 Hi friends and poor va and today we will work out the following question find the magnitude of two vectors vector a and vector b Having the same magnitude and such that the angle between them is 60 degree and Their scalar product is one upon two Let us now begin with the solution Now we are given that magnitude of vector a and vector b are same so since magnitude of vector a and vector b is same Therefore we have mod of vector a is equal to mod of vector b Now we are also given that the angle between them is 60 degree and this scalar product is one upon two so We are also given that theta is equal to 60 degree and Vector a dot vector b is equal to one upon two Now we know that Cos theta is equal to vector a dot vector b Upon mod of vector a into mod of vector b Now we are given that theta is equal to 60 degree. So we have this implies Cos 60 degree is equal to Now vector a dot vector b is given to be one upon two. So we have one upon two into mod of vector a into mod of vector b This further implies Cos 60 degree is equal to one upon two into mod of vector a into mod of vector a Since we are given that mod of vector a is equal to mod of vector b So we can write mod of vector a in place of mod of vector b This implies Now cos 60 degree is equal to one upon two. So we have one upon two is equal to one upon two into mod of vector a square Canceling one upon two from both the sides we get this implies mod of vector a square is equal to one which further implies mod of vector a is equal to one Now since we are given that mod of vector a is equal to mod of vector b So we have mod of vector b is also equal to one Because we have mod of vector a is equal to one Hence we have got our answer as Mod of vector a is equal to one and mod of vector b is equal to one Hope you have understood the solution. Bye and take care