 Hi and welcome to the session. I'm Shashi and I'm going to help you with the following question. Question says, find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i minus j plus 4k and is in the direction i plus 2j minus k. First of all, let us understand that vector equation of a line passing through a point and parallel to a given vector b is given by r vector is equal to a vector plus lambda multiplied by b vector where lambda is a parameter here in this equation r vector is a position vector of any arbitrary point p on the line. A vector is the position vector of the given point on the line and b vector is the vector which is parallel to the line. Now Cartesian equation of a line passing through a point x1, y1, z1 and parallel to a given vector b is given by x minus x1 upon a is equal to y minus y1 upon b is equal to z minus z1 upon c. Here x1, y1, z1 are coefficients of i, j and k of vector a and a, b, c are direction ratios or we can say a, b, c are coefficients of i, j, k in vector b. Now we will use these two equations as our key idea to solve the given question. Let us now start with the solution. Now we know given line passes through the point whose position vector is 2i minus j plus 4k so we can write vector a is equal to 2i minus j plus 4k. We are also given that line is in the direction i plus 2j minus k so we can say vector b is equal to i plus 2j minus k line is in the direction of this vector implies that line is parallel to this vector so we can denote this vector by b vector. Now we know vector equation of the given line is r vector is equal to a vector plus lambda multiplied by b vector where lambda is a real number. Now substituting the components of vector a and vector b in this equation we get r vector is equal to 2i minus j plus 4k plus lambda multiplied by i plus 2j minus k. We know in this equation r vector is the position vector of an arbitrary point p on the line a vector is the position vector of the given point and vector b is parallel to the line or we can say line is parallel to vector b. So this is the required vector equation of the given line. Now we will find the equation of the line in Cartesian form. Now using key idea we know x1 y1 z1 is equal to coefficients of unit vector i unit vector j and unit vector k in vector a so we get x1 y1 z1 is equal to 2 minus 1 4. Now we will find the values of abc now we know abc is equal to coefficients of unit vector i unit vector j and unit vector k in vector b. So we get abc is equal to 1 2 minus 1. Now we know required Cartesian equation is x minus x1 upon a is equal to y minus y1 upon b is equal to z minus z1 upon c. Now substituting corresponding values of x1 y1 z1 and abc in this equation we get x minus 2 upon 1 is equal to y plus 1 upon 2 is equal to z minus 4 upon minus 1. So this equation represents the required equation in vector form and this equation represents equation in Cartesian form. This completes the session hope you understood the solution take care and have a nice day.