AIEEE 2004 (Offline)Question 1
Distance between two parallel planes $2x+y+2z=8$ and $4x+2y+4z+5=0$ is :
3D Geometry
Solution
Second plane: $2x+y+2z+\frac52=0$. Distance $=\frac{8+\frac52}{3}=\frac72$.Solution
Second plane: $2x+y+2z+\frac52=0$. Distance $=\frac{8+\frac52}{3}=\frac72$.