 Hi friends I am Purva and today we will discuss the following question, find mod of vector A and mod of vector B if dot product of vector A plus vector B and vector A minus vector B is equal to 8 and mod of vector A is equal to 8 into mod of vector B. Let us now begin with the solution. Now we are given dot product of vector A plus vector B and vector A minus vector B is equal to 8. We can write this as this implies mod of vector A square minus vector A dot vector B plus vector B dot vector A minus mod of vector B square is equal to 8. Opening the bracket on left hand side we get mod of vector A square minus vector A dot vector B plus vector B dot vector A minus mod of vector B square is equal to 8 and we get this implies now cancelling out minus vector A dot vector B and vector B dot vector A we get mod of vector A square minus mod of vector B square is equal to 8. Now we are also given that mod of vector A is equal to 8 into mod of vector B. So putting the value of mod of vector A here we get this implies 8 square into mod of vector B square minus mod of vector B square is equal to 8 and this implies 64 into mod of vector B square minus mod of vector B square is equal to 8. This further implies 63 into mod of vector B square is equal to 8 which further implies mod of vector B square is equal to 8 upon 63. This implies mod of vector B is equal to under root of 8 upon 63 which implies mod of vector B is equal to 2 root 2 upon 3 root 7. So we have got mod of vector B is equal to 2 root 2 upon 3 root 7. Now we will find mod of vector A and this is equal to 8 into mod of vector B this is given to us. So we get this implies mod of vector A is equal to 8 into 2 root 2 upon 3 root 7 and this is equal to 16 root 2 upon 3 root 7. So we have got mod of vector A is equal to 16 root 2 upon 3 root 7. Hence we write our answer as mod of vector A is 16 root 2 upon 3 root 7 and mod of vector B is 2 root 2 upon 3 root 7. Hope you have understood the solution. Bye and take care.