 Good morning friends. I am Poojwa and today we will discuss the following question. Find the projection of the vector i cap plus 3 j cap plus 7 k cap on the vector 7 i cap minus j cap plus 8 k cap. Let us now begin with the solution. Now let vector a is equal to i cap plus 3 j cap plus 7 k cap and vector b is equal to 7 i cap minus j cap plus 8 k cap. Now first we will find the dot product of vector a and vector b. So now vector a dot vector b is equal to i cap plus 3 j cap plus 7 k cap dot 7 i cap minus j cap plus 8 k cap and we have this is equal to 1 into 7 plus 3 into minus 1 plus 7 into 8 and we get this is equal to 1 into 7 gives 7 3 into minus 1 gives minus 3 plus 7 into 8 gives 56 and we get this is equal to 60. So we have got vector a dot vector b is equal to 60. Now we will find mod of vector b. So mod of vector b is equal to under root of 7 square plus minus 1 square plus 8 square and we get this is equal to under root of now 7 square is equal to 49 plus minus 1 square is equal to 1 plus 8 square is equal to 64 and we get this is equal to under root 114. So we have got mod of vector b is equal to root 114. Now the projection of vector a on vector b is given by vector a dot vector b upon mod of vector b this is equal to vector a dot vector b which is equal to 60. So we have 60 upon mod of vector b which is equal to root 114. So we have got our answer as 60 upon root 114. Hope you have understood the solution. Bye and take care.