 Hello and welcome to the session. In this session, we are going to discuss the following question which says that Find the area of the parmalogram whose adjacent sides are determined by the vectors Vector a is equal to i cap minus j cap plus twice of j cap And vector b is equal to twice of i cap plus twice of j cap minus j cap We know that Area of the parmalogram whose adjacent sides are given by vector a and vector b is equal to modulus of vector a plus vector b With this key idea, let us proceed with the solution Here we are given Vector a is equal to i cap minus j cap plus twice of j cap And vector b is given by twice of i cap plus twice of j cap minus of j cap So vector a cross vector b is equal to the determinant containing elements i cap j cap j cap 1 minus 1 2 2 3 minus 1 On solving this we get i cap into minus 1 into minus 1 that is 1 minus of 3 into 2 that is 6 Minus of j cap into 1 into minus 1 that is minus 1 minus of 2 into 2 that is 4 Let us take out into 1 into 3 that is 3 minus of 2 into minus 1 that is minus 2 We get i cap into 1 minus 6 that is minus 5 minus of j cap into minus 1 minus 4 that is minus 5 plus j cap into 3 minus of minus 2 that is 5. Therefore vector a cross vector b is equal to minus 5 into i cap plus 5 into j cap plus 5 into j cap. Now, modulus of vector a cross vector b is equal to square root of minus 5 b whole square plus 5 square plus 5 square which is equal to square root of minus 5 b whole square is 25 plus 5 square is 25 plus 5 square is 25 which is equal to square root of 75. That is square root of 75 is equal to 5 into square root of 3 and from the key idea we know that area of the parallelogram with adjacent sides by vector a and vector b is given by modulus of vector a cross vector b. Hence, the required area of the parallelogram which is given by modulus of vector a cross vector b 5 into square root of 3 square units. This is the required answer. This completes our session. Hope you enjoyed this session.