vector c, where vector a is equal to a1 i cap plus a2 j cap plus a3 k cap and vector b plus vector c is equal to b1 plus c1 into i cap plus b2 plus c2 into j cap plus b3 plus c3 into k cap can be written in the determinant form as the determinant containing elements i cap j cap j cap a1 a2 a3 b1 plus c1 b2 plus c2 b3 plus c3 and solving this we get a2 into b3 plus c3 minus of a3 into b2 plus c2 multiplied by i cap plus a3 into b1 plus c1 minus of a1 into b3 plus c3 multiplied by j cap plus a1 into b2 plus c2 minus of a2 into b1 plus c1 multiplied by k cap and mark this equation as 1 now vector a cross vector b where vector a is given by a1 i cap plus a2 j cap plus a3 k cap and vector b is given by b1 i cap plus b2 j cap plus b3 k cap can be written in the determinant form as the determinant containing elements i cap j cap k cap a1 a2 a3 b1 b2 b3 and solving this we get a2 into b3 minus a3 into b2 into i cap plus a3 into b1 minus a1 into b3 into j cap plus a1 into b2 minus a2 into b1 into k cap and mark this equation as 2 now vector a cross vector c where vector a is given by a1 i cap plus a2 j cap plus a3 k cap and vector c is given by c1 i cap plus c2 j cap plus c3 k cap can be written in the determinant form as the determinant containing elements i cap j cap k cap a1 a2 a3 c1 c2 c3 and solving this we get into j cap plus a1 into c2 minus a2 into c1 into k cap and mark this equation as 3 now adding equation 2 and equation 3 we get vector a cross vector b plus vector a cross vector c is equal to 2 into b3 plus c3 minus of a3 into b2 plus c2 multiplied by i cap plus a3 into b1 plus c1 minus of a1 into b3 plus c3 multiplied by j cap plus a1 into b2 plus c2 minus of a2 into b1 plus c1 multiplied by k cap and mark this equation as 4 since right hand side of equation 1 and equation 4 are equal so the left hand sides are also equal therefore we can say vector a cross vector b plus vector c is equal to vector a cross vector b plus vector a cross vector c this completes our question hope you enjoyed this question