 Hi and welcome to the session. I am Shrishi and I am going to help you with the following question. Question says show that the line through points 4, 7, 8, 2, 3, 4 is parallel to the line through the points minus 1, minus 2, 1 and 1, 2, 5. First of all let us understand the direction cosines of a line joining two points p having coordinates x1, y1, z1 and point q having coordinates x2, y2, z2 are x2 minus x1 upon pq, y2 minus y1 upon pq, z2 minus z1 upon pq where pq is equal to square root of x2 minus x1 whole square plus y2 minus y1 whole square plus z2 minus z1 whole square. Also if l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines and theta is the acute angle between the two lines then cos theta is equal to absolute value of l1, l2 plus m1, m2 plus n1, n2. Now we will use this discussion as our key idea to solve the given question. Let us now start with the solution. First of all let us assume that a and b represent the points having coordinates 4, 7, 8 and 2, 3, 4. So we can write that a and b represent points having coordinates 4, 7, 8 and 2, 3, 4. Using key idea we know ab is equal to square root of x2 minus x1 that is 2 minus 4 whole square plus y2 minus y1 whole square that is 3 minus 7 whole square plus 4 minus 8 whole square that is z2 minus z1 whole square. Now simplifying further we get ab is equal to square root of 4 plus 16 plus 16. This further implies ab is equal to square root of 36 which is further equal to 6. We know ab cannot be negative so we have neglected negative value that is minus 6 and we get ab is equal to 6. Now direction cosines of line abr x2 minus x1 upon ab that is 2 minus 4 upon 6, y2 minus y1 upon ab that is 3 minus 7 upon 6, z2 minus z1 upon ab that is 4 minus 8 upon 6. Now simplifying these three terms we get minus 1 upon 3 minus 2 upon 3 minus 2 upon 3. Now let us assume that p and q represent points having coordinates minus 1 minus 2, 1 and 1 to 5. Now pq is equal to square root of x2 minus x1 whole square that is 1 plus 1 whole square. We know 1 minus minus 1 is equal to 1 plus 1 so here we will write 1 plus 1 whole square plus x2 minus x1 whole square that is 2 plus 2 whole square plus z2 minus z1 whole square that is 5 minus 1 whole square. Now simplifying further we get pq is equal to square root of 4 plus 16 plus 16. This further implies pq is equal to square root of 36 which is further equal to 6. Now we will find out direction cosines of line pq. So we can write direction cosines of line pq are x2 minus x1 upon pq that is 1 plus 1 upon 6 y2 minus y1 upon 6 that is 2 plus 2 upon 6 z2 minus z1 upon 6 that is 5 minus 1 upon 6 or we can simply write direction cosines of line pq are 1 upon 3, 2 upon 3, 2 upon 3. Now if theta is the angle between the lines AB and pq then cos theta is equal to absolute value of minus 1 upon 3 multiplied by 1 upon 3 plus minus 2 upon 3 multiplied by 2 upon 3 plus minus 2 upon 3 multiplied by 2 upon 3. From key idea we know cos theta is equal to l1 l2 plus m1 m2 plus n1 n2 where l1 m1 n1 and l2 m2 n2 are direction cosines of two lines and theta is the angle between the two lines. Now simplifying further we get cos theta is equal to absolute value of minus 1 upon 9 minus 4 upon 9 minus 4 upon 9 now this further implies cos theta is equal to absolute value of minus 9 upon 9 this further implies cos theta is equal to absolute value of minus 1 now we get cos theta is equal to 1 we know absolute value of minus 1 is 1 only now this further implies theta is equal to 0 degrees we know cos 0 is equal to 1 so theta must be equal to 0 so that cos theta is equal to 1. Now if theta is 0 degrees this implies that angle between AB and pq is 0 degrees or we can say the given two lines are parallel we know if the angle between the two lines is 0 degrees they are parallel to each other so we can write thus the given two lines are parallel hence proved this completes the session hope you understood the solution take care and have a nice day