 Good morning friends. I am Pura and today we will discuss the following question. The scalar product of the vector i cap plus j cap plus k cap with a unit vector along the sum of vectors 2 i cap plus 4 j cap minus 5 k cap and lambda i cap plus 2 j cap plus 3 k cap is equal to 1. Find the value of lambda. Let us begin with the solution now. Let vector a is equal to 2 i cap plus 4 j cap minus 5 k cap and vector b is equal to lambda i cap plus 2 j cap plus 3 k cap. Now we have to find the sum of vector a and vector b. So we have vector a plus vector b is equal to vector a is equal to 2 i cap plus 4 j cap minus 5 k cap plus vector b is equal to lambda i cap plus 2 j cap plus 3 k cap. This is equal to 2 plus lambda i cap plus 4 plus 2 gives 6 j cap minus 5 plus 3 gives minus 2 k cap. Now the unit vector along vector a plus vector b is n cap and it is given by n cap is equal to vector a plus vector b upon mod of vector a plus vector b. This is equal to now vector a plus vector b is equal to 2 plus lambda i cap plus 6 j cap minus 2 k cap upon mod of 2 plus lambda i cap plus 6 j cap minus 2 k cap. This is equal to 2 plus lambda i cap plus 6 j cap minus 2 k cap upon under root of 2 plus lambda whole square plus 6 square plus minus 2 whole square and this is for the equal to 2 plus lambda i cap plus 6 j cap minus 2 k cap upon under root of 2 plus lambda whole square plus 36 plus 4. And this is equal to 2 plus lambda i cap plus 6 j cap minus 2 k cap upon under root of 2 plus lambda whole square plus 36 plus 4 is equal to 40. So we get n cap is equal to this. Now we are given that the scalar product of i cap plus j cap plus k cap with n cap is equal to 1 that is we are given that 2 plus lambda i cap plus 6 j cap minus 2 k cap upon under root of 2 plus lambda whole square plus 40 dot i cap plus j cap plus k cap is equal to 1. Now this implies 2 plus lambda into 1 gives 2 plus lambda 6 into 1 gives 6. So we get plus 6 and minus 2 into 1 gives minus 2 upon under root of 2 plus lambda whole square plus 40 is equal to 1. This implies now in the numerator we get lambda plus 6 is equal to now we multiply 1 with under root of 2 plus lambda whole square plus 40. So we get under root of 2 plus lambda whole square plus 40. Squaring both the sides we get lambda plus 6 whole square is equal to 2 plus lambda whole square plus 40. This implies now lambda plus 6 whole square is equal to lambda square plus 12 lambda plus 36 and this is equal to now 2 plus lambda whole square is equal to 4 plus 4 lambda plus lambda square plus 40. Canceling lambda square on both the sides we get this implies 12 lambda minus 4 lambda is equal to 4 plus 40 is equal to 44 minus 36 and this implies 8 lambda is equal to 8 and this further implies lambda is equal to 1. Hence we write our answer as lambda is equal to 1. Hope you have understood the solution. Bye and take care.