 Hi friends, I am Purva and today we will work out the following question. The Cartesian equation of a line is x minus 5 upon 3 is equal to y plus 4 upon 7 is equal to z minus 6 upon 2. Write its vector form. Let the Cartesian equation of the line be x minus x1 upon a is equal to y minus y1 upon b is equal to z minus z1 upon c. Where we have x1, y1 and z1 are coefficients of i cap, j cap and k cap in vector a and a, b and c are the direction ratios. That is coefficients of i cap, j cap and k cap in vector b. Then the vector equation of this line is given by vector r is equal to vector a plus lambda times vector b, where lambda is the parameter. And this vector r is the position vector of an arbitrary point p on the line. So, this is the key idea behind our question. Let us begin with the solution now. Now we are given that the Cartesian equation of the line is x minus 5 upon 3 is equal to y plus 4 upon 7 is equal to z minus 6 upon 2. We mark this as 1. Now by key idea we know that the standard equation of a line in Cartesian form is given by x minus x1 upon a is equal to y minus y1 upon b is equal to z minus z1 upon c. We mark this as 2. Now here we have x1, y1 and z1 are coefficients of i cap, j cap and k cap in vector a and a, b and c are the direction ratios that is coefficients of i cap, j cap and k cap in vector b. Now on comparing 1 and 2 we see that the line is passing through the point 5 minus 4 and 6. Now comparing this equation 1 and 2 we can clearly see that x1 is equal to 5, y1 is equal to minus 4 and z1 is equal to 6. And this x1, y1 and z1 are coefficients of i cap, j cap and k cap in vector a. So we get thus vector a is equal to 5 i cap minus 4 j cap plus 6 k cap. Also the line is parallel to the vector whose direction ratios are 3, 7 and 2. Again by comparing 1 and 2 we can clearly see that here a is equal to 3, b is equal to 7 and c is equal to 2. And this a, b, c are the direction ratios that is coefficients of i cap, j cap and k cap in vector b. So we get thus equation of vector b is equal to 3 i cap plus 7 j cap plus 2 k cap. So in this figure this is the line l whose Cartesian equation is given to us and this is the vector b which is parallel to this line l. Now by key idea we know that the vector equation of the line is given by vector r is equal to vector a plus lambda times vector b where lambda is the parameter that is we have. Now vector a is equal to 5 i cap minus 4 j cap plus 6 k cap plus lambda times vector b is equal to 3 i cap plus 7 j cap plus 2 k cap. Thus we get the vector equation of the line is vector r is equal to 5 i cap minus 4 j cap plus 6 k cap plus lambda times 3 i cap plus 7 j cap plus 2 k cap. This is our answer. Hope you have understood the solution. Bye and take care.