 Hello friends I am Purva and today we will discuss the following question. Find a vector in the direction of vector phi by cap minus j cap plus 2 k cap which has magnitude 8 units. Let us now begin with the solution. Now let us denote phi by cap minus j cap plus 2 k cap by vector a. So we have vector a is equal to phi by cap minus j cap plus 2 k cap. Therefore we have mod of vector a is equal to under root of 5 square plus minus 1 square plus 2 square and this is equal to under root of 25 plus 1 plus 4 and this is equal to root 30. Therefore we have got mod of vector a is equal to root 30. Now the vector with magnitude 8 is vector b is equal to a cap into mod 8 where we have a cap is equal to vector a upon mod of vector a and we mark this as equation 1. Now first we will find a cap. So we have a cap is equal to vector a upon mod of vector a and this is equal to now vector a is equal to 5 i cap minus j cap plus 2 k cap upon mod of vector a is equal to root 30. So we have root 30 and this is equal to 5 upon root 30 i cap minus 1 upon root 30 j cap plus 2 upon root 30 k cap. So we have got a cap is equal to this. Now from equation 1 we have vector b is equal to a cap into mod 8 and we have this is equal to now a cap is equal to 5 upon root 30 i cap minus 1 upon root 30 j cap plus 2 upon root 30 k cap into 8 this is equal to now 5 into 8 is 40. So we have 40 upon root 30 i cap minus 1 into 8 is 8. So 8 upon root 30 j cap plus 2 into 8 is 16. So 16 upon root 30 k cap. Hence a vector in the direction of vector 5 i cap minus j cap plus 2 k cap which has magnitude 8 units is 40 upon root 30 i cap minus 8 upon root 30 j cap plus 16 upon root 30 k cap. This is our answer. Hope you have understood the solution. Bye and take care.