JEE Main2014MathematicsThree Dimensional GeometryActual

A symmetrical form of the line of intersection of the planes x=a y+b and z=c y+d is

Options

Ax-b a = y-1 1 = z-d c
Bx-b-a a = y-1 1 = z-d-c c
Cx - a b = y -0 1 = z - c d
Dx - b - a b = y -1 0 = z - d - c d

Correct answer

B. x-b-a a = y-1 1 = z-d-c c

Step-by-step solution

Given two planes: x-a y-b=0 and c y-z+d=0 Let, l, m, n be the direction ratio of the required line. Since the required line is perpendicular to normal of both the plane, therefore l-a m=0 and c m-n=0 l-a m+0 . n=0 and 0 . l+c m-n=0 l a-0 = m 0+1 = n c-0 Hence, d.R of the required line are a, 1 , c . Hence, options (c) and (d) are rejected. Now, the point (a+b, 1, c+d) satisfy the equation of the two given planes. Option (b) is correct.

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