 Hi friends I am Purva and today we will discuss the following question. Find the shortest distance between the lines vector r is equal to i cap plus 2 j cap plus k cap plus lambda into i cap minus j cap plus k cap and vector r is equal to 2 i cap minus j cap minus k cap plus mu into 2 i cap plus j cap plus 2 k cap. Now the shortest distance between two 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 k cap plus lambda into i cap minus j cap plus k cap and vector r is equal to 2 i cap minus j cap minus k cap plus mu into 2 i cap plus j cap plus 2 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 k cap vector b1 is equal to i cap minus j cap plus k cap vector a2 is equal to 2 i cap minus j cap minus k cap and vector b2 is equal to 2 i cap plus j cap plus 2 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 2 i cap minus j cap minus k cap minus vector a1 is equal to i cap plus 2 j cap plus k cap and this is equal to 2 i cap minus i cap is i cap minus j cap minus 2 j cap is minus 3 j cap and minus k cap minus k cap is minus 2 k cap. Now cross product of vector b1 and vector b2 is equal to determinant of i cap j cap k cap. Now below i cap we write 1 here we write minus 1 and below k cap we write 1 below 1 we write 2 below minus 1 we write 1 and below 1 we write 2 and this is equal to i cap into minus 2 minus 1 minus j cap into 2 minus 2 plus k cap into 1 plus 2 and we get this is equal to minus 3 i cap plus 3 k cap. Now mod of vector b1 cross vector b2 is equal to under root of minus 3 whole square plus 3 square this is equal to under root of minus 9 plus 9 which is equal to 3 root 2 we mark this as 2 and finally 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 3 i cap plus 3 k cap dot vector a2 minus vector a1 is equal to i cap minus 3 j cap minus 2 k cap this is equal to minus 3 into 1 is minus 3 here we have 0 j cap so 0 into minus 3 is 0 and 3 into minus 2 is minus 6 so we get this is equal to minus 9 we mark this as 3 now putting 2 and 3 in 1 we get d is equal to mod of minus 9 upon 3 root 2 this is equal to 9 upon 3 root 2 which is equal to 3 root 2 upon 2 thus we have got our answer as 3 root 2 upon 2 hope you have understood the solution bye and take care