 Hi friends I am Purva and today we will discuss the following question, find the vector and the Cartesian equations of the lines that passes through the origin and 5,-2,3. 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. In the vector equation of the line is given by vector R is equal to vector A plus lambda times vector B minus vector A where lambda is the parameter. Now eliminating the parameter 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 where x1, y1 and z1 are the coefficients of i cap, j cap and k cap in vector A and x2, y2 and z2 are coefficients of i cap, j cap and k cap in vector B. So this is the key idea behind our question. Let us begin with the solution now. Now let the line passes through the points A that has coordinates 0, 0, 0 and B which has coordinates 5, minus 2, 3. Then we have vector A is equal to 0 i cap plus 0 j cap plus 0 k cap and vector B is equal to 5 i cap minus 2 j cap plus 3 k cap. Now by key idea we know that x1, y1 and z1 are coefficients of i cap, j cap and k cap in vector A and x2, y2 and z2 are the coefficients of i cap, j cap and k cap in vector B. So here we have x1 is equal to 0, y1 is equal to 0 and z1 is equal to 0 and x2 is equal to 5, y2 is equal to minus 2 and z2 is equal to 3. Now the vector equation is given by vector R is equal to vector A plus lambda times vector B minus vector A that is we have vector R is equal to now vector A is equal to 0 i cap plus 0 j cap plus 0 k cap plus lambda times vector B is 5 i cap minus 2 j cap plus 3 k cap minus vector A is 0 i cap plus 0 j cap plus 0 k cap and this implies vector R is equal to lambda times 5 i cap minus 2 j cap plus 3 k cap. Now the Cartesian equation 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 putting all the values from this equation 1 we get x minus 0 upon 5 minus 0 is equal to y minus 0 upon minus 2 minus 0 is equal to z minus 0 upon 3 minus 0 or we can write this as x upon 5 is equal to y upon minus 2 is equal to 0 upon 3. Thus we get our answer as vector equation of the line is vector R is equal to lambda times 5 i cap minus 2 j cap plus 3 k cap and the Cartesian equation of the line is x upon 5 is equal to y upon minus 2 is equal to z upon 3. This is our answer. Hope you have understood the solution. Bye and take care.