Question

The symmetric form of the line of intersection of the two planes 4x + 4y − 5z − 12 = 0 and 8x + 12y − 13z − 32 = 0 is

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

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

(c) x/2 = (y - 2)/3 = z/4

(d) (x + 1)/2 = (y + 2)/3 = z/4

Answer 1

Correct option is (a) (x - 1)/2 = (y - 2)/3 = z/4

Explanation :

Let L be the line of intersection. We know that vector n₁ = (4,4,-5) and vector n₂ = (8,12,-13) are the normals of the given planes. Therefore, the DRs of the line L is  

Therefore, the DRs of L is (2, 3, 4). Also, substituting z = 0 in both plane equations, we get

4x + 4y = 12 and 8x + 12y = 32

Solving these equations, we get x = 1 and y = 2. Hence, (1, 2, 0) is a point on the line L. Thus, the equation of the line is 

(x - 1)/2 = (y - 2)3 = (z - 0)/4