 Hello friends, I am Purva and today we will work out the following question. Find the direction cosines of the vector joining the points A, 1, 2, minus 3 and B, minus 1, minus 2, 1 directed from A to B. Let us now begin with the solution. We are given the points A, 1, 2, minus 3 and B, minus 1, minus 2, 1. Now let O be the origin. Then we have vector OA is equal to i cap plus 2 j cap minus 3 k cap and vector OB is equal to minus i cap minus 2 j cap plus k cap. Now by trying the law we know that vector AB is equal to vector OB minus vector OA. This is equal to, now vector OB is equal to minus i cap minus 2 j cap plus k cap minus vector OA is equal to i cap plus 2 j cap minus 3 k cap. And we have this is equal to minus 1 minus 1 i cap plus minus 2 minus 2 j cap plus 1 plus 3 k cap and this is equal to minus 2 i cap minus 4 j cap plus 4 k cap. Now to find a unit vector in the direction of vector AB we need to find mod of vector AB. So we have mod of vector AB is equal to under root of minus 2 square plus minus 4 square plus 4 square and this is equal to under root of 4 plus 16 plus 16 which is equal to under root of 36. So we get mod of vector AB is equal to 6. Now the unit vector in the direction of vector AB that is n cap is equal to vector AB upon mod of vector AB. We have this is equal to now vector AB is equal to minus 2 i cap minus 4 j cap plus 4 k cap upon mod of vector AB is equal to 6. So we have 6. So we have n cap is equal to minus 1 upon 3 i cap minus 2 upon 3 j cap plus 2 upon 3 k cap. Hence we get the direction cosines as minus 1 upon 3 minus 2 upon 3 and 2 upon 3. Hence we write our answer as minus 1 upon 3 comma minus 2 upon 3 comma 2 upon 3. Hope you have understood the solution. Bye and take care.