Question

Which of the following equations represent the line of intersection of the planes 4x − 4y − z + 11 = 0 = x + 2y − z − 1? 

(A) (x - 2)/1 = y/1 = (z - 3)/4

(B) (2x - 13)/4 = (4y - 15)/4 = z/4

(C) x/2 = (y - 2)/1 = (z - 3)/4

(D) (x - 4)/-2 = (y - 4)/2 = (z -11)/2

Answer 1

Correct option is (B) and (C) 

(B) (2x - 13)/4 = (4y - 15)/4 = z/4

(C) x/2 = (y - 2)/1 = (z - 3)/4

Explanation :

We have 

|(vector i, vector j, vector k),(4,-4,-1),(1,2,-1)| = 6 vector i - 3 vector j + 12 vector k

Hence, the DRs of the line are (2, 1, 4). In the given plane equations, substitute x = 0 so that we have 4y + 8 = 11, 2y - z = 1. On solving these equations, we have (0, 2, 3) is a point on this line. Hence, the line equation is

x/2 = (y - 2)/1 = (z - 1)/4

Similarly, taking z = 0 and solving the two equations for x and y, we have x = -13/2 and y = -15/4 so that (-13/2,-15/4,0) is a point on the line. Hence, the equation of the line is

(x + (-13/2))/2 = (y + (-15/4))/1 = z/4

(2x - 13)/4 = (4y - 15)/4 = z/4

It is clear that (2, 0, 3) is not a point on the line and (−2, 2, 2) is not parallel to (2, 1, 4).