 Hello and welcome to the session. In this session we discussed the following question that says prove that the following lines do not intersect. Vector r is equal to 8 i cap minus 7 j cap plus 10 k cap plus lambda into i cap minus 3 j cap plus 6 k cap and vector r is equal to minus 2 i cap plus 2 j cap plus 5 k cap plus mu into 3 i cap minus 2 j cap minus k cap Now given two lines say vector r equal to vector a1 plus lambda into vector b1 let this be line 1 and vector r equal to vector a2 plus mu into vector b2 let this be equation 2 then we say that the lines 1 and 2 intersect at a point if the shortest distance between them is 0 Now the shortest distance that is sd or say d is equal to modulus of vector b1 cross vector b2 dot vector a2 minus vector a1 upon magnitude of vector b1 cross vector b2 so this is the formula to find the shortest distance between the two given lines and if the shortest distance between the two given lines is 0 then we say that the two given lines intersect so this is the key idea to be used in this question Now we proceed with the solution We are given the two lines as vector r equal to 8 i cap minus 7 j cap plus 10 k cap plus lambda into i cap minus 3 j cap plus 6 k cap let this be equation 1 and the other line is vector r equal to minus 2 i cap plus 2 j cap plus 5 k cap plus mu into 3 i cap minus 2 j cap minus k cap let this be equation 2 So when we compare these equations of line with these equations of line so we say like when we compare this equation 1 with this equation 1 so we say where we get vector a1 is equal to 8 i cap minus 7 j cap plus 10 k cap and vector b1 is equal to i cap minus 3 j cap plus 6 k cap and when we compare this equation 2 with this equation 2 we get vector a2 is equal to minus 2 i cap plus 2 j cap plus 5 k cap and vector b2 is equal to 3 i cap minus 2 j cap minus k cap Now let's find out vector a2 minus vector a1 this is equal to minus 2 i cap plus 2 j cap plus 5 k cap minus 8 i cap minus 7 j cap plus 10 k cap and so we get this is equal to minus 10 i cap plus 9 j cap minus 5 k cap so this is vector a2 minus vector a1 Next we find out vector b1 cross vector b2 and this is equal to determinant i cap j cap k cap Now 1 minus 3 is 6 that is the components of vector b1 then components of vector b2 3 minus 2 minus 1 and so we get vector b1 cross vector b2 is equal to i cap into minus 3 into minus 1 that is 3 minus minus 2 into 6 which is minus 12 and so this becomes plus 12 then minus j cap into 1 into minus 1 minus 1 minus 3 into 6 18 plus k cap into minus 2 into 1 minus 2 minus minus 3 into 3 minus 9 so this becomes plus 9 and so we get vector b1 cross vector b2 is equal to 15 i cap plus 19 j cap plus 7 k cap Now the magnitude of vector b1 cross vector b2 is equal to square root of 15 square plus 19 square plus 7 square which is equal to square root of 635 So this is the magnitude of vector b1 cross vector b2 now we find out the shortest distance between the lines 1 and 2 that is equal to SD so we have SD is equal to now using the formula given in the key idea we substitute the values for vector b1 cross vector b2 vector a2 minus vector a1 and magnitude of vector b1 cross vector b2 So we get the shortest distance SD is equal to the modulus of vector b1 cross vector b2 which is equal to 15 i cap plus 19 j cap plus 7 k cap dot vector a2 minus vector a1 which is minus 10 i cap plus 9 j cap minus 5 k cap this upon the magnitude of vector b1 cross vector b2 which is square root of 635 And so this is equal to modulus of minus 150 plus 171 minus 35 upon square root of 635 this is the shortest distance so this is equal to modulus of minus 14 upon square root of 635 or you can say this is equal to 14 upon the square root of 635 this is the shortest distance which is not equal to zero therefore we say that the lines 1 and 2 do not intersect as the shortest distance between them is not equal to zero so this completes the session hope you have understood the solution of this question