 Hello and welcome to the session. In this session we are going to discuss vector area of plane region bounded by a closed curve. Let us consider the boundary of a closed curve in anti-clockwise and clockwise directions. Here the direction of the curve A1, A2, A3 is anti-clockwise with respect to A1. And in this the direction of A1, A3, A2 is clockwise with respect to the point A1. Let a unit vector n cap be perpendicular to the plane of the curve in the direction in which the curve is anti-clockwise. Let A be the plane area of the region bounded by A1, A2, A3 or A1, A3, A2 in anti-clockwise direction A1, A2, A1, A3 and unit vector n cap form a vector triad in the right handed system for the closed curve A1, A2, A3. The vector area for this plane region is given by A into n cap and in clockwise direction A3, A1, A2 and unit vector minus n cap form a vector triad in the right handed system. Now for closed curve A1, A3, A2 the vector area of this plane bounded region is given by A into minus of unit vector n cap. Now we are going to discuss vector area of a triangle and first we have for triangle ABC the vector area is given by 1 by 2 into vector AB plus vector AC let delta be the area of the triangle ABC let n cap be a unit vector perpendicular to the plane of the triangle considered in the anti-clockwise direction from vector AB to vector AC. Now vector AB, vector AC and unit vector n cap form a vector triad in the right handed system. Then the vector area of the triangle is given by delta is equal to area of triangle ABC that is 1 by 2 into B into C into sin of angle A which can be written as 1 by 2 into C into B into sin of angle A which is equal to 1 by 2 into C that is modulus of vector AB into B into sin of angle A therefore vector area of triangle ABC which is given by delta into n cap is equal to 1 by 2 into modulus of vector AB into modulus of vector AC into sin of angle A into n cap and which is equal to 1 by 2 into vector AB plus vector AC. Secondly we have the vector area of a triangle ABC with vertices having position vectors vector A vector B vector C respectively is given by 1 by 2 into vector B cross vector C plus vector C cross vector A plus vector A cross vector B. We know that vector area of triangle ABC is given by 1 by 2 into vector AB cross vector AC since position vectors of A B and C are vector A, vector B and vector C respectively. Therefore vector AB is equal to position vector of B minus position vector of A that is vector B minus vector A and vector AC is given by position vector of C minus position vector of A that is vector C minus vector A therefore 1 by 2 into vector AB cross vector AC can be written as 1 by 2 into vector B minus vector A cross vector C minus vector A which is equal to 1 by 2 into vector B cross vector C minus of vector B cross vector A minus of vector A cross vector C plus vector A cross vector A and we know that vector A cross vector A is equal to 0 therefore we get 1 by 2 into vector B cross vector C minus of vector B cross vector A minus of vector A cross vector C and we know that vector A cross vector B can also be written as minus of vector B cross vector A therefore we get 1 by 2 into vector B cross vector C plus vector A cross vector B plus vector C cross vector A therefore vector area of triangle ABC is equal to 1 by 2 into vector B cross vector C plus vector C cross vector A plus vector A cross vector B we should also note that if AB and C are collinear points then the vector area of the triangle formed by these points must be 0 that is vector A cross vector B plus vector B cross vector C plus vector C cross vector A should be equal to 0 now we are going to discuss vector area of a parallelogram with vector A and vector B as adjacent sides let OP RQ be our parallelogram let vector OP be equal to vector A vector OP be equal to vector B and N cap be our unit vector perpendicular to the plane of vector A and vector B here OP RQ is the parallelogram where vector OP is equal to vector A and vector OP is equal to vector B here vector A vector B and unit vector N cap form a vector triad in the right handed system now area of triangle OPQ 1 by 2 into modulus of vector OP cross modulus of vector OP that is we have 1 by 2 into modulus of vector A cross vector B therefore area of parallelogram OP RQ is equal to twice of area of triangle OPQ which is equal to 2 into area of triangle OPQ that is 1 by 2 into modulus of vector A cross vector B which is equal to modulus of vector A cross vector B here vector A cross vector B represents the vector area of the parallelogram OP RQ whose adjacent sides are vector A and vector B such that vector A vector B and unit vector N cap form a vector triad in the right handed system and area of parallelogram OP RQ is equal to modulus of vector A cross vector B now we are going to learn vector area of a quadrilateral PQRS is given by 1 by 2 into vector PR cross vector QF let PQRS be a quadrilateral where minus PR and QS intersect at the point O then vector PR is equal to vector PO plus vector OR and vector QS is equal to vector QO plus vector OS therefore 1 by 2 into vector PR cross vector QS is equal to 1 by 2 into vector PO plus vector OR cross vector QO plus vector OS which is equal to 1 by 2 into vector PO cross vector QS plus 1 by 2 into vector PO cross vector OS plus 1 by 2 into vector OR cross vector QO as 1 by 2 into vector OR cross vector OS and we know that 1 by 2 into vector PO cross vector QO is equal to vector area of triangle POQ similarly 1 by 2 into vector PO cross vector OS is equal to vector area of triangle PO S and therefore we get vector area of triangle POQ plus vector area of triangle PO S plus 1 by 2 into vector OR cross vector QO which is equal to vector area of triangle ROQ plus 1 by 2 into vector OR cross vector OS which is equal to vector area of triangle ROS which is equal to vector area of quadrilateral PQRS this completes our session hope you enjoyed this session