JEE Main2013MathematicsThree Dimensional GeometryActual

The equation of a plane through the line of intersection of the planes x+2 y=3, y-2 z +1=0 , and perpendicular to the first plane is :

Options

A2 x-y-10 z=9
B2 x-y+7 z=11
C2 x-y+10 z=11
D2 x-y-9 z=10

Correct answer

C. 2 x-y+10 z=11

Step-by-step solution

Equation of a plane through the line of intersection of the planes x+2 y=3, y-2 z+1=0 is (x+2 y-3)+ (y-2 z+1)=0 x+(2+ ) y-2 (z)-3+ =0 Now, plane (i) is to x+2 y=3 Their dot product is zero i.e. 1+2(2+ )=0 =- 5 2 Thus, required plane is aligned & x+ (2- 5 2 ) y-2 -5 2 (z)-3- 5 2 =0 & x- y 2 +5 z- 11 2 =0 & 2 x-y+10 z-11=0 aligned

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