 Hello and welcome to the session. In this session, we discuss the following question which says the points A with coordinates 0, 5, 10, B with coordinates 2, 3, 4 and C with coordinates 1, 2, minus 1 are three vertices of a paradigmogram ABCD. Find the vector equations of the sides AB and BC and also find the coordinates of point D. Suppose we have two points with position vectors vector A and vector B, then the vector equation of the line which passes through the given two points is equal to vector R is equal to vector A plus lambda into vector B minus vector A. R equal to vector A plus lambda into vector B minus vector A, the whole is the vector equation of the line which passes through the two points with position vectors vector A and vector B. This is the key idea that we use for this question. Let's move on to the solution now. We have a paradigmogram ABCD where the point A has coordinates 4, 5, 10, B has coordinates 2, 3, 4 and C has coordinates 1, 2, minus 1. Now, with reference 2 as origin, let the position vector or say position vectors of the points AB, C and D be given as vector A, vector B, vector C and vector D respectively. First of all, we assume the coordinates of the point D as x, y, z. So, we are supposed to find x, y and z as point A has coordinates 4, 5, 10. So, vector A minus the position vector of the point A would be equal to 4 i cap plus 5 j cap plus 10 k cap, vector B minus the position vector of the point B would be 2 i cap plus 3 j cap plus 4 k, 3 and 4 are the coordinates of the point B. The vector C, which is the position vector of the point C that coordinates 1, 2 and minus 1 would be given as i cap plus 2 j cap minus k cap, vector D, which is the position vector of the point D, would be x i cap plus y j cap plus z k cap, where x, y and z are the coordinates of the point D. Now, we are supposed to find the vector equations of the lines AB and DC. Now, the vector equation of the site AB would be the vector equation of AB that passes through the points A and B. This is the vector equation of the line, which passes through the two points. Vector equation of the site AB that is this side, which passes through the points A and B would be given by equal to vector A, which is the position vector of the point A plus lambda into vector B, which is the position vector of the point B minus vector A the home. Thus, we have vector R equal to 4 i cap plus 5 j cap plus 10 k cap the home plus lambda into vector B, which is 2 i cap plus 3 j cap plus 4 k cap minus vector A minus 4 i cap minus 5 j cap minus 10 k cap the home, which means vector R is equal to 4 i cap plus 5 j cap plus 10 k cap the home plus lambda into 2 i cap minus 4 i cap is minus 2 i cap, 3 j cap minus 5 j cap is minus 2 j cap, 4 k cap minus 10 k cap is minus 6 k cap. This gives us vector R is equal to 4 i cap plus 5 j cap plus 10 k cap 2 lambda into i cap plus j cap the home can also be written as vector R is equal to 4 i cap plus 5 j cap plus 10 k cap i cap plus j cap plus 3 k cap the home where we have taken lambda 1 as minus 2 lambda. Vector equation A B are equal to 4 i cap plus 5 j cap plus the home plus lambda 1 into i cap plus j cap the home. The vector equation BC is given as vector R equal to vector V plus lambda into vector C minus vector B the home where vector B is the position vector of the point B vector C is the position vector of the point C putting the values of vector B and vector C we get vector R is equal to vector B which is a cap plus 3 j cap plus 4 k cap to vector C which is i cap plus 2 j cap minus k cap minus vector B that is minus 2 i cap minus 3 j cap minus 4 k cap the home. So this gives us vector R is equal to 2 i cap plus 3 j cap plus 4 k cap the home plus lambda into i cap minus 2 i cap minus i cap 2 j cap minus 3 j cap minus j cap minus k cap minus 4 k cap is minus 5 k cap the home. So where further we have vector R is equal to 2 i cap plus 3 j cap plus 4 k cap the home minus lambda into i cap plus j cap plus 5 k cap the home. So we have the vector equation inside V C is given as vector R equal to 2 i cap plus 3 j cap plus 4 k cap plus lambda 2 that is in case of minus lambda we take lambda 2 into i cap plus j cap plus 5 k cap B and here we have lambda 2 as minus lambda and the coordinates of the point D, BD and AC are the diamonds of the parallelogram the diamonds of a parallelogram each other of the diagonal BD would be equal to the midpoint of the diamond AC. This means the midpoint of the diagonal BD would be given as vector OB or OB this whole upon 2 equal to the midpoint of diagonal AC which is given as vector OA plus vector OC this whole upon 2. This means that vector OB plus vector OB is equal to vector OA plus vector OC the OB means the position vector as the point B with reference to the origin O that is vector B plus OB that is vector OB is the position vector of the point D that is vector D is equal to vector OA which is vector A that is the position vector of the point A plus vector OC that is vector C which is the position vector of the point C. Competing the respective values of the vectors A, B, C and D we get 2 i cap plus 3 j cap plus 4 k cap plus x i cap plus y j cap plus z k cap is equal to 2 k cap plus i cap plus 2 j cap minus 2 this gives us plus y j cap plus z k cap plus i cap is 5 j cap plus 2 j cap is plus 7 j cap 10 k cap minus k cap is plus 9 k cap 3 j cap plus this gives us x i cap plus y j cap plus z k cap is equal to 3 i cap 7 j cap minus 3 j cap is 4 j cap 9 k cap minus 4 k cap is plus 5 k cap. This is the position vector of the point D for where we have x as 3 y as 4 point D has coordinate 60 session hope you have understood the solution of this question.