 Hi friends, I am Purva and today we will work out the following question. Find the shortest distance between the lines whose vector equations are vector r is equal to i cap plus 2 j cap plus 3 k cap plus lambda into i cap minus 3 j cap plus 2 k cap and vector r is equal to 4 i cap plus 5 j cap plus 6 k cap plus mu into 2 i cap plus 3 j cap plus k cap. Now the shortest distance between two given lines vector r is equal to vector a1 plus lambda into vector b1 and vector r is equal to vector a2 plus mu into vector b2 is given by d is equal to mod of cross product of vector b1 and vector b2 dot vector a2 minus vector a1 upon mod of cross product of vector b1 and vector b2. So this is the key idea behind our question. Let us begin with the solution now. Now we are given the lines as vector r is equal to i cap plus 2 j cap plus 3 k cap plus lambda into i cap minus 3 j cap plus 2 k cap and vector r is equal to 4 i cap plus 5 j cap plus 6 k cap plus mu into 2 i cap plus 3 j cap plus k cap. Now comparing these two equations with these two equations in the key idea we can clearly see that here vector a1 is equal to i cap plus 2 j cap plus 3 k cap vector b1 is equal to i cap minus 3 j cap plus 2 k cap vector a2 is equal to 4 i cap plus 5 j cap plus 6 k cap and vector v2 is equal to 2 i cap plus 3 j cap plus k cap Now by key idea we know that the shortest distance between two lines is given by d is equal to mod of cross product of vector b1 and vector b2 dot vector a2 minus vector a1 upon mod of cross product of vector b1 and vector b2 we mark this as 1 Now vector a2 minus vector a1 is equal to vector a2 is equal to 4 i cap plus 5 j cap plus 6 k cap minus vector a1 is equal to i cap plus 2 j cap plus 3 k cap and this is equal to 4 i cap minus i cap is equal to 3 i cap 5 j cap minus 2 j cap is equal to 3 j cap and 6 k cap minus 3 k cap is equal to 3 k cap Now cross product of vector b1 and vector b2 is equal to determinant of i cap j cap k cap 1 minus 3 2 and 2 3 1 This is equal to i cap into minus 3 into 1 is minus 3 minus 3 into 2 is 6 minus j cap into 1 into 1 is 1 minus 2 into 2 is 4 plus k cap into 1 into 3 is 3 minus into minus becomes plus 2 into 3 is 6 and this is equal to minus 9 i cap plus 3 j cap plus 9 k cap Now mod of vector b1 cross vector b2 is equal to under root of minus 9 whole square plus 3 square plus 9 square under root of 81 plus 9 plus 81 and this is equal to 3 under root 19. We mark this as 2 Now cross product of vector b1 and vector b2 dot vector a2 minus vector a1 is equal to cross product of vector b1 and vector b2 is equal to minus 9 i cap plus 3 j cap plus 9 k cap dot vector a2 minus vector a1 is equal to 3 i cap plus 3 j cap plus 3 k cap This is equal to minus 9 into 3 is minus 27 3 into 3 is 9 plus 9 into 3 is 27 and this is equal to 9 We mark this as 3 Now putting the value of 2 and 3 in 1 we get d is equal to mod of 9 upon 3 under root 19 and this is equal to 9 upon 3 under root 19 which is further equal to 3 upon under root 19. So we have got our answer as 3 upon under root 19. Hope you have understood the solution Bye and take care.