Questions
The equation to the plane through the line of intersection of such that each plane is at a distance of 2 unit from the origin is
a
x+y+2z+13=0,x+y+z−3=0
b
2x+y−2z+3=0,x−2y−2z−3=0
c
15x−12y+16z+50=0, x+2y+2z−6=0
d
x-y+2z-13=0, x+y-z-3=0
detailed solution
Correct option is C
Equation of the plane is (2x+y+3z−2)+k(x−y+z+4)=0p=|−2+4k|(2+k)2+(1−k)2+(3+k)2=2⇒k=13,−1∴ Planes are (2x+y+3z−2)+13(x−y+z+4)=0,(2x+y+3z−2)−1(x−y+z+4)=0⇒15x−12y+16z+50=0,x+2y+2z−6=0Talk to our academic expert!
Similar Questions
If the equation of the plane passing through the line of intersection of the planes and perpendicular to the XY-plane is then S=
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