 Hi and welcome to the session, let us discuss the following question, question says, find the shortest distance between lines r vector is equal to 6i plus 2j plus 2k plus lambda multiplied by i minus 2j plus 2k and r vector is equal to minus 4i minus k plus mu multiplied by 3i minus 2j minus 2k. First of all, let us understand that shortest distance between these two lines is given by modulus of b1 vector cross b2 vector dot a2 vector minus a1 vector upon modulus of b1 vector cross b2 vector. This is the key idea to solve the given question. Let us now start with the solution. How given lines are? r vector is equal to 6i plus 2j plus 2k plus lambda multiplied by i minus 2j plus 2k and other line is r vector is equal to minus 4i minus k plus mu multiplied by 3i minus 2j minus 2k. Now comparing these two equations of the lines with the equations of the lines given in key idea, we get a1 vector is equal to 6i plus 2j plus 2k b1 vector is equal to i minus 2j plus 2k vector a2 is equal to minus 4i minus k and vector b2 is equal to 3i minus 2j minus 2k. Now from key idea we know shortest distance between two given lines is modulus of b1 vector cross b2 vector dot a2 vector minus a1 vector upon modulus of b1 vector cross b2 vector. Now first upon let us find out difference of a2 vector and a1 vector. It is given by minus 4i minus k minus 6i plus 2j plus 2k. Now subtracting corresponding components of these two vectors we get minus 10i minus 2j minus 3k is equal to vector a2 minus vector a1. Now we will find cross product of b1 vector and b2 vector it is given by determinant of unit vector i unit vector j unit vector k 1 minus 2 2 3 minus 2 minus 2. Now expanding this determinant with respect to first row we get 8i plus 8j plus 4k is equal to b1 vector cross b2 vector. Now we will find out modulus of b1 vector cross b2 vector it is given by square root of 8 square plus 8 square plus 4 square. Now simplifying further we get square root of 144 adding these three terms we get 144. Now square root of 144 is equal to 12. Now we will find b1 vector cross b2 vector dot a2 vector minus a1 vector. Now we know b1 vector cross b2 vector is equal to 8i plus 8j plus 4k dot a2 vector minus a1 vector is equal to minus 10i minus 2j minus 3k. Now we will find dot product of these two vectors by multiplying corresponding components. Multiplying 8i by minus 10i we get minus 18. Similarly multiplying these two components we get minus 16 and multiplying 4k by minus 3k we get minus 12. Now simplifying further we get this is equal to minus 108. Now we get shortest distance between the given two lines is equal to modulus of minus 108 upon 12. We know shortest distance between two given lines is given by this formula and clearly we can see we have just found numerator of this formula is equal to minus 108 and denominator is equal to 12. As shown in these two expressions now simplifying further we get shortest distance between two lines is equal to 9. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.