 Hello and welcome to the session. In this session we discussed the following question which says use the vector method to find the values for lambda and mu so that the points A with coordinates 1, 1, minus 3, B with coordinates 7, 6, minus 5 and C with coordinates lambda, mu, minus 1 are collinear. Before moving on to the solution, let's recall the vector equation of a line that passes through two points with position vectors vector A and vector B is given by vector R is equal to vector A plus lambda into vector B minus vector A for some scalar lambda. This is the key idea for this question. Now we proceed with the solution. We are given points A, B and C. A has coordinates 1, 1, minus 3, point B has coordinates 7, 6, minus 5 and point C has coordinates lambda, mu, minus 1. We take let vector A, vector B, vector C be the position vectors of points A, B and C respectively. Then we have vector A is equal to i cap plus j cap minus 3 k cap vector B is equal to 7 i cap plus 6 j cap minus 5 k cap and vector C is equal to lambda i cap plus mu j cap minus k cap. So now the vector equation of the line A, B is given as vector R equal to vector A plus T into vector B minus vector A for some scalar T. That is we have used this key idea where we have vector equation of a line that passes through two points with given position vectors vector A and vector B is given as vector R is equal to vector A plus lambda into vector B minus vector A where lambda is some scalar. So in this way we have found the vector equation of the line A, B. Now substituting for vector A and vector B we get vector R is equal to i cap plus j cap minus 3 k cap plus T into 7 i cap plus 6 j cap minus 5 k cap minus i cap plus j cap minus 3 k cap. And so we get vector R equal to i cap plus j cap minus 3 k cap plus T into 6 i cap plus 5 j cap minus 2 k cap. Further vector R is equal to 6 T plus 1 i cap plus 5 T plus 1 j cap plus minus 2 T minus 3 k cap. So this is the vector equation of the line A, B. Now if we have that the line A, B passes through the point C then we have vector C is equal to 6 T plus 1 i cap plus 5 T plus 1 j cap plus minus 2 T minus 3 k cap where we have T is some scalar. Now substituting for vector C which is lambda i cap is mu j cap minus k cap. So we get lambda i cap plus mu j cap minus k cap is equal to 6 T plus 1 i cap plus 5 T plus 1 j cap plus minus 2 T minus 3 k cap. Equating the components on both the sides we get lambda is equal to 6 T plus 1 mu is equal to 5 T plus 1 and minus 1 is equal to minus 2 T minus 3. Now that we have minus 1 is equal to minus 2 T minus 3 this means that 2 T plus 3 is equal to 1 or you say that 2 T is equal to minus 2. So from here we get T is equal to minus 1. Now putting T equal to minus 1 in this lambda and mu we get lambda is equal to 6 into minus 1 plus 1 which is equal to minus 5 mu is equal to 5 into minus 1 plus 1 which is equal to minus 4. So that we get value for lambda is minus 5 and value for mu is minus 4. This is our final answer. This completes the session. Hope you have understood the solution of this question.