Answer to Question #310467 in Analytic Geometry for Alex

Question

There are two planes:

x+4y-3z+12=0 and 2x+8y-6z+4=0

a) Explain how I know these planes are parallel and distanced?

b) How far apart are these lines?

Answer 1

The distance between planes is defined if the planes are parallel or, equivalently, the normal vectors of these planes are collinear. The distance between the planes is equal to the length of the perpendicular dropped from one plane to another.

For two planes A₁x + B₁y + C₁z + D₁ = 0 and A₂y + B₂y + C₂z + D₂ = 0 to be parallel, their normal vectors $\vec{n}_{1}$ and $\vec{n}_{2}$ must be collinear, i.e. $\vec{n}_{1}=\lambda\vec{n}_{2}$ , where λ≠0. If none of the coordinates of the vectors $\vec{n}_{1}$ and $\vec{n}_{2}$ is equal to zero, then it follows from the last equality that A₁/A₂ = B₁/B₂ = C₁/C_2. In our case A₁ = 1, A₂ = 2 B₁ = 4, B₂ = 8, C₁ = -3, C₂ = -6 and A₁/A₂ = B₁/B₂ = C₁/C₂ =1/2 which means that the planes are parallel.

The distance between the planes could be found using the following formula:

$$d=\frac{\left|D_{2} -D_{1}\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$$

For given planes the distance is $d=\frac{\left|2 -12\right|}{\sqrt{1^{2}+4^{2}+(-3)^{2}}}=\frac{\left|10\right|}{\sqrt{1+16+9}}\approx 1.96.$