 Good morning friends. I am Purva and today we will discuss the following question. Find the shortest distance between the lines whose vector equations are vector r is equal to 1 minus t i cap plus t minus 2 j cap plus 3 minus 2 t k cap and vector r is equal to s plus 1 i cap plus 2 s minus 1 j cap minus 2 s plus 1 k cap. Now the shortest distance between the two lines in the vector form, that is vector r is equal to vector a1 plus lambda times vector b1 and vector r is equal to vector a2 plus mu times 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 equation of the lines as vector r is equal to 1 minus t i cap plus t minus 2 j cap plus 3 minus 2 t k cap and vector r is equal to s plus 1 i cap plus 2 s minus 1 j cap minus 2 s plus 1 k cap. Now separating the parameters we get. The first equation becomes vector r is equal to i cap minus 2 j cap plus 3 k cap plus t into minus i cap plus j cap minus 2 k cap and the second equation becomes vector r is equal to i cap minus j cap minus k cap plus s into i cap plus 2 j cap minus 2 k cap. Now by 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 minus 2 j cap plus 3 k cap vector b1 is equal to minus i cap plus j cap minus 2 k cap vector a2 is equal to i cap minus j cap minus k cap and vector b2 is equal to i cap plus 2 j cap minus 2 k cap Now vector a2 minus vector a1 is equal to vector a2 is equal to i cap minus j cap minus k cap minus vector a1 is equal to i cap minus 2 j cap plus 3 k cap and this comes out to be equal to here i cap minus i cap becomes 0 minus 1 and minus into minus becomes plus so plus 2 gives 1 j cap and minus 1 minus 3 gives minus 4 k cap. Now the cross product of vector b1 and vector b2 is equal to determinant of i cap j cap k cap. Now vector b1 is equal to minus i cap plus j cap minus 2 k cap so we get here minus 1 1 minus 2 and vector b2 is equal to i cap plus 2 j cap minus 2 k cap so we get here 1 2 minus 2 this is equal to i cap into 1 into minus 2 is minus 2 minus 2 into minus 2 is minus 4 minus into minus becomes plus so we get plus 4 minus j cap into minus 1 into minus 2 is 2 minus 1 into minus 2 is minus 2 minus into minus becomes plus so we get plus 2 plus k cap into minus 1 into 2 is minus 2 minus 1 into 1 is 1 and this is equal to 2 i cap minus 4 j cap minus 3 k cap. Now mod of cross product of vector v1 and vector v2 is equal to under root of 2 square plus minus 4 square plus minus 3 square. Since we have cross product of vector v1 and vector v2 equal to 2 i cap minus 4 j cap minus 3 k cap. So we get its mod equal to under root of 2 square plus minus 4 square plus minus 3 square this is equal to under root of 4 plus 16 plus 9 and this is equal to under root 29. We mark this as 1. Now cross product of vector v1 and vector v2 dot vector a2 minus vector a1 is equal to 2 i cap minus 4 j cap minus 3 k cap dot j cap minus 4 k cap this is equal to 2 into now here the coefficient of i cap is 0. So 2 into 0 is 0 minus 4 into 1 is minus 4 and minus 3 into minus 4 is plus 12 this is equal to 8 we mark this as 2. Now by key idea we know that the shortest distance between the two lines is given by D is equal to mod of cross product of vector v1 and vector v2 dot vector a2 minus vector a1 upon mod of cross product of vector v1 and vector v2 we mark this as 3. Now putting 1 and 2 in 3 we get D is equal to 8 upon under root 29. So we get our answer as 8 upon under root 29. Hope you have understood the solution. Bye and take care.