 Hi and welcome to the session. Let's work out the following question. The question says, find the vector equation of the line passing through the point A, 2 minus 1, 1 and parallel to the line joining the points B, minus 1, 4, 1 and C, 1, 2, 2. Also find the Cartesian equation of the line. So let us start with the solution to this question. Here, point A is 2 minus 1, 1, point B is minus 1, 4, 1 and point C is 1, 2, 2. Therefore, vector BC is equal to vector OC minus vector OB. Vector OC will be i cap plus 2 j cap plus 2 k cap minus vector OB is minus i cap plus 4 j cap plus k cap. This is equal to 2 i cap minus 2 j cap plus k cap. Now vector equation of a line passing through the point A that is 2 minus 1, 1 and parallel BC is vector R is equal to 2 i cap minus 2 j cap plus k cap plus lambda times 2 i cap minus 2 j cap plus k cap. We call this equation 1. Let the Cartesian equation B vector R is equal to x i cap plus y j cap plus z k cap. We call this equation 2. Sorry, here we have minus j cap only. Now from 1 and 2 we have x i cap plus y j cap plus z k cap is equal to 2 i cap minus j cap plus k cap plus lambda times 2 i cap minus 2 j cap plus k cap. This implies that x i cap plus y j cap plus z k cap is equal to 2 plus 2 lambda i cap minus 1 plus 2 lambda j cap plus 1 plus lambda k cap. Now this implies that x is equal to 2 plus 2 lambda y is equal to minus 1 minus 2 lambda and z is equal to 1 plus lambda. So this implies that the Cartesian equation is x minus 2 divided by 2 is equal to y plus 1 divided by minus 2 is equal to z minus 1 divided by 1 is equal to lambda. So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.