 Hello students, let's work out the following problem. It says the Cartesian equations of a line are 3x plus 1 is equal to 6y minus 2 is equal to 1 minus z. Find the fixed point through which it passes its direction ratios and also its vector equation. So let's now move on to the solution. The equation of lines are given as 3x plus 1 is equal to 6y minus 2 is equal to 1 minus z. Now this can be written as taking 3 common. We have 3 into x plus 1 by 3 is equal to taking 6 common. We have 6 into y minus 2 by 6 is equal to taking 1 common minus 1 common we have z minus 1. Now again, this can be written as x plus 1 by 3 upon 1 by 3 and this can be written as y minus 2 by 6 and 2 by 6 is same as 1 by 3 upon 1 by 6 is equal to z minus 1 upon minus 1. Again, this can be written as x plus 1 by 3 upon 1 by 3 is equal to y minus 1 by 3 upon 1 by 6 is equal to z minus 1 upon minus 1. Now again, this can be written as x plus 1 by 3 upon 2 is equal to y minus 1 by 3 upon 1 is equal to z minus 1 upon minus 6. This is by multiplying all the denominators by 6. We have this. Now the direction ratios ratio of the liner 1 minus 6 and the points through which it passes that is the fixed point through which it passes r minus 1 by 3 1 by 3 and 1. As we know that the Cartesian equation of the line passing through x 1, y 1, z 1 and having direction ratios a, b, c it is x minus x 1 upon a is equal to y minus y 1 upon v is equal to z minus z 1 upon c. So here, this can be written as x minus of minus 1 by 3. So the fixed point through which the line passes minus 1 by 3 1 by 3 1. Now we have to find the vector equation of the line. Now the vector equation of the line passing through the points x 1, y 1, z 1 and direction ratios a, b, c is given by x 1 i cap plus y 1 j cap plus z 1 k cap lambda times a i cap plus b j cap plus c k cap. So here, the direction ratios are 2 1 minus 6 and the fixed point is minus 1 by 3 1 by 3 and 1. So the vector equation of the line is minus 1 by 3 i cap plus 1 by 3 j cap plus 1 k cap lambda times 2 i cap plus 1 j cap plus minus minus 6 k cap. Therefore, the direction ratios are 2 1 minus 6 and the fixed point through which the line passes is minus 1 by 3 1 by 3 and 1 and this is the required vector equation of the line. So this completes the question and the session. Bye for now. Take care. Have a good day.