 Hi friends, I am Purva and today we will work out the following question. Show that the line through the points 1 minus 1, 2 and 3, 4 minus 2 is perpendicular to the line through the points 0, 3, 2 and 3, 5, 6. Now the direction cosines of line passing through two points P whose coordinates are X1, Y1, Z1 and Q whose coordinates are X2, Y2, Z2 are X2 minus X1 upon PQ comma Y2 minus Y1 upon PQ comma Z2 minus Z1 upon PQ where we have PQ is equal to under root of X2 minus X1 whole square plus Y2 minus Y1 whole square plus Z2 minus Z1 whole square. Now let the direction cosines of two lines L1 and L2 be L1, M1, N1 and L2, M2, N2 respectively. And let theta be the angle between them then cos theta is equal to mod of L1, L2 plus M1, M2 plus N1, N2 and these two lines are perpendicular if theta is equal to 90 degrees. So this is the key idea behind a question. Let us begin with the solution now. Now suppose line L1 is passing through the points 1 minus 1, 2 and 3, 4 minus 2. Then let coordinates of point AB 1 minus 1, 2 and coordinates of point BB 3, 4 minus 2. Then AB is equal to under root of now by key idea we know that if X1, Y1, Z1 and X2, Y2, Z2 are coordinates of two points P and Q then PQ is equal to under root of X2 minus X1 whole square plus Y2 minus Y1 whole square plus Z2 minus Z1 whole square. So here we get AB is equal to under root of 3 minus 1 whole square plus 4 minus of minus 1 that is 4 plus 1 whole square plus minus 2 minus 2 whole square which is equal to under root of 4 plus 25 plus 16 which is equal to 300 root 5. Thus the direction cosines of line AB are 3 minus 1 upon 300 root 5 comma 4 plus 1 upon 300 root 5 comma minus 2 minus 2 upon 300 root 5. By key idea we know that the direction cosines of a line passing through two points P and Q are given by X2 minus X1 upon PQ comma Y2 minus Y1 upon PQ comma Z2 minus Z1 upon PQ. So here we get these are the direction cosines of line AB that is we get 2 upon 300 root 5 comma 5 upon 300 root 5 comma minus 4 upon 300 root 5. Now suppose line L2 is passing through the points 0, 3, 2 and 3, 5, 6. Then let the coordinates of point PR 0, 3, 2 and coordinates of point QR 3, 5, 6. Then we have PQ is equal to under root of 3 minus 0 whole square plus 5 minus 3 whole square plus 6 minus 2 whole square. This is equal to under root of 9 plus 4 plus 16 which is equal to under root 29. Thus the direction cosines of line PQ are 3 minus 0 upon under root 29 comma 5 minus 3 upon under root 29 comma 6 minus 2 upon under root 29. That is we have 3 upon under root 29 comma 2 upon under root 29 comma 4 upon under root 29. Now the angle theta between the lines AB and PQ is given by cos theta is equal to mod of. Now by key idea we know that the angle theta between two lines L1 and L2 whose direction cosines are L1 M1 N1 and L2 M2 N2 is given by cos theta is equal to mod of L1 L2 plus M1 M2 plus N1 N2. So here we have cos theta is equal to mod of 2 upon 300 root 5 into 3 upon under root 29 plus 5 upon 300 root 5 into 2 upon under root 29 plus minus 4 upon 300 root 5 into 4 upon under root 29. And this is equal to mod of 2 upon under root 145 plus 10 upon 3 under root 145 minus 16 upon 3 under root 145 which is equal to 6 plus 10 minus 16 upon 3 under root 145. And this implies cos theta is equal to 0 which further implies theta is equal to 90 degrees. Now since theta is equal to 90 degrees so by key idea we can say that thus the two lines are perpendicular. This is our answer hope you have understood the solution. Bye and take care.