 Good morning friends. I am Poojwa and today we will discuss the following question. Find the vector and the Cartesian equation of the line that passes through the points 3 minus 2 minus 5 and 3 minus 2, 6. Now, let vector A and vector B be the position vectors of the points given and let vector R be the position vector of an arbitrary point P on the line. Then the vector equation of the line is given by vector R is equal to vector A plus lambda into vector B minus vector A, where lambda is the parameter. And here we have vector A is equal to x1 i cap plus y1 j cap plus z1 k cap, vector B is equal to x2 i cap plus y2 j cap plus z2 k cap and vector R is equal to x i cap plus y j cap plus z k cap. Now, putting the values of vector A vector B and vector R in equation 1 and eliminating lambda, we get the Cartesian form as x minus x1 upon x2 minus x1 is equal to y minus y1 upon y2 minus y1 is equal to z minus z1 upon z2 minus z1. So, this is the key idea behind our question. Let us begin with the solution now. Now, we are given that the line passes through the points 3 minus 2 minus 5 and 3 minus 2 6. So, let point A has coordinates 3 minus 2 minus 5 and point B has coordinates 3 minus 2 6. Then we have the position vector of the points is given by vector A is equal to 3 i cap minus 2 j cap minus 5 k cap and vector B is equal to 3 i cap minus 2 j cap plus 6 k cap. Now, by key idea we know that vector A is equal to x1 i cap plus y1 j cap plus z1 k cap. So, here we have x1 is equal to 3 y1 is equal to minus 2 and z1 is equal to minus 5 and we also know that vector B is equal to x2 i cap plus y2 j cap plus z2 k cap. So, we have here x2 is equal to 3 y2 is equal to minus 2 and z2 is equal to 6. Now, the vector equation of a line is given by vector R is equal to vector A plus lambda into vector B minus vector A. Now, putting the value of vector A and vector B in this equation we get that is vector R is equal to now vector A is equal to 3 i cap minus 2 j cap minus 5 k cap plus lambda into vector B is equal to 3 i cap minus 2 j cap plus 6 k cap minus vector A is equal to 3 i cap minus 2 j cap minus 5 k cap. This is equal to 3 i cap minus 2 j cap minus 5 k cap plus lambda into 2. Now, 3 i cap minus 3 i cap gives 0 minus 2 j cap minus into minus becomes plus. So, minus 2 j cap plus 2 j cap gives 0 and 6 k cap minus into minus becomes plus. So, plus 5 k cap gives 11 k cap and this is equal to 3 i cap minus 2 j cap minus 2 j cap minus 5 k cap plus lambda into 11 k cap. Now, by key idea we know that the Cartesian equation of the line is given by x minus x1 upon x2 minus x1 is equal to y minus y1 upon y2 minus y1 is equal to z minus z1 upon z2 minus z1. Now, here we mark this as 1. So, putting all the values from 1 in this equation we get x minus x1 that is x minus 3 upon x2 minus x1 that is 3 minus 3 is equal to y minus y1. Now, y1 is equal to minus 2. So, we get y plus 2 upon y2 minus y1 that is minus 2 plus 2 is equal to z minus z1. Now, z1 is equal to minus 5. So, we get z plus 5 upon z2 minus z1 that is 6 plus 5 or we can write this as x minus 3 upon 0 is equal to y plus 2 upon 0 is equal to z plus 5 upon 11. Thus we have got vector equation of the line is vector r is equal to 3 i cap minus 2 i cap minus 5 k cap plus lambda into 11 k cap and Cartesian equation of the line is x minus 3 upon 0 is equal to y plus 2 upon 0 is equal to z plus 5 upon 11. This is our answer. Hope you have understood the solution. Bye and take care.